6023 x10 23 particles of chlorine molecule  1 mole  710 g


Molar mass Nitrogen moleculeN2 relative molecular mass 140x 2g 280g


6023 x10 23 particles of Nitrogen molecule  1 mole  280 g


iia compound has mass equal to relative formular mass RFM in gramsof the molecule Relative formular mass is the sum of the relative atomic masses of the elements making the compound eg


iMolar mass WaterH2O  relative formular mass 10 x 2   160g 180g


6023 x10 23 particles of Water molecule  1 mole  180 g


6023 x10 23 particles of Water molecule has


2 x 6023 x10 23 particles of Hydrogen atoms


1 x 6023 x10 23 particles of Oxygen atoms


iiMolar mass sulphuricVIacidH2SO4  relative formular mass


10 x 2   320  160 x 4g 980g


6023 x10 23 particles of sulphuricVIacidH2SO4  1 mole  980g


6023 x10 23 particles of sulphuricVIacidH2SO4 has


1 x 6023 x10 23 particles of Sulphur atoms


4 x 6023 x10 23 particles of Oxygen atoms


iiiMolar mass sodium carbonateIVNa2CO3  relative formular mass


230 x 2   120  160 x 3g 1060g


6023 x10 23 particles of sodium carbonateIVNa2CO3  1 mole  1060g


6023 x10 23 particles of sodium carbonateIVNa2CO3 has


2 x 6023 x10 23 particles of Sodium atoms


1 x 6023 x10 23 particles of Carbon atoms


3 x 6023 x10 23 particles of Oxygen atoms


ivMolar mass Calcium carbonateIVCaCO3  relative formular mass


400 120  160 x 3g 1000g


6023 x10 23 particles of Calcium carbonateIVCaCO3  1 mole  1000g


6023 x10 23 particles of Calcium carbonateIVCaCO3 has


1 x 6023 x10 23 particles of Calcium atoms


vMolar mass WaterH2O  relative formular mass


2 x 10  160 g 180g


6023 x10 23 particles of WaterH2O  1 mole  180g


6023 x10 23 particles of WaterH2O has


2 x 6023 x10 23 particles of Hydrogen atoms


2 x 6023 x10 23 particles of Oxygen atoms


Practice


1 Calculate the number of moles present in


i023 g of Sodium atoms


Molar mass of Sodium atoms  23g


Moles  mass in grams   023g  001moles


Molar mass 23


ii 023 g of Chlorine atoms


Molar mass of Chlorine atoms  355 g


Moles  mass in grams   023g  00065moles 65 x 103 moles


Molar mass 355


iii 023 g of Chlorine molecules


Molar mass of Chlorine molecules  355 x 2  710 g


Moles  mass in grams   023g  00032moles 32 x 103 moles


Molar mass 71


iv 023 g of dilute sulphuricVIacid


Molar mass of H2SO4  2 x 1  32  4 x14  980g


Moles  mass in grams   023g  00023moles 23 x 103 moles


Molar mass 98


2 Calculate the number of atoms present inAvogadros constant L  60 x 10 23


i 023 g of dilute sulphuric VIacid


Method I


1 mole has 60 x 10 23 atoms


23 x 103 moles has 23 x 103 x 60 x 10 23  138 x 10 21 atoms


Method II


980g  1 mole has 60 x 10 23 atoms


023 g therefore has 023 g x 60 x 10 23   138 x 10 21 atoms


ii023 g of sodium carbonateIVdecahydrate


Molar mass of Na2CO310H2 O


2 x 23  12  3 x16  10 x 10  10 x 16  2760g


Moles  mass in grams   023g  000083moles


Molar mass 276 83 x 104 moles


83 x 104 moles has 83 x 104 moles x 60 x 10 23  498 x 10 20 atoms


2760g  1 mole has 60 x 10 23 atoms


023 g therefore has 023 g x 60 x 10 23   498 x 10 20 atoms


iii023 g of Oxygen gas


Molar mass of O2  2 x16  320 g


Moles  mass in grams   023g  000718moles


Molar mass 32 718 x 103 moles


1 mole has 2 x 60 x 10 23 atoms in O2


718 x 103moles has 718 x 103moles x 2 x 60 x 10 23 8616 x 10 21atoms


Method II


320g  1 mole has 2 x 60 x 10 23 atoms in O2


023 g therefore has 023 g x 2 x 60 x 10 23   8616 x 10 21atoms


iv023 g of CarbonIVoxide gas


Molar mass of CO2  12  2 x16  440 g


Method I


Moles  mass in grams   023g  000522moles


Molar mass 44 522 x 103 moles


1 mole has 3 x 60 x 10 23 atoms in CO2


718 x 103moles has 522 x 103moles x 3 x 60 x 10 23 9396 x 10 21atoms


440g  1 mole has 3 x 60 x 10 23 atoms in CO2


023 g therefore has 023 g x 3 x 60 x 10 23   9409 x 10 21atoms


cEmpirical and molecular formula


1The empirical formula of a compound is its simplest formula It is the simplest whole number ratios in which atoms of elements combine to form the compound 2It is mathematically the lowest common multiple LCM of the atoms of the elements in the compound


3Practically the empirical formula of a compound can be determined as in the following examples


To determine the empirical formula of copper oxide


aMethod 1From copper to copperIIoxide


Procedure


Weigh a clean dry covered crucibleM1Put two spatula full of copper powder into the crucible Weigh again M2Heat the crucible on a strong Bunsen flame for five minutes Lift the lid and swirl the crucible carefully using a pair of tong Cover the crucible and continue heating for another five minutes Remove the lid and stop heating Allow the crucible to cool When cool replace the lid and weigh the contents again M3


Sample results


Sample questions


1 Calculate the mass of copper powder used


Mass of crucible  copper before heatingM2  184


Less Mass of crucibleM1   156g


Mass of copper 28 g


2 Calculate the mass of Oxygen used to react with copper


Method I


Mass of crucible  copper after heatingM3  191g


Mass of crucible  copper before heatingM2   184g


Mass of Oxygen  07 g


Method II


Mass of crucible   156g


Mass of copperIIOxide  35 g


Mass of copper   28 g


3 Calculate the number of moles of


i copper used Cu  635


number of moles of copper  mass used  28  00441moles


Molar mass 635


ii Oxygen used O  160


number of moles of oxygen  mass used  07  00441moles


Molar mass 160


4 Determine the mole ratio of the reactants


Moles of copper  00441moles  1  Mole ratio Cu O  11


Moles of oxygen 00441moles 1


5What is the empirical formula of copper oxide formed


CuO copperIIoxide


6 State and explain the observations made during the experiment


Observation


Colour change from brown to black


Explanation


Copper powder is brown On heating it reacts with oxygen from the air to form black copperIIoxide


7 Explain why magnesium ribbonshavings would be unsuitable in a similar experiment as the one above


Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesium oxide and magnesium nitride This causes experimental mass errors


bMethod 2From copperIIoxide to copper


Procedure


Weigh a clean dry porcelain boat M1 Put two spatula full of copperIIoxide powder into the crucible Reweigh the porcelain boat M2Put the porcelain boat in a glass tube and set up the apparatus as below


Pass slowlyto prevent copperIIoxide from being blown awaya stream of either dry Hydrogen ammonialaboratory gas carbonIIoxide gas for about two minutes from a suitable generator


When all the in the apparatus set up is driven out heat the copperIIoxide strongly for about five minutes until there is no further change Stop heating


Continue passing the gases until the glass tube is cool


Turn off the gas generator


Carefully remove the porcelain boat form the combustion tube


Reweigh M3


Sample results


Sample questions


1 Calculate the mass of copperIIoxide used


Mass of boat before heatingM2  191


Mass of empty boatM1   156g


Mass of copperIIOxide 35 g


2 Calculate the mass of


i Oxygen


Mass of boat after heating M3   184g


Mass of oxygen  07 g


iiCopper


Mass of copperIIOxide  35 g


Mass of oxygen  28 g


3 Calculate the number of moles of


i Copper used Cu  635


Molar mass 635


ii Oxygen used O  160


Molar mass 160


Moles of oxygen 00441moles 1


CuO copperIIoxide


Observation


Colour change from black to brown


Explanation


CopperIIoxide powder is black On heating it is reduced by a suitable reducing agent to brown copper metal


7 Explain why magnesium oxide would be unsuitable in a similar experiment as the one above


Magnesium is high in the reactivity series None of the above reducing agents is strong enough to reduce the oxide to the metal


8 Write the equation for the reaction that would take place when the reducing agent is


i Hydrogen


CuOs  H2g  Cus  H2Ol


Black brown colourless liquid form


on cooler parts


iiCarbonIIoxide


CuOs  CO g  Cus  CO2g


Black brown colourless gas form


white ppt with lime water


iiiAmmonia


3CuOs  2NH3g  3Cus  N2 g  3H2Ol


9 Explain why the following is necessary during the above experiment


iA stream of dry hydrogen gas should be passed before heating copper II Oxide


Air combine with hydrogen in presence of heat causing an explosion


iiA stream of dry hydrogen gas should be passed after heating copper II Oxide has been stopped


Hot metallic copper can be reoxidized back to copperIIoxide


iii A stream of excess carbon IIoxide gas should be ignited to burn


Carbon IIoxide is highly poisonoustoxic On ignition it burns to form less toxic carbon IVoxide gas


10 State two sources of error in this experiment


iAll copperIIoxide may not be reduced to copper


iiSome copperIIoxide may be blown out the boat by the reducing agent


4Theoreticaly the empirical formula of a compound can be determined as in the following examples


aA oxide of copper contain 80 by mass of copper Determine its empirical formula Cu  635 160


of Oxygen  100   of Copper  100 80  20 of Oxygen


Empirical formula is CuO


b160g of an oxide of Magnesium contain 084g by mass of Magnesium Determine its empirical formulaMg  240 160


Mass of Oxygen  160  084  056 g of Oxygen


Empirical formula is MgO


cAn oxide of Silicon contain 47 by mass of Silicon What is its empirical formulaSi  280 160


Mass of Oxygen  100  47  53 of Oxygen


Empirical formula is SiO2


dA compound contain 70 by mass of Iron and 30 Oxygen What is its empirical formulaFe  560 160


2During heating of a hydrated copper IIsulphateVI crystals the following readings were obtained


Mass of evaporating dish 3000g


Mass of evaporating dish  hydrated salt  3050g


Mass of evaporating dish  anhydrous salt  3032g


Calculate the number of water of crystallization molecules in hydrated copper IIsulphateVI


Cu 645 S  320O160 H  10


Working


Mass of Hydrated salt  3050g 3000g  50g


Mass of anhydrous salt  3032 g 3000g  32 g


Mass of water in hydrated salt  50g 32 g  18g


Molar mass of waterH2O  180g


Molar mass of anhydrous copper IIsulphateVI CuSO4  1605g


The empirical formula of hydrated salt  CuSO45H2O


Hydrated salt has five5 molecules of water of crystallizations


4 The molecular formula is the actual number of each kind of atoms present in a molecule of a compound


The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds the empirical formula may not be the same as the chemical formula


The molecular formula is a multiple of empirical formula It is determined from the relationship


i n  Relative formular mass


Relative empirical formula


where n is a whole number


ii Relative empirical formula x n  Relative formular mass where n is a whole number


Practice sample examples


1 A hydrocarbon was found to contain 923 carbon and the remaining Hydrogen


If the molecular mass of the compound is 78 determine the molecular formulaC120 H 10


Mass of Hydrogen  100  923  77 of Oxygen


Empirical formula is CH


The molecular formular is thus determined


n  Relative formular mass  78  6


Relative empirical formula 13


The molecular formula is C H  x 6  C6H6


2 A compound of carbon hydrogen and oxygen contain 5455 carbon 909 and remaining 3636 oxygen


If its relative molecular mass is 88 determine its molecular formulaC120 H 10 O 160


Empirical formula is C2H4O


The molecular formula is thus determined


n  Relative formular mass  88  2


Relative empirical formula 44


The molecular formula is C2H4O  x 2  C4H8O2


4A hydrocarbon burns completely in excess air to form 528 g of carbon IV oxide and 216g of water


If the molecular mass of the hydrocarbon is 84 draw and name its molecular structure


Since a hydrocarbon is a compound containing Carbon and Hydrogen only Then


Mass of carbon in CO2  Mass of C in CO2 x mass of CO2


Molar mass of CO2


12 x 528  144g


Mass of Hydrogen in H2O  Mass of C in H2O x mass of H2O


Molar mass of H2O


2 x 216  024g


Empirical formula is CH2


n  Relative formular mass  84  6


Relative empirical formula 14


The molecular formula is CH2  x 6  C6H12


molecular name HexeneHex1ene or any position isomer of Hexene


Molecular structure


H H H H H H


H C C C C C C H


H H H H


5 Compound A contain 52 by mass of Nitrogen The other elements present are Carbon hydrogen and Oxygen On combustion of 0085g of A in excess Oxygen0224g of carbonIVoxide and 00372g of water was formed Determine the empirical formula of A N140 O160  C120  H10


