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Acids and Bases

Unit 14 Acids and alkalis

Unit 15 Molarity, pH scale and strengths of acids and alkalis

Unit 16 Salts and neutralization

Unit 17 Concentration of solutions and volumetric analysis

Topic 4

KeyC o ncepts

Acids and Bases

Acids and alkalis• Characteristics of acids and

alkalis• Roleofwater for acids and

alkalis

Molarity, pH scale and strengths of acids and

alkalis• Conceptofmolarity• pH scale• Strengthsof acids and alkalis

Salts and neutralization• Acid-baseneutralization• Preparationof salts• Usesofneutralization

Concentration of solutions and volumetric

analysis• Concentrationof solutions

(in gdm–3)• Preparing standard solutions

of acids and alkalis• Calculations involving

volumetric analysis

Topic 4� Acids and Bases �Unit 14 Acids and alkalis

14.1 – 14.5

Summary

1 Anacidisahydrogen-containingsubstancethatgiveshydrogenions(H+(aq))astheonly typeofpositive ionswhendissolved inwater.

2 Concentratedacidscontainacidsdissolvedinasmallamountofwater.Diluteacidscontain acidsdissolved in a large amountofwater.

3 Characteristics of dilute acids include the following:

a) Mostdilute acidshave a sour taste.

b)Coloursof indicators in acids:

IndicatorColour in

hydrochloric acid sulphuric acid nitric acid

Litmus solution red

Methylorange red

Phenolphthalein colourless

c) Dilute acids reactwith reactivemetals to givehydrogengas.

metal +dilute acid salt +hydrogen

e.g.

Mg(s) + 2HCl(aq) MgCl2(aq) +H2(g)

Mg(s) +H2SO4(aq) MgSO4(aq) +H2(g)

d)Dilute acids reactwith carbonates to give carbondioxide gas.

carbonate+dilute acid salt +water + carbondioxide

e.g.

CaCO3(s) + 2HCl(aq) CaCl2(aq) +H2O(l) +CO2(g)

CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) +H2O(l) +CO2(g)

e) Dilute acids reactwithhydrogencarbonates to give carbondioxide gas.

hydrogencarbonate+dilute acid salt +water + carbondioxide

e.g.

NaHCO3(s or aq) +HCl(aq) NaCl(aq) +H2O(l) +CO2(g)

2NaHCO3(sor aq)+H2SO4(aq) Na2SO4(aq)+2H2O(l)+2CO2(g)

f) Oxides and hydroxides of metals are called bases. When an acid reacts with abase, a salt andwater are formed.The reaction is calledneutralization.

acid+metal oxide salt +water

acid+metalhydroxide salt +water

e.g.

H2SO4(aq) +CuO(s) CuSO4(aq) +H2O(l)

2HCl(aq) +Mg(OH)2(s) MgCl2(aq) + 2H2O(l)

14.1 Acids inourdaily lives

14.2 Acids in the laboratory

14.3 Characteristics of dilute acids

14.4 The roleofwater for acids

14.5 Basicity of an acid

14.6 Bases and alkalis

14.7 Alkalis in thehome

14.8 Alkalis in the laboratory

14.9 Characteristics of solutionsof alkalis

14.10 The roleofwater for alkalis

14.11 An introduction to analytical chemistry

14.12 Concentrated acids

14.13 Corrosive natureof concentrated acids and alkalis

14.14 Hygroscopic anddeliquescent substances

Unit 14 Acids and alkalis


ExamtipsExamtipsExamtipsExamtips ♦ Solid acids do NOT contain hydrogen ions. Hydrogen ions are formedonlywhen acidmolecules dissociate inwater.

e.g.

Solid citric aciddoesNOT containhydrogen ions. So, itCANNOT turndry blue litmuspaper red.

♦ Questions often ask about citric acid.

– It is aweak acid.

– It is an electrolyte.

– It exists as a solid at room conditions.

– It contains ionizable hydrogen atoms.

– When citric acid is dissolved in water, citric acid molecules becomemobile.

– Aqueous citric acid reacts with magnesium to give hydrogen whilesolid citric acid doesNOT.

♦ Questions may ask students to give the observable change(s) of achemical reaction.

e.g.

Dilute nitric acid is added to copper(II) oxide.

The black copper(II) oxide dissolves in dilute nitric acid to give a bluesolution.

CuO(s) + 2HNO3(aq) Cu(NO3)2(aq)+H2O(l)

– Whenaskedaboutobservablechanges,doNOTjustgivethenamesof theproducts.

– Both oxides and hydroxides do NOT give a gas when reacted withacids.

g) All dilute acids conduct electricitydue to thepresenceofmobile ions.

4 Watermust bepresent for an acid to show its acidicproperties.

5 Themaximumnumberofhydrogenionsproducedbyanacidmoleculeiscalledthebasicityof the acid.

Example

Consider the following information on two reactions involving magnesium ribbonsof the same shape.

Reacting mixture

Reaction 1 1.5gofMg+100cm3of 1.0moldm–3HCl(aq)

Reaction 2 1.5gofMg+100cm3of 1.0moldm–3H2SO4(aq)

a) Calculatewhethermagnesiumorhydrochloric acid is in excess in Reaction 1. (3marks)

b)Would you expect the initial rate of Reaction 2 to be higher, lower or the same asthatof Reaction 1? Explainyour answer. (3marks)

Answer

a) Mg(s) + 2HCl(aq) MgCl2(aq) +H2(g)

NumberofmolesofMgpresent = 1.5g24.3gmol–1

= 0.062mol (1)

NumberofmolesofHClused =1.0moldm–3 x 1001000

dm3

= 0.10mol (1)

Accordingtotheequation,1moleofMgreactswith2molesofHCl.DuringReaction 1, 0.10mole of HCl reacted with 0.050mole of magnesium. Therefore magnesiumwas in excess. (1)

b)The initial rateofReaction 2washigher than thatof Reaction 1. (1)

Duringthereactionbetweenmagnesiumandtheacids,magnesiumwouldreactwithhydrogen ions in the acids.

Sulphuric acid is adibasic acidwhilehydrochloric acid is amonobasic acid. (1)

Both are strong acids anddissociate completely inwater.

Therefore 1.0moldm–3 sulphuric acid has a higher concentration of hydrogenions. (1)

Thus, the initial rateof Reaction 2 ishigher than thatof Reaction 1.

➤Questions often ask students to compare strong acids withweak acids.

e.g.

– Comparing20cm3of1moldm–3CH3COOH(aq)with20cm3of1moldm–3HCl(aq)

✧they contain different number of hydrogen ions;

✧they havedifferent pH values;

✧they havedifferent electrical conductivity;

✧they reactwithmagnesium at different rates;

✧they give the same amount of hydrogen gas with equal mass ofmagnesium;

✧they require the same number of moles of NaOH for completeneutralization;

✧theygivedifferent colour changewith the samequantityofuniversalindicator;

✧whencompletelyneutralizedby1moldm–3NaOH(aq),HCl(aq)givesalarger temperature rise than CH3COOH(aq).

RemarksRemarks*


– Comparing20cm3of1moldm–3CH3COOH(aq)with10cm3of1moldm–3H2SO4(aq)

✧they havedifferent pH values;

✧they havedifferent electrical conductivity;

✧they react withmagnesium at different rates;

✧they give the same amount of hydrogen gas with equal mass ofmagnesium;

✧they require the same number of moles of NaOH for completeneutralization.

Adding NaOH(aq) to solution containing

Colour of precipitate

formed

Precipitate dissolves in excess

NaOH(aq)?