Mass of N in A  52 x 0085  000442 g


Mass of C in A  12 x 0224  00611g


Mass of H in A  2 x 00372  00041g


Mass of O in A  0085g  0004442g  00806g Mass of CHO


00611g  00041g  00652g Mass of CH


00806g Mass of CHO 00652g Mass of CH  00154 g


Empirical formula  C16H13NO3


dMolar gas volume


The volume occupied by one mole of all gases at the same temperature and pressure is a constantIt is


i 24dm324litres24000cm3 at room temperature25oC298Kand pressurertp


ie 1mole of all gases 24dm324litres24000cm3 at rtp


Examples


1mole of O2  32g 60 x1023 particles 24dm324litres24000cm3 at rtp


1mole of H2  2g 60 x1023 particles 24dm324litres24000cm3 at rtp


1mole of CO2  44g  60 x1023 particles 24dm324litres24000cm3 at rtp


1mole of NH3  17g 60 x1023 particles  24dm324litres24000cm3 at rtp


1mole of CH4  16g 60 x1023 particles 24dm324litres24000cm3 at rtp


ii224dm3224litres22400cm3 at standard temperature0oC273K and pressurestp


ie 1mole of all gases 224dm3224litres22400cm3 at stp


1mole of O2  32g 60 x1023 particles 224dm3224litres22400cm3 at stp


1mole of H2  2g 60 x1023 particles  224dm3224litres22400cm3 at stp


1mole of CO2  44g  60 x1023 particles  224dm3224litres22400cm3 at stp


1mole of NH3  17g 60 x1023 particles 224dm3224litres22400cm3 at stp


1mole of CH4  16g 60 x1023 particles  224dm3224litres22400cm3 at stp


The volume occupied by one mole of a gas at rtp or stp is commonly called the molar gas volume Whether the molar gas volume is at rtp or stp must always be specified


From the above therefore a less or more volume can be determined as in the examples below


Practice examples


1 Calculate the number of particles present in


Avogadros constant 60 x1023mole1


i 224dm3 of Oxygen


224dm3  60 x1023


224dm3  224 x 60 x1023


60 x1022 molecules  2 x 60 x1022  12 x1023 atoms


ii 224dm3 of CarbonIVoxide


60 x1022 molecules  CO2  3 x 60 x1022  18 x1023 atoms


2 0135 g of a gaseous hydrocarbon X on complete combustion produces 041g of carbonIVoxide and 0209g of water029g of X occupy 120cm3 at room temperature and 1 atmosphere pressure Name X and draw its molecular structureC120O 16OH101 mole of gas occupies 24dm3 at rtp


Molar mass CO2 44 gmole1 Molar mass H2O  18 gmole1


Molar mass X  029 x 24 x 1000cm3  58 gmole1


Since a hydrocarbon is a compound containing Carbon and Hydrogen only Then


Mass of carbon in CO2  Mass of C in CO2 x mass of CO2


Molar mass of CO2


12 x 041  01118g


Mass of Hydrogen in H2O  Mass of C in H2O x mass of H2O


Molar mass of H2O


2 x 0209  00232g


Empirical formula is C2H5


The molecular formular is thus determined


n  Relative formular mass  58  2


Relative empirical formula 29


The molecular formula is C2H5  x 2  C4H10


Molecule name Butane


Molecula structure


H H H H


H C C C C H


eGravimetric analysis


Gravimetric analysis is the relationship between reacting masses and the volumes and or masses of products All reactants are in mole ratios to their products in accordance to their stoichiometric equation Using the mole ration of reactants and products any volume andor mass can be determined as in the examples


1 Calculate the volume of carbonIVoxide at rtp produced when 50 g of calcium carbonate is strongly heatedCa400 C 120O  1601 mole of gas 224 at rtp


Chemical equation


CaCO3s  CaOs  CO2g


Mole ratios 1 1 1


Molar Mass CaCO3 100g


Method 1


100g CaCO3s  24dm3 CO2g at rtp


50 g CaCO3s  50 g x 24dm3  12dm31200cm3


Method 2


Moles of 50 g CaCO3s  50 g  005 moles


100 g


Mole ratio 11


Moles of CO2g  005moles


Volume of CO2g  005 x 24000cm3 1200cm3 12dm3


2 10g of an alloy of aluminium and copper were reacted with excess hydrochloric acid If 840cm3 of hydrogen at stp was produced calculate the  of copper in the alloyAl 270one mole of a gas at stp 224dm3


Copper does not react with hydrochloric acid


2Als  6HClaq  2AlCl3aq  3H2g


3H2g  3 moles x 224 x 1000cm3  2 x 27 g Al


840cm3  840cm3 x 2 x 27  0675g of Aluminium


3 x 224 x 1000


Total mass of alloy  mass of aluminium  mass of copper


10g  0675g 0325g of copper


copper  mass of copper x100  325


Mass of alloy


Mole ratio 2Al 3H2  23


Moles of Hydrogen gas  volume of gas  840cm3  00375moles


Molar gas volume 22400cm3


Moles of Al  23 moles of H2  23x 00375moles  0025moles


Mass of Al  moles x molar mass 0025moles x 27  0675g


10g  0675g  0325 g of copper


fGay Lussacs law


Gay Lussacs law states that when gases combinereact they do so in simple volume ratios to each other and to their gaseous products at constantsame temperature and pressure


Gay Lussacs law thus only apply to gases


Given the volume of one gas reactant the other gaseous reactants can be deduced thus


Examples


1 Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen


Chemical equation 2H2 g  O2 g  2H2Ol


Volume ratios 2  1  0


Reacting volumes 50cm3  25cm3


50cm3 of Oxygen is used


2 Calculate the volume of air required to completely reacts with 50cm3 of Hydrogenassume Oxygen is 21 by volume of air


21  25cm3


100  100 x 25


3If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of CarbonIVoxide and 10cm3 of water vapoursteam obtain the equation for the reaction and hence find the value of x and y in CxHy


Chemical equation CxHy g  O2 g  H2Og  CO2g


Volumes 5cm3  15cm3  10cm3  10cm3


Volume ratios 5cm3  15cm3  10cm3  10cm3 divide by lowest volume 5 5 5 5


Reacting volume ratios 1volume 3 volume 2 volume 2 volume


Balanced chemical equation CxHy g  3O2 g  2H2Og  2CO2g


If 4H are in 2H2Og the y4


If 2C are in 2CO2 g the x2


Thusi chemical formula of hydrocarbon  C2H4


ii chemical name of hydrocarbon  Ethene


4100cm3 of nitrogen IIoxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen All volumes measured at the same temperature and pressure Obtain the equation for the reaction and name the gaseous product


Chemical equation NO g  O2 g  NOx


Volumes 100cm3  50cm3  100


Volume ratios 100cm3  50cm3  100cm3 divide by lowest volume 50 50 50


Reacting volume ratios 2volume 1 volume 2 volume


Balanced chemical equation 2 NO g  O2 g  2NO xg


Thusi chemical formula of the nitrogen compound  2 NO2


ii chemical name of compound  NitrogenIVoxide


5When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen the resulting gaseous mixture occupied70cm3 at room temperature and pressure When the gaseous mixture was passed through potassium hydroxide its volume decreased to 25cm3


aWhat volume of Oxygen was used during the reaction1mk


Volume of Oxygen used 10025 75cm3


P was completely burnt


bDetermine the molecular formula of the hydrocarbon2mk


CxHy  O2  xCO2  yH2O


15cm3  75cm3


15 15


1  3


1 atom of C react with 6 3x2atoms of Oxygen


Thus x  1 and y  2  P has molecula formula CH4


g Ionic equations


An ionic equation is a chemical statement showing the movement of ions cations and anions  from reactants to products


Solids gases and liquids do not ionizedissociate into free ions Only ionic compounds in aqueoussolution or molten state ionizedissociate into free cations and anions ions


An ionic equation is usually derived from a stoichiometric equation by using the following guidelines


Guidelines for writing ionic equations


1Write the balanced stoichiometric equation


2Indicate the state symbols of the reactants and products


3Split into cations and anions all the reactants and products that exist in aqueous state


4Cancel out any cation and anion that appear on both the product and reactant side


5 Rewrite the chemical equation It is an ionic equation


Practice


aPrecipitation of an insoluble salt


All insoluble salts are prepared in the laboratory from double decomposition precipitation This involves mixing two soluble salts to form one soluble and one insoluble salt


1 When silver nitrateV solution is added to sodium chloride solutionsodium nitrateV solution and a white precipitate of silver chloride are formed


Balanced stoichiometric equation


AgNO3aq  NaClaq  AgCl s  NaNO3 aq


Split reactants product existing in aqueous state as cationanion


Agaq  NO3 aq  Naaq  Claq  AgCls  Naaq NO3 aq


Cancel out ions appearing on reactant and product side


Rewrite the equation


Agaq  Claq  AgCls ionic equation


2 When barium nitrateV solution is added to copperIIsulphateVI solution copperII nitrateV solution and a white precipitate of barium sulphateVI are formed


BaNO32aq  CuSO4aq  BaSO4 s  CuNO3 2 aq


Ba2aq  2NO3 aq  Cu2aq  SO42aq  BaSO4 s  2NO3 aq Cu2aq


Ba2aq  2NO3 aq Cu2 aq  SO42aq BaSO4s  2NO3 aq  Cu2aq


Ba2aq  SO42aq  BaSO4s ionic equation


3A yellow precipitate of Potassium Iodide is formed from the reaction of LeadIInitratev and potassium iodide


PbNO32aq  2KI aq  PbI2 s  2KNO3 aq


Pb2aq  2NO3 aq  2K aq  2I  aq  PbI2 s  2NO3 aq 2K aq


Pb2aq  2I  aq  PbI2 s ionic equation


bNeutralization


Neutralization is the reaction of an acid with a soluble basealkali or insoluble base


iReaction of alkalis with acids


1Reaction of nitricVacid with potassium hydroxide


Balanced stoichiometric equation


HNO3aq  KOH aq  H2O l  KNO3 aq


Haq  NO3 aq  K aq  OH  aq  H2O l  NO3 aq K aq


Rewrite the equation


H aq  OH  aq  H2O l ionic equation


2Reaction of sulphuricVIacid with ammonia solution


H2SO4aq  2NH4OH aq  H2O l  NH4 2SO4 aq


2Haq  SO42 aq  2NH4 aq 2OH  aq 2H2O l SO42 aq 2NH4  aq


2H aq  2OH  aq  2H2O l


3Reaction of hydrochloric acid with Zinc hydroxide


2HClaq  ZnOH2 s  2H2O l  ZnCl 2 aq


2Haq  2Cl aq  ZnOH2 s 2H2O l  2Cl aq Zn 2 aq


2Haq  ZnOH2 s 2H2O l  Zn 2 aq ionic equation


hMolar solutions


A molar solution is one whose concentration is known The SI unit of concentration is Molarity denoted M


Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution


One cubic decimeter is equal to one litre and also equal to 1000cm3


The higher the molarity the higher the concentration and the highermore solute has been dissolved in the solvent to make one cubic decimeter litre1000cm3 solution


Examples


2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter litre1000cm3 uniform solution mixture of sodium hydroxide and water


002M sodium hydroxide means 002 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter litre1000cm3 uniform solution mixture of sodium hydroxide and water


2M is more concentrated than002M


Preparation of molar solution


Procedure


Weigh accurately 40 g of sodium hydroxide pellets into a 250cm3 volumetric flask


Using a wash bottle add about 200cm3 of distilled water


Stopper the flask


Shake vigorously for three minutes


Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved


Add more water slowly upto exactly the 250 cm3 mark


Sample questions


1Calculate the number of moles of sodium hydroxide pellets present in


i 40 g


Molar mass of NaOH  23  16  1  40g


Moles  Mass  40  01  10 x 10 1 moles


Molar mass 40


ii 250 cm3 solution in the volumetric flask


Moles in 250 cm3  01  10 x 10 1 moles


iii one decimeter of solution


Method 1


Moles in decimeters  Molarity  Moles x 1000cm31dm3


Volume of solution


10 x 10 1 moles x 1000cm3


04 M  04 molesdm3


Method 2


250cm3 solution contain 10 x 10 1 moles


1000cm3 solution  Molarity contain 1000 x 10 x 10 1 moles


250 cm3


Theoretical sample practice


1 Calculate the molarity of a solution containing


i 40 g sodium hydroxide dissolved in 500cm3 solution


Molar mass of NaOH  23  16  1  40g


Moles  Mass  40  01  10 x 10 1 moles


Molar mass 40


02 M  02 molesdm3


500 cm3 solution contain 10 x 10 1 moles


500 cm3


ii 53 g anhydrous sodium carbonate dissolved in 50cm3 solution


Molar mass of Na2CO3  23 x 2  12  16 x 3  106 g


Moles  Mass  53  005  5 0 x 102 moles


Molar mass 106


Method 1


Volume of solution


10 moles x 1000cm3


10 M


Method 2


50 cm3 solution contain 50 x 10 2 moles


1000cm3 solution  Molarity contain 1000 x 50 x 10 2 moles


50 cm3


10M  10 molesdm3


iii 53 g hydrated sodium carbonate decahydrate dissolved in 50cm3 solution


Molar mass of Na2CO310H2O  23 x 2  12  16 x 3  20 x 1  10 x 16 286g


Moles  Mass  53  00185  185 x 10 2 moles


Molar mass 286


185 x 10 2 moles x 1000cm3


037 M037 molesdm3


50 cm3 solution contain 185 x 10 2 moles


1000cm3 solution  Molarity contain 1000 x 185 x 10 2 moles


37 x 101 M  37 x 101 molesdm3


iv 71 g of anhydrous sodium sulphateVIwas dissolved in 200 cm3 solution Calculate the molarity of the solution