Ionic equation

Al3+(aq) white

✔

(a colourlesssolutionforms)

Al3+(aq) + 3OH–(aq) Al(OH)3(s)

Al(OH)3(s) +OH–(aq) [Al(OH)4]–(aq)

Pb2+(aq) white

✔


Pb2+(aq) + 2OH–(aq) Pb(OH)2(s)

Pb(OH)2(s) + 2OH–(aq) [Pb(OH)4]2–(aq)

Zn2+(aq) white

✔


Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)

Zn(OH)2(s) + 2OH–(aq) [Zn(OH)4]2–(aq)

Fe2+(aq) green ✘ Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

Fe3+(aq) reddishbrown ✘ Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

Cu2+(aq) paleblue ✘ Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

e) Adding dilute aqueous ammonia to solutions containing some metal ions givesprecipitates as shown in the following table:

Adding NH3(aq) to

solution containing


formed

Precipitate dissolves in excess NH3(aq)?

Ionic equation

Mg2+(aq) white ✘ Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)

Al3+(aq) white ✘ Al3+(aq) + 3OH–(aq) Al(OH)3(s)

Pb2+(aq) white ✘ Pb2+(aq) + 2OH–(aq) Pb(OH)2(s)

Zn2+(aq) white

✔


Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)

Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]

2+(aq) + 2OH–(aq)

Fe2+(aq) green ✘ Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

Fe3+(aq) reddishbrown ✘ Fe3+(aq) + 3OH–(aq) Fe(OH)3(s)

Cu2+(aq) paleblue

✔

(adeepbluesolutionforms)

Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]

2+(aq) + 2OH–(aq)

f) Heating solids or solutions of ammonium compounds with solutions of alkalisliberates ammonia gas.

e.g.

2NH4Cl(s or aq) +Ca(OH)2(aq) 2NH3(g) +CaCl2(aq) + 2H2O(l)

NH4Cl(s or aq) +NaOH(aq) NH3(g) +NaCl(aq) +H2O(l)

g) A solutionof alkali reactswith an acid to give a salt andwater only.

h)All solutionsof alkalis conduct electricitydue to thepresenceofmobile ions.

14.6 – 14.10

Summary

1 Abaseisacompoundwhichreactswithanacidtogiveasaltandwateronly.Basesareusuallyoxidesorhydroxidesofmetals.

acid+base salt +water

2 Alkalis arebases that are soluble inwater.

3 The properties of solutions of alkalis depend on the presence of mobile hydroxideions (OH–(aq)).

4 Characteristics of solutionsof alkalis include the following:

a) Solutionsof alkalishave abitter taste.

b)Dilute solutionsof alkalishave a ‘slippery’ feel.

c) Coloursof indicators in solutionsof alkalis:

Indicator

Colour in

potassium hydroxide solution

sodium hydroxide solution aqueous ammonia

Litmus solution blue

Methylorange yellow

Phenolphthalein red

d)Addingdilutesodiumhydroxidesolutiontosolutionscontainingsomemetalionsgivesprecipitates as shown in the following table:

Adding NaOH(aq) to solution containing


formed

Precipitate dissolves in excess

NaOH(aq)?

Ionic equation

Ca2+(aq) white ✘ Ca2+(aq) + 2OH–(aq) Ca(OH)2(s)

Mg2+(aq) white ✘ Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)

Topic 410 Acids and Bases 11Unit 14 Acids and alkalis

ExamtipsExamtipsExamtipsExamtips ♦ Questions often ask about nickel(II) compounds.

– Ni2+(aq) is green in colour.

– Nickel(II) hydroxide can be precipitated by adding dilute sodiumhydroxide solution to a solution of nickel(II) salt.

Ni2+(aq)+2OH–(aq) Ni(OH)2(s) green precipitate

– Nickel(II) carbonate can be precipitated by mixing Ni2+(aq) andCO3

2–(aq).

Ni2+(aq)+CO32–(aq) NiCO3(s)

♦ Be careful with the spelling of the chemicals ‘ammonium chloride’,‘ammonium sulphate’, etc.

♦ Cu2+(aq) ionscanbeseparatedfromamixturecontainingCu2+(aq)andK+(aq) ions by adding NaOH(aq). Cu2+(aq) ions give a precipitate withNaOH(aq)whileK+(aq) ions donot.

♦ Whenreadingquestions,doNOTmixupthetwowords‘separate’and‘distinguish’.

♦ DoNOT confuse the colour of Fe3+(aq) ions with that of Fe(OH)3(s).

➤Questions often ask about the chemical test for identifying an ammoniumcompound.Remember it is necessary towarm the sampleunder testwithsodium hydroxide solution / calcium hydroxide solution. The ammoniumcompound gives an alkaline gas,NH3.

RemarksRemarks*

Example

Foreachofthefollowingpairsofspecies,suggestachemicaltesttodistinguishbetweenthemandwrite the chemical equation(s) of the reaction(s) involved.

a) Mg2+(aq) andPb2+(aq) (4marks)

b)NH4NO3(s) andKNO3(s) (3marks)

Answer

a) Adddilute sodiumhydroxide solution to each species. (1)

OnlyPb2+(aq) gives awhiteprecipitatewhich is soluble in excessNaOH(aq). (1)

Pb2+(aq) + 2OH–(aq) Pb(OH)2(s) (1)

Pb(OH)2(s) + 2OH–(aq) [Pb(OH)4]2–(aq) (1)

b)Warmeach solidwithdilute sodiumhydroxide solution. (1)

OnlyNH4NO3(s) gives a gas that turnsmoist red litmuspaperblue. (1)

NH4NO3(s) +NaOH(aq) NH3(g) +NaNO3(aq) +H2O(l) (1)

ExamtipsExamtipsExamtipsExamtips ♦ Questionsoftenaskaboutthereagents (ormethods)thatcanbeusedto distinguish between two species. Examples:

– aluminium sulphate solutionandlead(II) ethanoate solution

distinguished by hydrochloric acid(only lead(II) ethanoate solutionproduces awhite precipitate, PbCl2)

– solid sodium carbonateandsolid calcium carbonate

distinguished by testing watersolubility

– iron(II) sulphate solutionandiron(III) sulphate solution

distinguished by colour / diluteaqueous ammonia / dilute sodiumhydroxide solution (they giveprecipitates of different colours withdilute aqueous ammonia / dilutesodium hydroxide solution)

– magnesium nitrate solutionandsilver nitrate solution

distinguished by potassium chloridesolution(only silver nitrate solution produces awhite precipitate, AgCl)

– solid ammonium chlorideandsolid potassium chloride

distinguished bywarming withsodium hydroxide solution / calciumhydroxide solution(only ammonium chloride gives analkaline gas,NH3)

– dilute sulphuric acidanddilute nitric acid

distinguished by heating with copper(only dilute nitric acid gives gasbubbles)

– dilute sulphuric acidanddilute nitric acid

distinguished by barium chloridesolution(only dilute sulphuric acid gives awhite precipitate, BaSO4)

– dilute hydrochloric acidanddilute nitric acid

distinguished by heating with copper(only dilute nitric acid gives gasbubbles)

– dilute hydrochloric acidanddilute nitric acid

distinguished by silver nitrate solution(only dilute hydrochloric acid gives awhite precipitate, AgCl)

14.11 – 14.14

Summary

1 Thechemical reactions learnt in thisunit canbeused to identify ionspresent inasample.

2 Concentrated acids and alkalis are corrosive.

3 Hygroscopicanddeliquescentsubstances(e.g.anhydrouscalciumchloride,silicageland concentrated sulphuric acid) areoftenused asdrying agents.

Topic 412 Acids and Bases 13Unit 14 Acids and alkalis

– dilute sodium hydroxidesolutionanddilute aqueous ammonia

distinguished by a solution containingaluminium ions / lead(II) ions(only dilute sodiumhydroxide solutiongives awhite precipitatewhichdissolves in excess alkali)

– solutionof cane sugarandsolution of sodium chloride

distinguished by electrical conductivity(only solution of sodium chloride canconduct electricity)

♦ Questions often ask about the hazard warning labels that should bedisplayedon thebottle of a certain chemical.

e.g.

腐蝕性氧化性有毒易燃腐蝕性

♦ Questions often ask about the set-ups and drying agents that can beused to dry a certain gas.

e.g.

– AmmoniaCANNOTbedriedbyconcentratedsulphuricacidasthereis a reaction between them. Ammonia can be dried by calciumoxide.