Method 1


200cm3 solution 71 g


1000cm3 solution  1000 x 71  3550 g dm3


Molar mass Na2SO4  142 g


Moles dm3  Molarity  Mass 3550  25 M molesdm3


Molar mass 142


Method 2


Moles  Mass  71  005  50 x 10 2 moles


Method 2a


Volume of solution


50 x 10 2 moles x 1000cm3


25 M25 molesdm3


Method 2b


20 cm3 solution contain 50 x 10 2 moles


20 cm3


iv The density of sulphuricVI is 184gcm3 Calculate the molarity of the acid


10cm3 solution 184 g


1000cm3 solution  1000 x 184  1840 g dm3


Molar mass H2SO4  98 g


Moles dm3  Molarity  Mass  1840


Molar mass 98


187755 M molesdm3


Moles  Mass  184  00188  188 x 10 2 moles


Moles in decimeters  Molarity  Moles x 1000cm31dm3


188 x 10 2 moles x 1000cm3


188M188 molesdm3


20 cm3 solution contain 188 x 10 2 moles


1000cm3 solution  Molarity contain 1000 x 188 x 10 2 moles


10 cm3


2 Calculate the mass of


i 25 cm3 of 02M sodium hydroxide solutionNa 230O 160 H10


Molar mass NaOH  40g


Moles in 25 cm3  Molarity x volume  02 x 25  000550 x 103moles


1000 1000


Mass of NaOH Moles x molar mass  50 x 103 x 40  02 g


ii 20 cm3 of 0625 M sulphuricVIacid S 320O 160 H10


Molar mass H2SO4  98g


Moles in 20 cm3  Molarity x volume 0625 x 20  001251250 x 103moles


Mass of H2SO4 Moles x molar mass  50 x 103 x 40  02 g


iii 10 cm3 of 25 M NitricVacid N 140O 160 H10


Molar mass HNO3  63 g


Moles in 1 cm3  Molarity x volume  25 x 1  00025  25 x 103moles


Mass of HNO3 Moles x molar mass  25 x 103 x 40  01 g


3 Calculate the volume required to dissolve


ai 025moles of sodium hydroxide solution to form a 08M solution


Volume in cm3  moles x 1000  025 x 1000  3125cm3


Molarity 08


ii 100cm3 was added to the sodium hydroxide solution above Calculate the concentration of the solution


C1 x V1  C2 x V2 where


C1  molarityconcentration before dilutingadding water


C2  molarityconcentration after dilutingadding water


V1  volume before dilutingadding water


V2  volume after dilutingadding water


08M x 3125cm3  C2 x 3125  100


C2  08M x 3125cm3  06061M


bii 001M solution containing 001moles of sodium hydroxide solution


Volume in cm3  moles x 1000  001 x 1000  1000 cm3


Molarity 001


ii Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0008M solution


001M x 1000 cm3  0008 x V2


V2  001M x 1000cm3  1250cm3


Volume added  1250  1000  250cm3


cVolumetric analysisTitration


Volumetric analysisTitration is the process of determining unknown concentration of one reactant from a known concentration and volume of another


Reactions take place in simple mole ratio of reactants and products


Knowing the concentration volume of one reactant the other can be determined from the relationship


M1V1  M2V2 where


n1 n2


M1  Molarity of 1st reactant


M2  Molarity of 2nd reactant


V1  Volume of 1st reactant


V1  Volume of 2nd reactant


n1  number of moles of 1st reactant from stoichiometric equation


n2  number of moles of 2nd reactant from stoichiometric equation


Examples


1Calculate the molarity of MCO3 if 50cm3 of MCO3 react with 250cm3 of 05M hydrochloric acidC120 O 160


Stoichiometric equation MCO3s  2HClaq  MCl2aq  CO2g  H2Ol


Method 1


M1V1  M2V2  M1 x 50cm3  05M x 250cm3


n1 n2 1 2


M1  05 x 250 x1  125M  125 moledm3


50 x 2


Method 2


Moles of HCl used  molarity x volume


05 x 250  00125 125 x 102moles


Mole ratio MCO3  HCl  12


Moles MCO3  00125 125 x 102moles  000625  625 x 103 moles


Molarity MCO3  moles x 1000  000625  625 x 103 x 1000


Volume 5


125M  125 moledm3


2 20cm3 of 05M hydrochloric acid react with 01M of M2CO3 Calculate the volume of 01M M2CO3 used


Stoichiometric equation M2CO3 aq  2HClaq  2MCl aq  CO2g  H2Ol


M1V1  M2V2  05 x 20cm3  01M x V2 cm3


n1 n2 2 1


V2  05 x 20 x1  125M  125 moledm3


01 x 2


05 x 20  00125 125 x 102moles


Mole ratio M2CO3  HCl  12


Moles M2CO3  00125 125 x 102moles  000625  625 x 103 moles


Molarity M2CO3  moles x 1000  000625  625 x 103 x 1000


3 50cm3 of 01M sodium iodide react with 01M of LeadIInitrateV Calculatei the volume of LeadIInitrateV used


iithe mass of LeadIIIodide formed


Pb2070 I 1270


Stoichiometric equation 2NaIaq  PbNO32aq  2NaNO3aq  PbI2s


iVolume of LeadIInitrateV used


Method 1


M1V1  M2V2  5 x 01cm3  01M x V2 cm3


V2  01 x 50 x 1  125M  125 moledm3


Method 2


Moles of HCl used  molarity x volume


01 x 50  00125 125 x 102moles


Volume 5


125M  125 moledm3


4 0388g of a monobasic organic acid B required 465 cm3 of 0095M sodium hydroxide for complete neutralization Name and draw the structural formula of B


Moles of NaOH used  molarity x volume


0095 x 465  00044175 44175 x 103moles


Mole ratio B  NaOH  11


Moles B  00044175 44175 x 103moles


Molar mass B  mass  0388


moles 00044175 44175 x 103moles


878324 gmole1


XCOOH  878324 where X is an alkyl group


X 878324 42  42832443


By elimination CH3  15 CH3CH2  29 CH3CH2 CH2  43


Molecula formula  CH3CH2 CH2COOH


Molecule name  Butan1oic acid


Molecular structure


H H H O


H C C C C O H


H H H H


5 105 g of an impure sample containing ammonium sulphate VI fertilizer was warmed with 250cm3 of o8M sodium hydroxide solutionThe excess of the alkali was neutralized by 85cm3 of 05M hydrochloric acid Calculate the  of impurities in the ammonium sulphate VIfertilizer N140S320O160 H10


Equation for neutralization


NaOHaq  HClaq  NaOHaq  H2Ol


Mole ratio NaOHaqHClaq 11


Moles of HCl  Molarity x volume  05 x 85  00425 moles


1000 1000


Excess moles of NaOHaq 00425 moles


Equation for reaction with ammonium salt


2NaOHaq  NH4 2SO4aq  Na 2SO4aq  2NH3 g 2H2Ol


Mole ratio NaOHaq NH4 2SO4aq 21


Total moles of NaOH  Molarity x volume  08 x 250  02 moles


Moles of NaOH that reacted withNH4 2SO4  02  00425  01575moles


Moles NH4 2SO4   x 01575moles  0 07875moles


Molar mass NH4 2SO4 132 gmole1


Mass of in impure sample  moles x molar mass 0 07875 x 132  10395 g


Mass of impurities  105 10395  0105 g


impurities  0105 x 100  10


Practically volumetric analysis involves titration


Titration generally involves filling a burette with knownunknown concentration of a solution then adding the solution to unknownknown concentration of another solution in a conical flask until there is complete reaction If the solutions used are both colourless an indicator is added to the conical flask When the reaction is over a slightlittle excess of burette contents change the colour of the indicator This is called the end point


Set up of titration apparatus


The titration process involve involves determination of titre The titre is the volume of burette contentsreading before and after the end point Burette contentsreading before titration is usually called the Initial burette reading Burette contentsreading after titration is usually called the Final burette reading The titre value is thus a sum of the Final less Initial burette readings


To reduce errors titration process should be repeated at least once more


The results of titration are recorded in a titration table as below


Sample titration table


As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration


For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council


As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided


Calculate the average volume of solution used


240  240  240  240 cm3


As evidence of understanding the degree of accuracy of burettes  all readings must be recorded to a decimal point


As evidence of accuracy in carrying the out the titration  candidates value should be within 02 of the school value


The school value is the teachers readings presented to the examining bodycouncil based on the concentrations of the solutions she presented to herhis candidates


Bonus mark is awarded for averaged reading within 01 school value as Final answer


Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the


irelationship among the mole molar mass mole ratios concentration molarity


ii mathematical application of 1st principles


Very useful information which candidates forget appears usually in the beginning of the question paper as


You are provided with


All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point


Candidates are expected to use a non programmable scientific calculator


aSample Titration Practice 1 Simple Titration


01M sodium hydroxide solution A


Hydrochloric acid solution B


You are required to determine the concentration of solution B in moles per litre


Procedure


Fill the burette with solution B Pipette 250cm3 of solution A into a conical flask Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1


Sample results Titration table 1


Sample worked questions


1 Calculate the average volume of solution B used


Average titre  Titre 1  Titre 2 Titre 3   200 200 200   200cm3


3 3


2 How many moles of


isolution A were present in 25cm3 solution


Moles of solution A  Molarity x volume  01 x 25  25 x 103 moles


1000 1000


iisolution B were present in the average volume


Chemical equation NaOHaq  HClaq  NaClaq  H2Ol


Mole ratio 11  Moles of A  Moles of B  25 x 103 moles


iii solution B in moles per litre


Moles of B per litre  moles x 1000  25 x 103 x 1000  01M


Volume 20


bSample Titration Practice 2 Redox Titration


You are provided with


Acidified Potassium manganateVII solution A


01M of an iron IIsalt solution B


85g of ammonium ironIIsulphateVI crystalsNH42 SO4FeSO4xH2O solid C


You are required to


istandardize acidified potassium manganateVII


iidetermine the value of x in the formula NH42 SO4FeSO4xH2O


Procedure 1


Fill the burette with solution A Pipette 250cm3 of solution B into a conical flask Titrate solution A with solution B until a pink colour just appears


Record your results to complete table 1


Table 1Sample results


Sample worked questions


1 Calculate the average volume of solution A used


3 3


2 How many moles of


isolution B were present in 25cm3 solution


1000 1000


iisolution A were present in the average volume Assume one mole of B react with five moles of B


Mole ratio A  B  15


Moles of A  Moles of B  25 x 103 moles  50 x 10 4 moles


5 5


iii solution B in moles per litre


Moles of B per litre  moles x 1000  25 x 103 x 1000


Volume 20


0025 M moles per litre moles l1


Procedure 2


Place all the solid C into the 250cm3 volumetric flask carefully Add about 200cm3 of distilled water Shake to dissolve Make up to the 250cm3 of solution by adding more distilled water Label this solution C Pipette 25cm3 of solution C into a conical flask Titrate solution C with solution A until a permanent pink colour just appears Complete table 2


Table 2Sample results


Sample worked questions


Average titre  Titre 1  Titre 2 Titre 3   200 200 200   200cm3


3 3


2 How many moles of


isolution A were present inin the average titre


Moles of solution A  Molarity x volume  0025 x 20  50 x 104 moles


1000 1000


iisolution C in 25cm3 solution given the equation for the reaction


MnO4 aq  8Haq  5Fe2 aq  Mn2aq  5Fe3aq  4H2Ol


Mole ratio MnO4 aq 5Fe2 aq  15  Moles of 5Fe2 aq  Moles of MnO4 aq  50 x 104 moles  10 x 10 4 moles


5 5


iii solution B in 250cm3


Moles of B per litre  moles x 250  10 x 10 4 x 250  10 x 10 3 moles Volume 25


3 Calculate the molar mass of solid C and hence the value of x in the chemical formula NH42SO4FeSO4xH2O


N140 S320 Fe560 H10 O160


Molar mass  mass perlitre  85  8500 g


Moles per litre 10 x 10 3 moles


NH42SO4FeSO4xH2O  8500


284  18x 8500


8500  284  8216  18x  4544444


18 18


x  454 whole number


cSample Titration Practice 3 Back titration


You are provided with


ian impure calcium carbonate labeled M


iiHydrochloric acid labeled solution N


iiisolution L containing 20g per litre sodium hydroxide


You are required to determine the concentration of N in moles per litre and the  of calcium carbonate in mixture M