– Hydrogen chloridegasCANNOTbedriedby calciumoxide as thereisa reactionbetween them.Hydrogenchloridegascanbedriedbyanhydrous calcium chloride / concentrated sulphuric acid.

– The set-up shown below CANNOT be used to dry a gas becausethe end of the delivery tube for the incoming gas should be putunderthedryingagenttoensurethatalltheincominggascanpassthrough thedrying agent.

dry gasmoist gas

concentratedsulphuric acid

Example

You are provided with three unlabelled bottles, each containing one of the whitepowders listedbelow:

KBr(s), SiO2(s) andglucose

a) Outline the physical tests you would use to distinguish the three substances fromone another. (4marks)

b)Describe a chemical test you would use to distinguish between KBr(s) fromglucose. (2marks)

Answer

a) Addwater to each solid. (1)

Only SiO2(s) is insoluble. (1)

Test the electrical conductivityof the solutionobtained. (1)

KBr(aq) conducts, but glucose solutiondoesnot. (1)

b)Heat each solid strongly. (1)

Only glucose chars. (1)

➤Part (a) of the question asks for physical tests, so do NOT give chemicaltests.

➤The solids CANNOT be distinguished by determining their boiling pointsbecause the boiling point of SiO2(s) is so high (2230°C) that it CANNOTbedetermined using apparatus commonly available in a school laboratory.

RemarksRemarks*

Topic 41� Acids and Bases 1�Unit 15 Molarity, pH scale and strengths of acids and alkalis

15.1 Concentration of a solution

15.2 The pH scale

15.3 Determining pHvaluesof solutions

15.4 Strong andweak acids

15.5 Comparing the strengthof acids

15.6 pH and the extentofdissociationof acids

15.7 Strong andweak alkalis

15.8 Comparing the strengthof alkalis

15.9 Strength versus concentration

Molarity, pH scale and strengths of acids and alkalisUnit 15 15.1 – 15.9

Summary

1 The molarity of a solution is the number of moles of solute dissolved in 1dm3 ofthe solution.

Molarityof a solution(moldm3)

= numberofmolesof solute (mol)volumeof solution (dm3)

2 ThepHof a solution is –log10of themolar concentrationofhydrogen ions in thatsolution.

pH= –log10[H+]

3 ThepHscale isusedtomeasure thedegreeofacidityoralkalinityofasolution. Itsvalue ranges from0 to14.

increasing acidity increasing alkalinity

4 Methodsused todetermine thepHvaluesof solutions include:

a) usinguniversal indicator;

b)using apHmeter; and

c) using adata-loggerwith apH sensor.

5 A strongacid is anacid that almost completelydissociates inwater.Aweakacid isan acid thatonlypartiallydissociates inwater.

Compared toaweakacidof the sameconcentration,a strongacidhasa lowerpH,ahigher electrical conductivity and ahigher rateof reactionwithmetals.

6 Astrongalkaliisanalkalithatalmostcompletelydissociatestogivehydroxideions(OH–(aq)) in water. A weak alkali is an alkali that only partially dissociates to givehydroxide ions inwater.

Compared to a weak alkali of the same concentration, a strong alkali has a higherpHand ahigher electrical conductivity.

7 Strengthofanacidoralkaliconcernstheextentofdissociationoftheacidoralkaliinwater.Concentrationofanacidoralkaliconcernstheamountoftheacidoralkaliin aunit volumeof solution.

ExamtipsExamtipsExamtipsExamtips ♦ Adding water to KOH(aq) decreases its degree of alkalinity. Thus, thepHofKOH(aq) decreases.

♦ Adding water to HCl(aq) decreases its degree of acidity. Thus, the pHofHCl(aq) increases.

♦ Concentrated ammonia solution is aweak alkali.

Topic 41� Acids and Bases 1�Unit 16 Salts and neutralization

Example

Explainwhether you agreewith the following statement.