Procedure 1


Pipette 250cm3 of solution L into a conical flask Add 23 drops of phenolphthalein indicator Titrate with dilute hydrochloric acid solution N and record your results in table 14mark


Sample Table 1


Sample questions


a Calculate the average volume of solution N used


65  65  65  65 cm3


b How many moles of sodium hydroxide are contained in 25cm3of solution L


Molar mass NaOH 40g


Molarity of L  mass per litre  20  05M


Molar mass NaOH 40


Moles NaOH in 25cm3  molarity x volume  05M x 25cm3  00125 moles


1000 1000


cCalculate


ithe number of moles of hydrochloric acidthat react with sodium hydroxide in babove


Mole ratio NaOH  HCl from stoichiometric equation 11


Moles HCl Moles NaOH  00125 moles


iithe molarity of hydrochloric acid solution N


Molarity  moles x 1000  00125 moles x 1000 19231Mmoledm3


65 65


Procedure 2


Place the 40 g of M provided into a conical flask and add 250cm3 of the dilute hydrochloric acid to it using a clean pipette Swirl the contents of the flask vigorously until effervescence stopUsing a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3Label this solution KUsing a clean pipettetransfer 250cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 23 drops of methyl orange indicatorRecord your observations in table 2


Sample Table 2


Sample calculations


aCalculate the average volume of solution L used1mk


245  245  245  245cm3


bHow many moles of sodium hydroxide are present in the average volume of solution L used


Moles  molarity x average burette volume  05 x 245


001225 1225 x 102 moles


c How many moles of hydrochloric acid are present in the original 200cm3 of solution K


Mole ratio NaOH HCl  11  moles of HCl  001225 1225 x 102 moles


Moles in 200cm3  200cm3 x 001225 1225 x 102moles


25cm3volume pipetted


049 49 x 101moles


dHow many moles of hydrochloric acid were contained in original 25 cm3 solution N used


Original moles  Original molarity x pipetted volume


19231Mmoledm3 x 25  0048074807 x 102 moles


eHow many moles of hydrochloric acid were used to react with calcium carbonate present


Moles that reacted  original moles moles in average titre


0048074807 x 102moles  001225 1225 x 102moles


0035823582 x 10 2 moles


fWrite the equation for the reaction between calcium carbonate and hydrochloric acid


CaCO3s  2HClaq  CaCl2aq  CO2g  H2Ol


gCalculate the number of moles of calcium carbonate that reacted with hydrochloric acid


From the equation CaCO3s2HClaq  12


Moles CaCO3s  12moles HCl


12 x 0035823582 x 10 2 moles


001791 1791 x 102moles


hCalculate the mass of calcium carbonate in 40g of mixture M Ca400O  160C120


Molar mass CaCO3  100g


Mass CaCO3  moles x molar mass  001791 1791 x 102moles x 100g


iDetermine the  of calcium carbonate present in the mixture


CaCO3  mass of pure x 100  1791g x 100  44775


Mass of impure 40


dSample titration practice 4 Multiple titration


You are provided with


isodium L containing 50g per litre of a dibasic organic acid H2X2H2O


iisolution M which is acidified potassium manganateVII


iiisolution N a mixture of sodium ethanedioate and ethanedioic acid


iv01M sodium hydroxide solution P


v10M sulphuricVI


You are required to


istandardize solution M using solution L


iiuse standardized solution M and solution P to determine the  of sodium ethanedioate in the mixture


Procedure 1


Fill the burette with solution M Pipette 250cm3 of solution L into a conical flask Heat this solution to about 70oCbut not to boilTitrate the hot solution L with solution M until a permanent pink colour just appears Shake thoroughly during the titration Repeat this procedure to complete table 1


Sample Table 1


Sample calculations


aCalculate the average volume of solution L used 1mk


240  240  240  240cm3


bGiven that the concentration of the dibasic acid is 005molesdm3determine the value of x in the formula H2X2H2O H10O160


Molar mass H2X2H2O mass per litre  50glitre  100g


Moleslitre 005molesdm3


H2X2H2O 100


X  100  2 x1  2 x 2 x1  2 x 16  100  34  66


c Calculate the number of moles of the dibasic acid H2X2H2O


Moles  molarity x pipette volume  05 x 25  00125125 x10 2 moles


1000 1000


dGiven the mole ratio manganateVIIMnO4 acid H2X is 25 calculate the number of moles of manganateVII MnO4 in the average titre


Moles H2X  25 moles of MnO4


25 x 00125125 x10 2 moles


000550 x 10 3moles


eCalculate the concentration of the manganateVIIMnO4 in moles per litre


Moles per litremolarity  moles x 1000


average burette volume


000550 x 10 3moles x 1000  02083 molesl1M


Procedure 2


With solution M still in the burette pipette 250cm3 of solution N into a conical flask Heat the conical flask containing solution N to about 70oCTitrate while hot with solution MRepeat the experiment to complete table 2


Sample Table 2


Sample calculations


125  125  125 125cm3


bCalculations


iHow many moles of manganateVIIions are contained in the average volume of solution M used


Moles  molarity of solution M x average burette volume


02083 molesl1 M x 125  00026  25 x 103 moles


iiThe reaction between manganateVIIions and ethanedioate ions that reacted with is as in the equation


2MnO4 aq  5C2O42 aq  16H aq  2Mn2aq  10CO2g  8H2Ol


Calculate the number of moles of ethanedioate ions that reacted with manganate VII ions in the average volume of solution M


From the stoichiometric equationmole ratio MnO4 aq C2O42 aq  25


moles C2O42  52 moles MnO4  52 x 00026  25 x 103 moles


00065 65 x103 moles


iiiCalculate the number of moles of ethanedioate ions contained in 250cm3 solution N


25cm3 pipette volume  00065 65 x103 moles


00065 65 x103 moles x 250  0065  65 x102 moles


Procedure 3


Remove solution M from the burette and rinse it with distilled water Fill the burette with sodium hydroxide solution P Pipette 25cm3 of solution N into a conical flask and add 23 drops of phenolphthalein indicator Titrate this solution N with solution P from the burette Repeat the procedure to complete table 3


Sample Table 2


Sample calculations


aCalculate the average volume of solution L used 1mk


249  249  249  249 cm3


bCalculations


iHow many moles of sodium hydroxide solution P were contained in the average volume


Moles  molarity of solution P x average burette volume


01 molesl1 x 249  000249  249 x 103 moles


iiGiven that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is


2NaOH aq  H2C2O4 aq  Na2C2O4g  2H2Ol


Calculate the number of moles of ethanedioic acid that were used in the reaction


From the stoichiometric equationmole ratio NaOHaq H2C2O4 aq  21


moles H2C2O4  12 moles NaOH  12 x 000249  249 x 103 moles


00012451245 x103 moles


iiiHow many moles of ethanedioic acid were contained in 250cm3 of solution N


25cm3 pipette volume  00012451245 x103 moles


00012451245 x103 moles x 250  0012451245 x102 moles


iiiDetermine the  by mass of sodium ethanedioate in the micture H 10O160C120 and total mass of mixture 20 g in 250cm3 solution


Molar mass H2C2O4  900g


Mass of H2C2O4 in 250cm3  moles in 250cm3 x molar mass H2C2O4


0012451245 x102 moles x 900


by mass of sodium ethanedioate


Mass of mixture  mass of H2C2O4 x 100


Mass of mixture


20  11205 g  43975


Note


i L is 005M Oxalic acid


ii M is 001M KMnO4


iii N is 003M oxalic acidwithout sodium oxalate


Practice example 5Determining equation for a reaction


You are provided with


01M hydrochloric acid solution A


05M sodium hydroxide solution B


You are to determine the equation for thereaction between solution A and B


Procedure


Fill the burette with solution AUsing a pipette and pipette filler transfer 250cm3 of solution B into a conical flaskAdd 23 drops of phenolphthalein indicatorRun solution A into solution B until a permanent pink colour just appearsRecord your results in Table 1Repeat the experiment to obtain three concordant results to complete Table 1


Table 1Sample results


Sample questions


Calculate the average volume of solution A used


125125125  125cm3


Theoretical Practice examples


1 10g of dibasic acid HOOCCH2xCOOH was dissolved in 250cm3 solution 250 cm3 of this solution reacted with 300cm3 of 006M sodium hydroxide solution Calculate the value of x in HOOCCH2xCOOH C120H10O16


Chemical equation


2NaOHaq  H2Xaq  Na2X aq  2H2Oaq


Mole ratio NaOHaq H2Xaq  21


Method 1


Ma Va  na  Ma x 250  1  Ma 006 x 300 x1


Mb Vb  nb 006 x 300 2 250 x 2


Molarity of acid  0036MMole l1


Mass of acid per lite  10 x1000  40 gl


0036M Mole l1  40 g l


1 mole molar mass of HOOCCH2xCOOH  40 x 1  1111111 g


Molar mass CH2x  1111111  HOOCCOOH  900  211111


CH2x  14x  211111  15  1 whole number


Method 2


Moles of sodium hydroxide  Molarity x volume  006 x 30  18 x 10 3moles


Moles of Hydrochloric acid  12 x 18 x 10 3moles  90 x10 4moles


Molarity of Hydrochloric acid  moles x 1000  90 x10 4moles x1000


Volume 25


2 200cm3 of 005 M acidified potassium manganateVIIsolution oxidized 250cm3 of Fe2aq ions in 400gl of impure Iron IIsulphateVI to Fe3aq ions Calculate the percentage impurities in the Iron IIsulphateVI


MnO4 aq  8Haq 5Fe2aq 5Fe3aq  Mn2aq  4H2Oaq


Fe560S 320 O160


Moles of MnO4 aq  Molarity x volume  005 x 200  0001 Moles


1000 1000


Mole ratio MnO4 aq 5Fe2aq 15


Moles 5Fe2aq  5 x0001  0005 Moles


Moles of 5Fe2aq per litremolarity  Moles x 1000  0005 x 1000


Volume 250


02 M Moleslitre


Molar mass FeSO4152 g


Mass of in the mixture  Moles x molar mass  02 x 152  304 g


Mass of impurity  400  304 96 g


impurity  96 g x100  240  impurity


397 g of a mixture of Potassium hydroxide and Potassium chloride was dissolved to make one litre solution200cm3 of this solution required 250cm3 of 012M hydrochloric acid for completed neutralization Calculate the percentage by mass of Potassium chlorideK390Cl 355


Chemical equation


KOHaq  HClaq  KClaq  H2Ol


Moles of HCl  Molarity x volume  012 x 250  000330 x 10 3 moles


Mole ratio KOHaq  HClaq  11


Moles KOH 000330 x 10 3 moles


Method 1


Molar mass KOH 560g


Mass KOH in 25cm3 000330 x 10 3 moles x560  0168g


Mass KOH in 1000cm31 litre  0168 x1000 84 gl


Mass of KCl  97g  84g  13 g


of KCl  13 x 100  134021


Method 2


Moles KOH in 1000cm3 1 litre  Moles in 20cm3 x 1000 0003 x 1000


20 20


015MMoles litre


Mass KOH in 10001 litre  015MMoles litre x 560  84gl


4A certain carbonate GCO3 reacts with dilute hydrochloric acid according to the equation given below


GCO3s  2HClaq  GCl2 aq  CO2 g  H2Ol


If 1 g of the carbonate reacts completely with 20 cm3 of 1 M hydrochloric acid calculate the relative atomic mass of G C  120  160


Moles of HCl  Molarity x volume 1 x20  002 moles


1000 1000


Mole ratio HCl GCO3  21


Moles of GCO3 002 moles  001moles


Molar mass of GCO3  mass  1  100 g


moles 001moles


G GCO3  CO3 100g  12 16 x3  60  40no units


5 460g of a metal carbonate MCO3 was dissolved 160cm3 of 01M excess hydrochloric acid and the resultant solution diluted to one litre250cm3 of this solution required 200cm3 of 01M sodium hydroxide solution for complete neutralization Calculate the atomic mass of M


Equation


Chemical equation


NaOHaq  HClaq  KClaq  H2Ol


Moles of NaOH  Molarity x volume 01 x20  0002 moles


Mole ratio HCl NaOH  11


Excess moles of HCl  0002 moles


25cm3  0002 moles


1000cm3  1000 x 0002  008moles


Original moles of HCl  Molarity x volume  1M x 1litre  10 moles


Moles of HCl reacted with MCO3  10  008 moles  092moles


MCO3s  2HClaq  MCl2 aq  CO2 g  H2Ol


Mole ratio MCO3s  HClaq 12


Moles of MCO3  092moles  046moles


Molar mass of MCO3 mass  46g  100 g


moles 046moles


M MCO3  CO3 100g  12 16 x3  60  40


6 250cm3 of a mixture of Fe2 and Fe3 ions in an aqueous salt was acidified with sulphuricVIacid then titrated against potassium manganateVIThe salt required 15cm3 ofe002M potassium manganateVI for complete reaction


A second 25cm3 portion of the Fe2 and Fe3 ion salt was reduced by Zinc then titrated against the same concentration of potassium manganateVI190cm3 of potassium manganateVIsolution was used for complete reaction Calculate the concentration of Fe2 and Fe3 ion in the solution on moles per litre


Mole ratio Fe2 Mn04  51


Moles Mn04 used  002 x 15  30 x 104 moles


Moles Fe2  30 x 104 moles  60 x 105 moles


Molarity of Fe2  60 x 104 moles x 1000  24 x 103 moles l1


Since Zinc reduces Fe3 to Fe2 in the mixture


Moles Mn04 that reacted with all Fe2 002 x 19  38 x 104 moles


Moles of all Fe2  38 x 104 moles  76 x 105 moles


Moles of Fe3  38 x 104  60 x 105  16 x 105 moles


Molarity of Fe3  16 x 105 moles x 1000  40 x 104 moles l1


CHEMISTRY FORM FOUR NOTES


ACIDS BASES AND SALTS


AACIDS AND BASES


At a school laboratory


iAn acid may be defined as a substance that turn litmus red


iiA base may be defined as a substance that turn litmus blue


Litmus is a lichen found mainly in West Africa It changes its colour depending on whether the solution it is in is basicalkaline or acidicIt is thus able to identifyshow whether


1 An acid is a substance that dissolves in water to form HH3O as the only positive ioncation This is called the Arrhenius definition of an acid From this definition an acid dissociateionize in water releasing H thus


HClaq  H aq  Cl aq


HNO3aq  H aq  NO3 aq


CH3COOHaq  H aq  CH3COOaq


H2SO4aq  2H aq  SO42aq


H2CO3aq  2H aq  CO32aq


H3PO4aq  3H aq  PO43aq


2A base is a substance which dissolves in water to form OH as the only negatively charged ionanion


This is called Arrhenius definition of a base


From this definition a base dissociateionize in water releasing OH thus


KOHaq  Kaq  OHaq


NaOHaq  Naaq  OHaq


NH4OHaq  NH4aq  OHaq


CaOH2aq  Ca2aq  2OHaq


MgOH2aq  Mg2aq  2OHaq


3 An acid is a proton donor


A base is a proton acceptor


This is called BronstedLowry definition of acids and bases


From this definition an acid donates H


H has no electrons and neutrons It contains only a proton


Examples


I From the equation


HClaq  H2Ol  H3Oaq  Cl aq


aiFor the forward reaction from left to right H2O gains a proton to form H3O and thus H2O is a proton acceptor It is a BronstedLowry base


ii For the backward reaction from right to left H3O donates a proton to form H2O and thus H3O is an opposite proton donor It is a BronstedLowry conjugate acid


biFor the forward reaction from left to right HCl donates a proton to form Cl and thus HCl is a proton donor


It is a BronstedLowry acid


ii For the backward reaction from right to left Cl gains a proton to form HCl and thus Cl is an opposite proton acceptor


It is a BronstedLowry conjugate base


Every base acid from BronstedLowry definition thus must have a conjugate productreactant


II From the equation


HClaq  NH3aq  NH4aq  Cl aq


aiFor the forward reaction from left to right NH3 gains a proton to form NH4 and thus NH3 is a proton acceptor


It is a BronstedLowry base


ii For the backward reaction from right to left NH4 donates a proton to form NH3 and thus NH4 is an opposite proton donor


It is a BronstedLowry conjugate acid


4 Acids and bases show acidic and alkaline propertiescharacteristics only in water but not in other solvents eg


aHydrogen chloride gas dissolves in water to form hydrochloric acid Hydrochloric acid dissociatesionizes in water to free HaqH3Oaq ions The free H3Oaq  Haq ions are responsible for


iturning blue litmus papersolution red


iishow pH value 123456


iiiare good electrolytesconductors of electricityundergo electrolysis


ivreact with metals to produce evolve hydrogen gas and a salt ie


Ionically


For a monovalent metal 2Ms  2Haq  2Maq  H2g


For a divalent metal Ms  2Haq  M2aq  H2g


For a trivalent metal 2Ms  6Haq  2M3aq  3H2g


Examples


For a monovalent metal 2Nas  2Haq  2Naaq  H2g


For a divalent metal Cas  2Haq  Ca2aq  H2g


For a trivalent metal 2Als  6Haq  2Al3aq  3H2g


vreact with metal carbonates and hhydrogen carbonates to produce evolve carbonIVoxide gas water and a salt ie


For a monovalent metal M2CO3s 2Haq  2Maq  H2O l CO2g


MHCO3s Haq  Maq  H2O l CO2g


For a divalent metal MCO3s 2Haq  M2aq  H2O l CO2g


MHCO3 2aq2Haq M2aq2H2Ol2CO2g


For a monovalent metal K2CO3s 2Haq  2Kaq  H2O l CO2g


NH4HCO3s Haq  NH4aq  H2O l CO2g


For a divalent metal ZnCO3s 2Haq  Zn2aq  H2O l CO2g


MgHCO3 2aq2Haq Mg2aq2H2Ol2CO2g


vineutralize metal oxideshydroxides to salt and water only ie


Ionically


For a monovalent metal M2Os  2Haq  2Maq  H2O l


MOHaq  Haq  Maq  H2O l


For a divalent metal MOs  2Haq  M2aq  H2O l


MOH 2s  2Haq  M2aq  2H2Ol


For a trivalent metal M2O3s  6Haq  2M3aq  3H2O l


MOH 3s  3Haq  M3aq  3H2Ol


Examples


For a monovalent metal K2Os  2Haq  2Kaq  H2O l


NH4OHaq  Haq  NH4aq  H2O l


For a divalent metal ZnO s  2Haq  Zn2aq  H2O l


PbOH 2s  2Haq  Pb2aq  2H2Ol


bHydrogen chloride gas dissolves in methylbenzene benzene but does not dissociate ionize into free ions


It exists in molecular state showing none of the above properties


cAmmonia gas dissolves in water to form aqueous ammonia which dissociateionize to free NH4 aq and OHaq ions


This dissociationionization makes aqueous ammonia to


iturn litmus papersolution blue


iihave pH 891011


iiibe a good electrical conductor


ivreact with acids to form ammonium salt and water only


NH4OHaq  HClaq  NH4Claq  H2Ol


dAmmonia gas dissolves in methylbenzenebenzene kerosene but does not dissociate into free ions therefore existing as molecules


6 Solvents are either polar or nonpolar


A polar solvent is one which dissolves ionic compounds and other polar solvents


Water is polar solvent that dissolves ionic and polar substance by surrounding the free ions as below


H  H  O


H  H


H  O  H  H


H


O  H O  O  Cl  O


H  O  H  H H


Beaker


Cl


Cl H water


Cl H Free ions


NoteWater is polar It is made up of


Oxygen atom is partially negative and two hydrogen atoms which are partially positive


They surround the free H and Cl ions


A non polar solvent is one which dissolved nonpolar substances and covalent compounds


If a polar ionic compound is dissolved in nonpolar solvent it does not ionizedissociate into free ions as below


HCl


HCl methyl benzene HCl HCl Covalent bond


7 Some acids and bases are strong while others are weak


aA strong acidbase is one which is fullywhollycompletely dissociated  ionized into many free H OH ions ie


I Strong acids exists more as free H ions than molecules eg


HClaq Haq  Cl aq molecules cation anion


HNO3aq Haq  NO3aq molecules cation anion


H2SO4aq 2Haq  SO42aq molecules cation anion


II Strong basesalkalis exists more as free OH ions than molecules eg


KOHaq Kaq  OH aq molecules cation anion


NaOHaq Naaq  OHaq molecules cation anion


b A weak baseacid is one which is partially partly dissociated ionized in water into free OH aq and Haq ions


I Weak acids exists more as molecules than as free H ions eg


CH3COOHaq Haq  CH3COO aq molecules cation anion


H3PO4aq 3Haq  PO43aq molecules cation anion


H2CO3aq 2Haq  CO32aq molecules cation anion


II Weak basesalkalis exists more as molecules than free OH ions eg


NH4OHaq NH4aq  OH aq molecules cation anion CaOH2aq Ca2aq  2OHaq molecules cation anion


MgOH2aq Mg2aq  2OHaq molecules cation anion


8 The concentration of an acidbasealkali is based on the number of moles of acidbases dissolved in a decimeterlitreof the solution


An acidbasealkali with more acidbasealkali in a decimeterlitre of solution is said to be concentrated while that with less is said to be dilute


9 a istrong acids have pH 123 while weak acids have high pH 456


iia neutral solution have pH 7


iiistrong alkalisbases have pH 121314 while weak basesalkalis have pH 1110 9  8


b pH is a measure of Haq concentration in a solution


The higher the Haqions concentration


the higher the acidity


the lower the pH


the lower the concentration of OHaq


the lower the alkalinity


At pH 7  a solution has equal concentration of Haq and OHaq


Beyond pH 7the concentration of the OHaq increases as the Haq ions decreases


10a When acids bases dissolve in water the ions present in the solution conduct electricity


The more the dissociation the higher the yield of ions and the greater the electrical conductivity of the solution


A compound that conducts electricity in an electrolyte and thus a compound showing high electrical conductivity is a strong electrolyte while a compound showing low electrical conductivity is a weak electrolyte


b Practically a bright light on a bulb a high voltage reading from a voltmeter high ammeter reading from an ammeter a big deflection on a galvanometer is an indicator of strong electrolyteacidbase and the opposite for weak electrolytesacidsbase


11 Some compounds exhibitshow both properties of acids and basesalkalis


A substance that reacts with both acids and bases is said to be amphotellic


The examples below show the amphotellic properties of


a Zinc IIoxideZnO and Zinc hydroxideZnOH2


iWhen  spatula full of ZincIIoxide is placed in a boiling tube containing 10cm3 of either 2M nitricVacid or 2M sodium hydroxide hydroxide solution it dissolves on both the acid and the alkalibase to form a colourless solution ie


i when reacting with nitricVacid the oxide shows basic properties by reacting with an acid to form a simple salt and water only


Basic oxide  Acid  salt  water


Examples


Chemical equation


ZnOs  2HNO3aq  ZnNO3 2 aq  H2Ol


ZnOs  2HClaq  ZnCl2 aq  H2Ol


ZnOs  H2SO4aq  ZnSO4 aq  H2Ol


Ionic equation


ZnOs  2H aq  Zn 2 aq  H2Ol


ii when reacting with sodium hydroxide the oxide shows acidic properties by reacting with a base to form a complex salt


Basic oxide  Basealkali  Water  Complex salt


1When Zinc oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincateII complex salt


ZnOs  2NaOHaq  H2Ol  Na2ZnOH 4aq


2When Zinc oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincateII complex salt


ZnOs  2KOHaq  H2Ol  K2ZnOH 4aq


ZnOs  2OHaq  H2Ol  2ZnOH 42 aq


iiWhen ZincIIhydroxide is placed in a boiling tube containing 10cm3 of either 2M nitricVacid or 2M sodium hydroxide hydroxide solution it dissolves on both the acid and the alkalibase to form a colourless solution ie


i when reacting with nitricVacid the hydroxide shows basic properties It reacts with an acid to form a simple salt and water only


Basic hydroxide  Acid  salt  water


Examples


Chemical equation


ZnOH 2 s  2HNO3aq  ZnNO3 2 aq  2H2Ol


ZnOH 2 s  2HClaq  ZnCl2 aq  2H2Ol


ZnOH 2 s  H2SO4aq  ZnSO4 aq  2H2Ol


Ionic equation


ZnOH 2 s  2H aq  Zn 2 aq  2H2Ol


ii when reacting with sodium hydroxide the hydroxide shows acidic properties by reacting with a base to form a complex salt


Basic hydroxide  Basealkali  Complex salt


1When Zinc hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincateII complex salt


ZnOH 2 s  2NaOHaq  Na2ZnOH 4aq


2When Zinc hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincateII complex salt


ZnOH 2 s  2KOHaq  K2ZnOH 4aq


ZnOH 2 s  2OHaq  2ZnOH 42 aq


b Lead IIoxidePbO and LeadII hydroxide PbOH2


iWhen  spatula full of LeadIIoxide is placed in a boiling tube containing 10cm3 of either 2M nitricVacid or 2M sodium hydroxide hydroxide solution it dissolves on both the acid and the alkalibase to form a colourless solution ie


i when reacting with nitricVacid the oxide shows basic properties by reacting with an acid to form a simple salt and water only All other Lead salts are insoluble