‘AisastrongeracidthanB,sothepHofanaqueoussolutionofAmustbelowerthanthatofB.’ (3marks)

Answer

Not agree

‘A is a stronger acid than B’ only means that the degree of dissociation of A is largerthan thatofB. (1)

However, the pH of an aqueous solution of an acid depends on both its degree ofdissociation and concentration. (1)

As such, an aqueous solution of the stronger acid A may have a higher pH than thatof the weaker acid B if the concentration of acid B is higher than that of acid A byan adequate amount. (1)

♦ Questionsoftenaskstudentstocomparethepropertiesofastrongacid/ alkaliwith thoseof aweak acid / alkali of the same concentration.

e.g.

0.5moldm–3NaOH(aq) 0.5moldm–3NH3(aq)

higher pH lower pH

higher electrical conductivity lower electrical conductivity

larger temperature rise when smaller temperature risewhen completely neutralized by completely neutralizedby 1moldm–3HCl(aq) 1moldm–3HCl(aq)

e.g.

Suppose two identical zinc strips are added to solutionsofmonobasicacidsX andYof the same concentration.

zinc strip

solution of monobasic acid X

solution of monobasic acid Y

Fromtherateof thereactionbetweeneachacidsolutionandthezincstrip, it canbededuced that acidX is weaker than acidY.

➤To compare the strength of a strong acid and a weak acid, remember touse the same concentration in order to fairly compare their pH values.

RemarksRemarks*

16.1 Acid-base reactions

16.2 Heat changeduringneutralization

16.3 Formation of salts

16.4 Naming of salts

16.5 Soluble and insoluble salts

16.6 Preparation of soluble salts

16.7 Preparation of insoluble salts

16.8 Uses ofneutralization

Unit 16 Salts and neutralization

Topic 41� Acids and Bases 1�Unit 16 Salts and neutralization

16.1 – 16.8

Summary

1 Neutralization is the combination of hydrogen ions and hydroxide ions (or oxideions) to formwatermolecules.

2 Inneutralization reactions, salt andwater are theonlyproducts.

acid+ alkali salt +water

acid+ insolublemetalhydroxide salt +water

acid+ insolublemetal oxide salt +water

e.g.

HCl(aq) +NaOH(aq) NaCl(aq) +H2O(l)

3HNO3(aq) + Fe(OH)3(s) Fe(NO3)3(aq) + 3H2O(l)

H2SO4(aq) +CuO(s) CuSO4(aq) +H2O(l)

3 Neutralization reactions are exothermic reactions. Heat is released in thesereactions.

4 Rules fornaming salts are as follows:

a) thenameof themetal or ammonium ions always comes first;

b)ifthemetalcanformmorethanonetypeofpositiveion,writeaRomannumeralinbrackets to show the chargeof the ion;

c) if the salt containsonly two elements, thename ends in –ide; and

d)ifthesaltcontainsthreeormoreelements,oneofwhichbeingoxygen,thenameends in –ite or –ate.

5 The following table summarizesmethods for thepreparationof salts.

Salt Method of preparation Example(s)

Solublesalts

actionsof acidonmetal / insolublebase / insolublecarbonate

Preparing zinc sulphate:

Zn(s) +H2SO4(aq) ZnSO4(aq) +H2(g)

ZnO(s) +H2SO4(aq) ZnSO4(aq) +H2O(l)

ZnCO3(s) +H2SO4(aq) ZnSO4(aq)+H2O(l) +CO2(g)

actionof acidon alkali /soluble carbonate(titration)

Preparingpotassiumchloride:

KOH(aq) +HCl(aq) KCl(aq) +H2O(l)

K2CO3(aq) + 2HCl(aq) 2KCl(aq) +H2O(l) +CO2(g)

Insolublesalts precipitation

Preparing silver chloride:

Ag+(aq) + Cl–(aq) AgCl(s)

from fromAgNO3(aq) NaCl(aq)

6 Usesofneutralization:

a) Acidic soil canbeneutralizedby addingquicklime (calciumoxide) to it.

b)Liquidwaste fromfactoriescontainingacidscanbeneutralizedbyaddingslakedlime (calciumhydroxide) to it.

ExamtipsExamtipsExamtipsExamtips ♦ Examinationquestionsoftenask students tocompare the temperaturerises upon mixing acids and alkalis.

* Forthereactionbetweenastrongacidandastrongalkali, theheatreleased is 57kJ for 1mole ofwater formed.

TheheatofneutralizationwillbefurtherdiscussedinTopic9ChemicalReactions and Energy.

ExampleTemperature

rise

Number of moles of water formed

Heat released*

Explanation

• 20cm3of1moldm–3HCl(aq) mixed with20cm3of1moldm–3NaOH(aq)

and


the same

0.02mole

0.04mole

1.14 kJ

2.28 kJ

The first mixture (totalvolume40cm3)absorbs1 .14 k J wh i l e t hesecond mixture (totalvolume80cm3)absorbs2.28kJ.Hencethetwomixturesgive the sametemperature rise.


and


temperaturerise ofsecond

mixture istwice thatof the first

mixture

0.02mole

0.04mole

1.14 kJ

2.28 kJ

The total volumes ofthe two mixtures arethe same. Hence thetemperature rise ofthe second mixture istwice that of the firstmixture.

• 20cm3of1moldm–3CH3COOH(aq) mixedw i t h 2 0 c m 3 o f1moldm–3NaOH(aq)

and


temperaturerise of firstmixture isless thanthat of

the secondmixture

0.02mole

0.02mole

<1.14 kJ

1.14 kJ

For neutralization inwhicheithertheacidoralkaliorbothareweak,the standard enthalpychangeofneutralizationw i l l be l e s s than–57.1kJmol–1.

CH 3COOH(aq) i s aweak acid. Hence thetemperature rise ofthe first mixture is lessthanthatofthesecondmixture.

c) Fertilizers such as ammonium nitrate and ammonium sulphate are prepared byneutralization reactions.

d)Antacidscontainingbasessuchasmagnesiumhydroxidecancurethepaincausedby excess acid in the stomach.

Topic 420 Acids and Bases 21Unit 16 Salts and neutralization

Example

Copper(II) sulphate crystals (CuSO4•5H2O(s)) can be prepared in a laboratory by thefollowing steps.

Step 1 Add excess copper(II) oxide todilute sulphuric acid andwarm themixture.

Step 2 Remove the remaining copper(II) oxide from the solutionobtained.

Step 3 Evaporate the solutionuntil it becomes saturated.

Step 4 Allow the saturated solution to cool down to obtain copper(II) sulphatecrystals.

Step 5 Separate the crystals from the saturated solution.

Step 6 Dry the crystals obtained.

a) For Step 1, explainwhy copper(II) oxide shouldbe in excess. (1mark)

b)For Step 4, explain why crystals would be obtained when the saturated solution iscooleddown. (1mark)

c) For Step 5, suggesthow the crystals is separated from the saturated solution. (1mark)

d)For Step 6, suggesthow the crystals canbedried. (1mark)

e) When CuSO4•5H2O(s) is heated slowly such that the temperature rises steadily, itwill lose fourwatermoleculesatabout110°C,and then the lastwatermoleculeatabout250°C.

i) Suggest a chemical test to show thatwater is being released. (2 marks)

♦ NOT all salts are formed fromneutralization.

♦ NOT all salt solutions are neutral.

e.g.

NaHSO4(aq)isacidic.TheHSO4–(aq)ionscandissociatetogivehydrogen

ions.

HSO4–(aq) H+(aq)+ SO4

2–(aq)

♦ Questions often ask about suitable method for preparing varioussalts.

– Lead(II) sulphate can be prepared by adding lead(II) nitrate solutionto dilute sulphuric acid.

Pb2+(aq) + SO42–(aq) PbSO4(s)

– Copper(II)sulphatecanbepreparedbyaddingexcesscopper(II)oxideto dilute sulphuric acid.