PbOs  2HNO3aq  PbNO3 2 aq  H2Ol


PbOs  2H aq  Pb 2 aq  H2Ol


Chemical equation


1When LeadII oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbateII complex salt


PbOs  2NaOHaq  H2Ol  Na2PbOH 4aq


2When LeadII oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbateII complex salt


PbOs  2KOHaq  H2Ol  K2PbOH 4aq


Ionic equation


PbOs  2OHaq  H2Ol  2PbOH 42 aq


iiWhen LeadIIhydroxide is placed in a boiling tube containing 10cm3 of either 2M nitricVacid or 2M sodium hydroxide hydroxide solution it dissolves on both the acid and the alkalibase to form a colourless solution ie


i when reacting with nitricVacid the hydroxide shows basic properties It reacts with the acid to form a simple salt and water only


PbOH 2 s  2HNO3aq  PbNO3 2 aq  2H2Ol


PbOH 2 s  2H aq  Pb 2 aq  2H2Ol


ii when reacting with sodium hydroxide the hydroxide shows acidic properties It reacts with a base to form a complex salt


1When LeadII hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbateII complex salt


PbOH 2 s  2NaOHaq  Na2PbOH 4aq


2When LeadII hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbateII complex salt


PbOH 2 s  2KOHaq  K2PbOH 4aq


PbOH 2 s  2OHaq  2PbOH 42 aq


cAluminiumIIIoxideAl2O3 and AluminiumIIIhydroxideAlOH3


iWhen  spatula full of AluminiumIIIoxide is placed in a boiling tube containing 10cm3 of either 2M nitricVacid or 2M sodium hydroxide hydroxide solution it dissolves on both the acid and the alkalibase to form a colourless solution ie


i when reacting with nitricVacid the oxide shows basic properties by reacting with an acid to form a simple salt and water only


Chemical equation


Al2O3 s  6HNO3aq  AlNO33 aq  3H2Ol


Al2O3 s  6HClaq  AlCl3 aq  3H2Ol


Al2O3 s  3H2SO4aq  Al2SO43 aq  3H2Ol


Ionic equation


Al2O3 s  3H aq  Al 3 aq  3H2Ol


ii when reacting with sodium hydroxide the oxide shows acidic properties by reacting with a base to form a complex salt


1When AluminiumIII oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminateIII complex salt


Al2O3 s  2NaOHaq  3H2Ol  2NaAlOH 4aq


2When AluminiumIII oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminateII complex salt


Al2O3 s  2KOHaq  3H2Ol  2NaAlOH 4aq


Al2O3 s  2OHaq  3H2Ol  2AlOH 4 aq


iiWhen AluminiumIIIhydroxide is placed in a boiling tube containing 10cm3 of either 2M nitricVacid or 2M sodium hydroxide hydroxide solution it dissolves on both the acid and the alkalibase to form a colourless solution ie


AlOH 3 s  3HNO3aq  AlNO33 aq  3H2Ol


AlOH3 s  3HClaq  AlCl3 aq  3H2Ol


2AlOH3 s  3H2SO4aq  Al2SO43 aq  3H2Ol


AlOH3 s  3H aq  Al 3 aq  3H2Ol


Chemical equation


1When aluminiumIII hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminateIII complex salt


AlOH 3 s  NaOHaq  NaAlOH 4aq


2When aluminiumIII hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminateIII complex salt


AlOH 3 s  KOHaq  KAlOH 4aq


Ionic equation


AlOH 3 s  OHaq  AlOH 4 aq


Summary of amphotellic oxideshydroxides


12a A salt is an ionic compound formed when the cation from a base combine with the anion derived from an acid


A salt is therefore formed when the hydrogen ions in an acid are replaced whollyfully or partiallypartly directly or indirectly by a metal or ammonium radical b The number of ionizablereplaceable hydrogen in an acid is called basicity of an acid


Some acids are therefore


imonobasic acids generally denoted HX eg


HCl HNO3HCOOHCH3COOH


iidibasic acids  generally denoted H2X eg


H2SO4 H2SO3 H2CO3HOOCOOH


iiitribasic acids  generally denoted H3X eg


c Some salts are normal salts while other are acid salts


iA normal salt is formed when all the ionizable replaceable hydrogen in an acid is replaced by a metal or metallic ammonium radical


iiAn acid salt is formed when partportion the ionizable replaceable hydrogen in an acid is replaced by a metal or metallic ammonium radical


Table showing normal and acid salts derived from common acids


The table below show shows some examples of salts


d Some salts undergo hygroscopy deliquescence and efflorescence


i Hygroscopic salts compounds are those that absorb water from the atmosphere but do not form a solution


Some salts which are hygroscopic include anhydrous copperIIsulphateVI anhydrous cobaltIIchloride potassium nitrateV common table salt


iiDeliquescent salts compounds are those that absorb water from the atmosphere and form a solution


Some salts which are deliquescent include Sodium nitrateVCalcium chloride Sodium hydroxide IronIIchloride Magnesium chloride


iiiEfflorescent saltscompounds are those that lose their water of crystallization to the atmosphere


Some salts which effloresces include sodium carbonate decahydrate IronIIsulphateVIheptahydrate sodium sulphate VIdecahydrate


eSome salts contain water of crystallizationThey are hydratedOthers do not contain water of crystallization They are anhydrous


Table showing some hydrated salts


fSome salts exist as a simple salt while some as complex salts Below are some complex salts


Table of some complex salts


gSome salts exist as two salts in one They are called double salts


Table of some double salts


hSome salts dissolve in water to form a solution They are said to be soluble Others do not dissolve in water They form a suspensionprecipitate in water


Table of solubility of salts


13 Salts can be prepared in a school laboratory by a method that uses its solubility in water


Soluble salts may be prepared by using any of the following methods


iDirect displacementreaction of a metal with an acid


By reacting a metal higher in the reactivity series than hydrogen with a dilute acida salt is formed and hydrogen gas is evolved


Excess of the metal must be used to ensure all the acid has reacted


When effervescencebubbling fizzing has stopped excess metal is filtered


The filtrate is heated to concentrate then allowed to crystallize


Washing with distilled water then drying between filter papers produces a sample crystal of the salt ie


Ms  H2X  MXaq  H2g


Examples


Mgs  H2SO4aq  MgSO4 aq  H2g


Zns  H2SO4aq  ZnSO4 aq  H2g


Pbs  2HNO3aq  PbNO3 2aq  H2g


Cas  2HNO3aq  CaNO3 2aq  H2g


Mgs  2HNO3aq  MgNO3 2aq  H2g


Mgs  2HClaq  MgCl 2aq  H2g


Zns  2HClaq  ZnCl 2aq  H2g


iiReaction of an insoluble base with an acid


By adding an insoluble base oxidehydroxide to a dilute acid until no more dissolves in the acida salt and water are formed Excess of the base is filtered off The filtrate is heated to concentrate allowed to crystallize then washed with distilled water before drying between filter papers eg


PbOs  2HNO3aq  PbNO3 2aq  H2O l


PbOH2s  2HNO3aq  PbNO3 2aq  2H2O l


CaO s  2HNO3aq  CaNO3 2aq  H2O l


MgO s  2HNO3aq  MgNO3 2aq  H2O l


MgO s  2HClaq  MgCl 2aq  H2O l


ZnO s  2HClaq  ZnCl 2aq  H2O l


ZnOH2s  2HNO3aq  ZnNO3 2aq  2H2O l


CuO s  2HClaq  CuCl 2aq  H2O l


CuO s  H2SO4aq  CuSO4aq  H2O l


Ag2Os  2HNO3aq  2AgNO3aq  H2O l


Na2Os  2HNO3aq  2NaNO3aq  H2O l


iiireaction of insoluble soluble carbonate hydrogen carbonate with an acid


By adding an excess of a soluble insoluble carbonate or hydrogen carbonate to adilute acid effervescence fizzingbubbling out of carbonIVoxide gas shows the reaction is taking place When effervescence fizzingbubbling out of the gas is over excess of the insoluble carbonate is filtered off The filtrate is heated to concentrate allowed to crystallize then washed with distilled water before drying between filter paper papers eg


PbCO3 s  2HNO3aq  PbNO3 2aq  H2O l CO2g


ZnCO3 s  2HNO3aq  ZnNO3 2aq  H2O l CO2g


CaCO3 s  2HNO3aq  CaNO3 2aq  H2O l CO2g


MgCO3 s  H2SO4aq  MgSO4aq  H2O l CO2g


Cu CO3 s  H2SO4aq  CuSO4aq  H2O l  CO2g


Ag2CO3 s  2HNO3aq  2AgNO3aq  H2O l  CO2g


Na2CO3 s  2HNO3aq  2NaNO3aq  H2O l  CO2g


K2CO3 s  2HClaq  2KClaq  H2O l  CO2g


NaHCO3 s  HNO3aq  NaNO3aq  H2O l  CO2g


KHCO3 s  HClaq  KClaq  H2O l  CO2g


ivneutralizationreaction of soluble basealkali with dilute acid


By adding an acid to a burette into a known volume of an alkali with 23 drops of an indicator the colour of the indicator changes when the acid has completely reacted with an alkali at the end point The procedure is then repeated without the indicator The solution mixture is then heated to concentrate  allowed to crystallize washed with distilled water before drying with filter papers eg


NaOH aq  HNO3aq  NaNO3aq  H2O l


KOH aq  HNO3aq  KNO3aq  H2O l


KOH aq  HClaq  KClaq  H2O l


2KOH aq  H2SO4aq  K2SO4aq  2H2O l


2 NH4OH aq  H2SO4aq  NH42SO4aq  2H2O l


NH4OH aq  HNO3aq  NH4NO3aq  H2O l


ivdirect synthesiscombination


When a metal burn in a gas jar containing a non metal  the two directly combine to form a salt eg


2Nas  Cl2g  2NaCls


2Ks  Cl2g  2KCls


Mgs  Cl2g  Mg Cl2 s


Cas  Cl2g  Ca Cl2 s


Some salts once formed undergo sublimation and hydrolysis Care should be taken to avoid watermoisture into the reaction flask during their preparationSuch salts include aluminiumIIIchlorideAlCl3 and iron IIIchlorideFeCl3


1 Heated aluminium foil reacts with chlorine to form aluminiumIIIchloride that sublimes away from the source of heating then deposited as solid again


2Als  3Cl2g  2AlCl3 sg


Once formed aluminiumIIIchloride hydrolysesreacts with water vapour  moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas


AlCl3s 3H2 Og  AlOH3 aq  3HClg


2 Heated iron filings reacts with chlorine to form ironIIIchloride that sublimes away from the source of heating then deposited as solid again


2Fes  3Cl2g  2FeCl3 sg


Once formed  aluminiumIIIchloride hydrolysesreacts with water vapour  moisture present to form aluminium hydroxide solution and highly acidic fumes of hydrogen chloride gas


FeCl3s 3H2 Og  FeOH3 aq  3HClg


bInsoluble salts can be prepared by reacting two suitable soluble salts to form one soluble and one insoluble This is called double decomposition or precipitation The mixture is filtered and the residue is washed with distilled water then dried


CuSO4aq  Na2CO3 aq  CuCO3 s  Na2 SO4aq


BaCl2aq  K2SO4 aq  BaSO4 s  2KCl aq


PbNO32aq  K2SO4 aq  PbSO4 s  2KNO3 aq


2AgNO3aq  MgCl2 aq  2AgCls  MgNO32 aq


PbNO32aq  NH4 2SO4 aq  PbSO4 s  2NH4NO 3aq


BaCl2aq  K2SO3 aq  BaSO3 s  2KCl aq


14 Salts may lose their water of crystallization  decompose melt or sublime on heating on a Bunsen burner flame


The following shows the behavior of some salts on heating gently or strongly in a laboratory school burner


aeffect of heat on chlorides


All chlorides have very high melting and boiling points and therefore are not affected by laboratory heating except ammonium chloride Ammonium chloride sublimes on gentle heating It dissociate into the constituent ammonia and hydrogen chloride gases on strong heating


NH4Cls NH4Clg NH3g  HClg


sublimation dissociation


beffect of heat on nitrateV


i Potassium nitrateVKNO3 and sodium nitrateVNaNO3 decompose on heating to form Potassium nitrateIIIKNO2 and sodium nitrateIIINaNO2 and producing Oxygen gas in each case


2KNO3 s  2KNO2s  O2g


2NaNO3 s  2NaNO2s  O2g


iiHeavy metal nitratesV salts decompose on heating to form the oxide and a mixture of brown acidic nitrogenIVoxide and oxygen gases eg


2CaNO32 s  2CaOs  4NO2g  O2g


2MgNO32s  2MgOs  4NO2g  O2g


2ZnNO32s  2ZnOs  4NO2g  O2g


2PbNO32s  2PbOs  4NO2g  O2g


2CuNO32s  2CuOs  4NO2g  O2g


2FeNO32s  2FeOs  4NO2g  O2g


iiiSilverInitrateV and mercuryII nitrateV are lowest in the reactivity series They decompose on heating to form the metalsilver and mercuryand the NitrogenIVoxide and oxygen gas ie


2AgNO3s  2Ag s  2NO2g  O2g


2HgNO32 s  2Hg s  4NO2g  O2g


ivAmmonium nitrateV and Ammonium nitrateIII decompose on heating to NitrogenIoxiderelightsrekindles glowing splint and nitrogen gas respectivelyWater is also formedie


NH4NO3s  N2O g  H2Ol


NH4NO2s  N2 g  H2Ol


c effect of heat on nitrateV


Only IronIIsulphateVI IronIIIsulphateVI and copperIIsulphateVI decompose on heating They form the oxide and produce highly acidic fumes of acidic sulphurIVoxide gas


2FeSO4 s  Fe2O3s  SO3g  SO2g


Fe2SO4 3s  Fe2O3s  SO3g


CuSO4 s  CuOs  SO3g


d effect of heat on carbonatesIV and hydrogen carbonateIV


iSodium carbonateIVand potassium carbonateIVdo not decompose on heating


iiHeavy metal nitrateIVsalts decompose on heating to form the oxide and produce carbonIVoxide gas Carbon IVoxide gas forms a white precipitate when bubbled in lime water The white precipitate dissolves if the gas is in excess eg CuCO3 s  CuOs  CO2g


CaCO3 s  CaOs  CO2g


PbCO3 s  PbOs  CO2g


FeCO3 s  FeOs  CO2g


ZnCO3 s  ZnOs  CO2g


iiiSodium hydrogen carbonateIV and Potassium hydrogen carbonateIVdecompose on heating to give the corresponding carbonate IV and form water and carbonIVoxide gas ie


2NaHCO 3s  Na2CO3s  CO2g  H2Ol


2KHCO 3s  K2CO3s  CO2g  H2Ol


iii Calcium hydrogen carbonate IV and Magnesium hydrogen carbonateIV decompose on heating to give the corresponding carbonate IV and form water and carbonIVoxide gas i e


CaHCO3 2aq  CaCO3s  CO2g  H2Ol


MgHCO3 2aq  MgCO3s  CO2g  H2Ol


15 Salts contain cationpositively charged ion and anionsnegatively charged ionWhen dissolved in polar solventswater


The cation and anion in a salt is determinedknown usually by precipitation of the salt using a precipitating reagent


The colour of the precipitate is a basis of qualitative analysis of a compound


16Qualitative analysis is the process of identifying an unknown compound salt by identifying the unique qualities of the saltcompound


It involves some of the following processes


aReaction of cation with sodiumpotassium hydroxide solution


Both sodiumpotassium hydroxide solutions are precipitating reagents


The alkalis produce unique colour of a precipitatesuspension when a fewthree drops is added and then excess alkali is added to unknown saltcompound solution


NB Potassium hydroxide is not commonly used because it is more expensive than sodium hydroxide


The table below shows the observations inferences  deductions and explanations from the following test tube experiments


Procedure


Put about 2cm3 of MgCl2 CaCl2 AlCl3 NaCl KCl FeSO4 Fe2SO4 3 CuSO4 ZnSO4NH4NO3 PbNO3 2 BaNO3 2 each into separate test tubes Add three drops of 2M sodium hydroxide solution then excess 23 the length of a standard test tube


bReaction of cation with aqueous ammonia


Aqueous ammonia precipitating reagent that can be used to identify the cations present in a salt


Like NaOHKOH the OH ion in NH4OH react with the cation to form a characteristic hydroxide


Below are the observations inferences and explanations of the reactions of aqueous ammonia with salts from the following test tube reactions


Put about 2cm3 of MgCl2 CaCl2 AlCl3 NaCl KCl FeSO4 Fe2SO4 3 CuSO4 ZnSO4NH4NO3 PbNO3 2 BaNO3 2 each into separate test tubes


Add three drops of 2M aqueous ammonia then excess 23 the length of a standard test tube


Note


i Only Zn2 ionssalts form a white precipitate that dissolve in excess of both 2M sodium hydroxide and 2M aqueous ammonia


ii Pb2 and Al3 ionssalts form a white precipitate that dissolve in excess of 2M sodium hydroxide but not in 2M aqueous ammonia


iii Cu2 ionssalts form a blue precipitate that dissolve to form a deeproyal blue solution in excess of 2M aqueous ammonia but only blue insoluble precipitate in 2M sodium hydroxide


cReaction of cation with Chloride Clions


All chlorides are soluble in water except Silver chloride and Lead IIchloride That dissolve in hot waterWhen a soluble chloride like NaCl KCl NH4Cl is added to about 2cm3 of a salt containing Ag or Pb2ions a white precipitate of AgCl or PbCl2 is formed The following test tube reactions illustrate the above


Experiment


Put about 2cm3 of silver nitrateV andLeadIInitrateVsolution into separate test tubes Add five drops of NaCl KCl  NH4ClHCl Heat to boil


Both Pb2 and Al3 ions forms an insoluble white precipitate in excess aqueous ammonia A white precipitate on adding Cl ionssalts shows Pb2


No white precipitate on adding Cl ionssalts shows Al3


Adding a chloride Cl ionssalts can thus be used to separate the identity of Al3 and Pb2


dReaction of cation with sulphateVISO42 and sulphateIVSO32 ions


All sulphateVI and sulphateIVSO32 ionssalts are solubledissolve in water except Calcium sulphateVICaSO4 Calcium sulphateIVCaSO3 Barium sulphateVIBaSO4 Barium sulphateIVBaSO3 LeadII sulphateVIPbSO4 and LeadII sulphateIVPbSO3When a soluble sulphateVISO42 salt like Na2SO4 H2SO4 NH42SO4 or Na2SO3 is added to a salt containing Ca2 Pb2 Ba2 ions a white precipitate is formed


The following test tube experiments illustrate the above


Procedure


Place about 2cm3 of CaNO32 BaNO32 BaCl2 and PbNO32 in separate boiling tubes Add six drops of sulphuricVIacid sodium sulphateVIammonium sulphateVIsolution Repeat with six drops of sodium sulphateIV


eReaction of cation with carbonateIVCO32 ions


All carbonate salts are insoluble except sodiumpotassium carbonateIV and ammonium carbonateIV


They dissociate ionize to release CO32 ions CO32 ions produce a white precipitate when the soluble carbonate salts is added to any metallic cation


Place about 2cm3 of CaNO32 BaNO32 MgCl2 PbNO32 andZnSO4 in separate boiling tubes


Add six drops of Potassium sodium carbonateIV ammonium carbonate IVsolution


Note


iIronIIIcarbonateIV does not exist


iiCopperIICarbonateIV exist only as the basic CuCO3CuOH 2


iiiBoth BaCO3 and BaSO3 are insoluble white precipitate If hydrochloric acid is added to the white precipitate


I BaCO3 produces CO2 gas When bubbleddirected into lime water solutiona white precipitate is formed


II I BaSO3 produces SO2 gas When bubbleddirected into orange acidified potassium dichromateVI solution it turns to greendecolorizes acidified potassium manganateVII


f Reaction of cation with sulphide  S2 ions


All sulphides are insoluble black solidsprecipitates except sodium sulphide Na2S potassium sulphideK2SWhen a few3drops of the soluble sulphide is added to a metal cationsalt a black precipitate is formed


Place about 2cm3 of CuNO32 FeSO4 MgCl2PbNO32 and ZnSO4 in separate boiling tubes


Add six drops of Potassium sodium sulphide solution


Sample qualitative analysis guide


You are provided with solid Yaluminium IIIsulphateVIhexahydrateCarry out the following tests and record your observations and inferences in the space provided


1a Appearance


Observations inference 1mark


White crystalline solid Coloured ions Cu2  Fe2 Fe3 absent


bPlace about a half spatula full of the solid into a clean dry boiling tube Heat gently then strongly


Colourless droplets formed on the cooler Hydrated compoundcompound


part of the test tube containing water of crystallization


Solid remains a white residue


cPlace all the remaining portion of the solid in a test tube Add about 10cm3 of distilled water Shake thoroughly Divide the mixture into five portions


Observation Inference 1mark


Solid dissolves to form Polar soluble compound


a colourless solution Cu2  Fe2 Fe3 absent


iTo the first portion add three drops of sodium hydroxide then add excess of the alkali


White ppt soluble in excess Zn2  Pb2  Al3


iiTo the second portion add three drops of aqueous ammonia then add excess of the alkali


White ppt insoluble in excess Pb2  Al3


iiiTo the third portion add three drops of sodium sulphateVIsolution


No white ppt Al3


ivITo the fourth portion add three drops of LeadIInitrateIVsolution Preserve


White ppt CO32 SO42 SO32 Cl


IITo the portion in iv I above  add five drops of dilute hydrochloric acid


White ppt persistremains SO42 Cl


IIITo the portion in iv II above heat to boil


White ppt persistremains SO42


Note that


iFrom test above it can be deduced that solid Y is hydrated aluminiumIIIsulphateVI solid


iiAny ion inferred from an observation below must be derived from previous correct observation and inferences above eg


Al3 in ciii must be correctly inferred in eitheror in cii or ciabove


SO42 in civIII must be correctly inferred in eitheror in civII or civI above


iiiContradiction in observations and inferences should be avoidedeg


White ppt soluble in excess to infer presence of Al3 Ba2 Pb3


ivSymbols of elementsions should be correctly capitalized eg


SO42 is wrong sO42 is wrong cu2 is wrong


Sample solutions of salt were labeled as III III and IV The actual solutions not in that order are lead nitrate zinc sulphate potassium chloride and calcium chloride


aWhen aqueous sodium carbonate was added to each sample separately a white precipitate was formed in I III and IV only Identify solution II


bWhen excess sodium hydroxide was added to each sample separately a white precipitate was formed in solutions III and I only


Identify solution I


17When solidssalts solutes are added to a solvent some dissolve to form a solution


Solute  Solvent  Solvent


If a solution has a lot of solute dissolved in a solvent it is said to be concentrated


If a solution has little solute dissolved in a solvent it is said to be dilute


There is a limit to how much solute can dissolve in a given specified amount of solventwater at a given specified temperature


The maximum mass of saltsolidsolute that dissolve in 100g of solventwater at a specified temperature is called solubility of a salt


When no more solute can dissolve in a given amount of solvent at a specified temperature a saturated solution is formed


For some salts on heating more of the saltsolidsolute dissolve in the saturated solution to form a super saturated solution


The solubility of a salt is thus calculated from the formula


Solubility  Mass of solutesaltsolid x 100


Massvolume of watersolvent


Practice examples


aCalculate the solubility of potassium nitrateV if 50 g of the salt is dissolved in 500cm3 of water


Solubility  Mass of solutesaltsolid x 100  50 x 100   100 g 100g H2O


Massvolume of watersolvent 500


bCalculate the solubility of potassium chlorateV if 500 g of the salt is dissolved in 2500cm3 of water


Solubility  Mass of solutesaltsolid x 100  500 x 100   200 g 100g H2O


Massvolume of watersolvent 2500


cIf the solubility of potassium chlorateV is 5g100g H2O at 80oChow much can dissolve in 5cm3 of water at 80oC


Mass of solutesaltsolid  Solubility x Massvolume of watersolvent


5 x 5  025g of KClO3 dissolve


dIf the solubility of potassium chlorateV is 72g100g H2O at 20oChow much can saturate 25g of water at 20oC


72 x 25  180g of KClO3 dissolvesaturate


e 22g of potassium nitrateV was dissolved in 400g of water at 10oC Calculate the solubility of potassium nitrateV at 10oC


Solubility  Mass of solutesaltsolid x 100  22 x 100   550 g 100g H2O


Massvolume of watersolvent 400


fWhat volume of water should be added to 220g of water at 10oC if the solubility of KNO3 at 10oC is 50g100g H2O


Solubility is mass100g H2O  220g  x  100cm3100g H2O


X 100  22  78 cm3 of H2O


18 A graph of solubility against temperature is called solubility curve


It shows the influence of temperature on solubility of different substancessolidssalts


Some substances dissolve more with increase in temperature while for others dissolve less with increase in temperature


Note


isolubility of KNO3 and KClO3 increase with increase in temperature


iisolubility of KNO3 is always higher than that of KClO3 at any specified temperature


iiisolubility of NaCl decrease with increase in temperature


ivNaCl has the highest solubility at low temperature while KClO3 has the lowest solubility at low temperature


vAt point A both NaCl and KNO3 are equally soluble


viAt point B both NaCl and KClO3 are equally soluble


vii An area above the solubility curve of the salt shows a saturated supersaturated solution


viii An area below the solubility curve of the salt shows an unsaturated solution


19a For salts whose solubility increases with increase in temperature crystals form when the salt solution at higher temperatures is cooled to a lower temperature


For salts whose solubility decreases with increase in temperature crystals form when the salt solution at lower temperatures is heated to a higher temperature


The examples below shows determination of the mass of crystals deposited with changes in temperature


1The solubility of KClO3 at 100oC is 60g100g water What mass of KClO3 will be deposited at


i75 oC if the solubility is now 39g100g water


At 100oC  600g


Less at 75oC   390g


Mass of crystallized out 210g


i35 oC if the solubility is now 28 g100g water


Less at 35oC   2800g


Mass of crystallized out 320g


2 KNO3 has a solubility of 42 g100g water at 20oCThe salt was heated and added 38g more of the solute which dissolved at100oC Calculate the solubility of KNO3 at 100oC


Solubility of KNO3 at 100oC  solubility at 20oC  mass of KNO3 added


42g  38g  80g KNO3 100g H2O


3 A salt solution has a mass of 65g containing 5g of solute The solubility of this salt is 25g per 100g water at 20oC 60g of the salt are added to the solution at 20oCCalculate the mass of the solute that remain undissolved


Mass of solvent at 20oC  mass of solution  mass of solute


65  5  60g


Solubility before adding salt  mass of solute x 100


Volume of solvent


5 x 100  83333g100g water


Mass of solute to equalize with solubility  25  83333g  166667g


Mass of solute undissolved  600  166667g  433333 g


4 Study the table below


iWhat happens when the two salts are dissolved in water then cooled from 50oC to 20oC


90  30  600 g of KNO3 crystals precipitate


20  6  140 g of KClO3 crystals precipitate


iiState the assumption made in i above


Solubility of one salt has no effect on the solubility of the other


5 100 g of hydrated potassium carbonate IV K2CO3xH2O on heating leave 793 of the hydrate


aCalculate the mass of anhydrous salt obtained


Hydrated on heating leave anhydrous  793 g


bCalculate the mass of water of crystallization in the hydrated salt


Mass of water of crystallization  hydrated  anhydrous


100  793  207 g


cHow many moles of anhydrous salt are there in 10of hydrate K 390C120O 160


Molar mass K2CO3 138


Moles K2CO3  mass of K2CO3  793  00515 moles


dHow many moles of water are present in the hydrate for every one mole of K2CO3  H10O 160


Molar mass H2O  18


Moles H2O  mass of H2O  207  0115 moles


Molar mass H2O 18


Mole ratio H2O  K2CO3  0115 moles 2  2


00515 moles 1


eWhat is the formula of the hydrated salt


K2CO3 2 H2O


6 The table below shows the solubility of Potassium nitrateV at different temperatures


aPlot a graph of mass of in 100g wateryaxis against temperature in oC


bFrom the graph show and determine


ithe mass of KNO3 dissolved at


I 20oC


From a correctly plotted graph  32g


II 35oC


From a correctly plotted graph  57g


III 55oC


From a correctly plotted graph  104g


iithe temperature at which the following mass of KNO3 dissolved


I 22g


From a correctly plotted graph 130oC


II 30g


From a correctly plotted graph 175oC


From a correctly plotted graph 545oC


cExplain the shape of your graph


Solubility of KNO3 increase with increase in temperatureMore KNO3 dissolve as temperature rises


dShow on the graph the supersaturated and unsaturated solutions


Above the solubility curve write supersaturated


Below the solubility curve write unsaturated


eFrom your graph calculate the amount of crystals obtained when a saturated solution of KNO3 containing 180g of the salt is cooled from 80oC to


I 20oC


Solubility before heating  180 g


Less Solubility after heatingfrom the graph  32 g


Mass of KNO3crystals  148 g


II 35oC


Less Solubility after heatingfrom the graph  58 g


Mass of KNO3crystals  122 g


III 55oC


Less Solubility after heatingfrom the graph  102 g


Mass of KNO3crystals  78 g


7 The table below shows the solubility of salts A and B at various temperatures


aOn the same axis plot a graph of solubility yaxis against temperature for each salt


bAt what temperature are the two salts equally soluble


The point of intersection of the two curves  24oC


cWhat happens when a mixture of 100g of salt B with 100g if water is heated to 80oC


From the graph the solubility of B at 80oC is 169g 100g water All the 100g crystals of B dissolve


dWhat happens when the mixture in c above is then cooled from 50oC to 20oC


Method I


Total mass before cooling at 50oC  1000 g


From graph Solubilitymass after cooling at 20oC  320 g


Mass of crystals deposited 680 g


Method II


Mass of soluble salt crystals at 50oC added  100 g


From graphSolubilitymass before cooling at 50oC  850 g


Mass of crystals that cannot dissolve at 50oC 150 g


From graph Solubilitymass before cooling at 50oC  850 g


Mass of crystals deposited after cooling 530 g


Total mass of crystals deposited  150  530  680 g


eA mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oCThe solution is then allowed to cool to 10oCDescribe clearly what happens


IFor salt A


Solubility of A before heating  mass of A x 100


Volume of water added


40 x 100  400g100g Water


TheoreticalSolubility of A before heating  400 g


Less From graph  Solubility of A after heating at 70oC  48g


Mass of crystals that can not dissolve at70oC  352 g


From graph  Solubility of A after heating at 70oC  48g


Less From graph  Solubility of A after cooling to 10oC  31g


Mass of crystals that crystallize out on cooling to10oC  17 g


Add Mass of crystals that crystallize out on cooling to10oC  17 g


Total mass of A that does not dissolvecrystallizeprecipitate  369 g


IFor salt B


Solubility of B before heating  mass of B x 100


60 x 100  600g100g Water


TheoreticalSolubility of B before heating  600 g


Less From graph  Solubility of B after heating at 70oC  138g


Mass of crystals that cannot dissolve at70oC  462 g


From graph  Solubility of B after heating at 70oC  138g


Less From graph  Solubility of B after cooling to 10oC  21g


Mass of crystals that crystallize out on cooling to10oC  117 g


Add Mass of crystals that crystallize out on cooling to10oC  117 g


Total mass of A that does not dissolvecrystallizeprecipitate  579 g


fState the assumption made in eabove


Solubility of one salt has no effect on the solubility of the other


8 When 50 g of potassium chlorate V was put in 10cm3 of water and heated the solid dissolves When the solution was cooled  the temperature at which crystals reappear was noted Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear The table below shows the the results obtained


aComplete the table to show the solubility of KclO3 at different temperatures


bPlot a graph of mass of KClO3 per 100g water against temperature at which crystals form


cFrom the graph show and determine


ithe solubility of KClO3 at


I 50oC


From a well plotted graph  145 g KClO3100g water


II 35oC


From a well plotted graph  90 g KclO3100g water


iithe temperature at which the solubility is


I10g100g water


From a well plotted graph  380 oC


II45g100g water


From a well plotted graph  775 oC


dExplain the shape of the graph


Solubility of KClO3 increase with increase in temperaturemore KclO3dissolve as temperature rises


eWhat happens when 100g per 100g water is cooled to 350 oC


Solubility before heating  1000


From the graph Solubility after cooling  90


Mass of salt precipitatedcrystallization  910 g


9 250cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below


aComplete the table


bPlot a solubility curve


cWhat happens when a saturated solution of ammonium chloride is cooled from 80oC to 40oC


From the graph Solubility at 80oC  240 g


Less From the graph Solubility at 40oC  168 g


Mass of crystallizedprecipitated  72 g


20 Solubility and solubility curves are therefore used


i to know the effect of temperature on the solubility of a salt


iito fractional crystallize two soluble salts by applying their differences in solubility at different temperatures


iiidetermine the mass of crystal that is obtained from crystallization


21Natural fractional crystallization takes place in KenyaEast Africa at


i Lake Magadi during extraction of soda ashSodium carbonate from Tronasodium sesquicarbonate


ii Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common saltsodium chloride


22Extraction of soda ash from Lake Magadi in Kenya


Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically


The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya


Temperatures around the lake are very high 3040oC during the day


The solubility of trona decrease with increase in temperature therefore solid crystals of trona grows on top of the lake upto or more than 30metres thick


A bucket dredger mines the trona which is then crushed mixed with lake liquor and pumped to washery plant where it is further refined to a green granular product called CRS


The CRS is then heated to chemically decompose trona to soda ashSodium carbonate


Chemical equation


2Na2CO3NaHCO32H2Os  3Na2CO3 s  CO2g  5H2Ol


Soda ashSodium carbonate is then stored It is called Magadi Soda Magadi Soda is used


make glass


for making soapless detergents


softening hard water


Common salt is colledcted at night because its solubility decreases with decrease in temperature It is used as salt lickfeed for animals


Summary flow diagram showing the extraction of Soda ash from Trona


23Extraction of common salt from Indian Ocean at Ngomeni in Kenya


Oceans are saltyThey contain a variety of dissolved salts about 77 being sodium chloride


During high tide water is collected into shallow pods and allowed to crystallize as evaporation takes placeThe pods are constructed in series to increase the rate of evaporation


At the final pod the crystals are scapped togetherpiled in a heap and washed with brine concentrated sodium chloride


It contains MgCl2 and CaCl2  MgCl2 and CaCl2are hygroscopic They absorb water from the atmosphere and form a solution


This makes table salt dampwet on exposure to the atmosphere


24Some water form lather easily with soap while others do not


Water which form lather easily with soap is said to be soft


Water which do not form lather easily with soap is said to be hard


Hardness of water is caused by the presence of Ca2 and Mg2 ions


Ca2 and Mg2 ions react with soap to form an insoluble grey white suspensionprecipitate called Scum curd Ca2 and Mg2 ions in water come from the water sources passing through rocks containing soluble salts of Ca2 and Mg2 eg Limestone or gypsum


There are two types of water hardness


atemporary hardness of water


bpermanent hardness of water


Temporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonateCaHCO32 and magnesium hydrogen carbonateMgHCO32


When rain water dissolve carbonIV oxide from the air it forms waek carbonicIV acid ie


CO2g  H2Ol  H2CO3aq


When carbonicIV acid passes through limestonedolomite rocks it reacts to form soluble salts ie


In limestone areas H2CO3aq  CaCO3s  CaHCO32 aq


In dolomite areas H2CO3aq  MgCO3s  MgHCO32 aq


Permanent hardness of water is caused by the presence of dissolved calcium sulphateVICaSO4 and magnesium sulphateVIMg SO4 Permanent hardness of water is caused by water dissolving CaSO4 and MgSO4 from ground rocks


Hardness of water can be removed by the following methods


aRemoving temporary hardness of water


iBoilingheating


Boiling decomposes insoluble calcium hydrogen carbonateCaHCO32 and magnesium hydrogen carbonateMgHCO32 to insoluble CaCO3 and MgCO3 that precipitate away ie


Chemical equation


CaHCO32aq  CaCO3 s  CO2g  H2Ol


MgHCO32aq  MgCO3 s  CO2g  H2Ol


iiAdding sodium carbonate IV Washing soda


Since boiling is expensive on a large scale a calculated amount of sodium carbonate decahydrate Na2CO310H2O precipitates insoluble Ca2aq and Mg2aq ions as carbonates to remove both temporary and permanent hardness of water This a double decomposition reaction where two soluble salts form an insoluble and soluble salt ie


iwith temporary hard water


Na2CO3 aq  CaHCO3 2 aq  NaHCO3aq  CaCO3 s


Na2CO3 aq  MgHCO3 2 aq  NaHCO3aq  MgCO3 s


Ionic equation


CO32 aq  Ca2 aq  CaCO3 s


CO32 aq  Mg2 aq  MgCO3 s


iiwith permanent hard water


Na2CO3 aq  MgSO4 aq  Na2SO4 aq  MgCO3 s


Na2CO3 aq  CaSO4 aq  Na2SO4 aq  MgCO3 s


iiiAdding calcium IIhydroxideLime water


Lime watercalcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonateIV


CaOH2 aq  CaHCO3 2 aq  2H2Ol  2CaCO3 s


Excess of Lime watercalcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonateIV causing the hardness again


ivAdding aqueous ammonia


Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonateIV and magnesium carbonateIV


Chemical equation


2NH3 aq  CaHCO3 2 aq  NH4 2CO3aq  CaCO3 s


2NH3 aq  MgHCO3 2 aq  NH4 2CO3aq  MgCO3 s


vUse of ionexchange permutit


This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit


The sodium permutit releases sodium ions that are exchanged with Mg2 and Ca2 ions in hard water making the water to be soft ie


Na2Xaq  Ca2 aq  Na aq  CaXs


Na2Xaq  Mg2 aq  Na aq  MgXs


Hard water containing Mg2 and Ca2


Ion exchange resin as


Sodium permutit


Na ions replace Mg2


and Ca2 to make the water soft


When all the Na ions in the resin is fully exchanged with Ca2 and Ng2 ions in the permutit column it is said to be exhausted


Brine concentrated sodium chloride solution is passed through the permutit column to regenerated recharge the column again


viDeionization demineralization


This is an advanced ion exchange method of producing deionized water Deionized water is extremely pure water made only of hydrogen and oxygen only without any dissolved substances


Deionization involve using the resins that remove all the cations by using


iA cation exchanger which remove absorb all the cations present in water and leave only H ions