CuO(s) +H2SO4(aq) CuSO4(aq)+H2O(l)

ii)Usingtheaxesbelow,sketchthechangeofmasswhenasampleofCuSO4•5H2O(s)isheated slowly. (2marks)

Temperature (°C)

Mas

s (a

rbit

rary

un

it)

100 150 200 250 300

Answer

a) To make sure that all the sulphuric acid has been reacted. / To make sure that theproduct isnot contaminatedwith sulphuric acid. (1)

b)Thesolubilityofcopper(II)sulphatedecreaseswhenthetemperatureofthesolutiondrops. (1)

c) By filtration (1)

d)Absorb thewater by filter paper. / Place in adesiccator. (1)

e) i) Treat the vapour evolvedwithdry cobalt(II) chloridepaper. (1)

A colour change fromblue topink shows thepresenceofwater. (1)

ii)

(2)Temperature (°C)

Mas

s (a

rbit

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Topic 422 Acids and Bases 23Unit 17 Concentration of solutions and volumetric analysis

➤Questions often ask about the procedure for preparing CuSO4•5H2Ocrystals.

e.g.

CuSO4•5H2O crystals can be obtained by heating excess copper withconcentrated sulphuric acid.

Cu(s) + 2H2SO4(l) CuSO4(aq)+ SO2(g) + 2H2O(l)

Addwater to the resulting mixture and filter off the excess copper.

Evaporate the filtrate to give a saturated copper(II) sulphate solution.

Allow the solution to cool to obtain CuSO4•5H2O crystals.

Dry the crystals in a desiccator.

RemarksRemarks*

17.1 Concentration of a solution

17.2 Dilution

17.3 Volumetric analysis

17.4 Preparing a standard solutionof an acid / alkali

17.5 Acid-alkali titration

17.6 pH changeduring a titration

17.7 Using an indicator in an acid-alkali titration

17.8 Equivalence pointdetectionby temperature change

17.9 Applications of acid-alkali titrations

17.10 Back titration

Concentration of solutions and volumetric analysisUnit 17

Topic 42� Acids and Bases 2�Unit 17 Concentration of solutions and volumetric analysis

17.1 – 17.4

Summary1 Concentrationof a solution

(gdm3) = massof solute (g)

volumeof solution (dm3)

2 Thenumbersofmolesof solute ina solutionbeforedilutionandafterdilutionarethe same, i.e.

(MV)beforedilution= (MV) afterdilution,whereM=molarity,V= volume.

ExamtipsExamtipsExamtipsExamtips ♦ Questions often ask about the procedure of diluting a concentratedacid / alkali of known concentration.

– To prepare 250.0cm3 of 0.100moldm–3 Na2CO3(aq) from1.00moldm–3 Na2CO3(aq), use a 25.0cm3 pipette and a 250.0cm3volumetric flask.

– Whenpreparingasolutionfromasolidacid,apipetteisNOTrequiredfor transferring the acid to the volumetric flask.

♦ Inthedilutionprocessusingavolumetricflask,distilledwater isaddedto the flask until the bottom of the meniscus reaches the graduatedmarkon the flask.

Example

20.0cm3of1.00moldm–3NaCl(aq)aremixedwith10.0cm3of2.00moldm–3Na2CO3(aq).What is the concentrationofNa+(aq) ions in the resulting solution? (3 marks)

Answer

NumberofmolesofNa+(aq) ions inNaCl(aq)

=1.00moldm–3 x 20.01000

dm3

=0.0200mol (1)

NumberofmolesofNa+(aq) ions inNa2CO3(aq)

=2 x 2.00moldm–3 x 10.01000

dm3

=0.0400mol (1)

ConcentrationofNa+(aq) ions in the resulting solution

= (0.0200+0.0400)mol20.0+10.0

1000dm3

=2.00moldm–3 (1)

➤In the above example, students need to use the correct volume of theresulting solution for calculations in order to arrive at the right answer.

RemarksRemarks*

17.5 – 17.10

Summary

1 Volumetric analysis is amethodof findingout the concentration (or amount)of asubstance inasample. It reliesontheprecisemeasurementofvolumesofsolutionsinvolved in chemical reactions.

2 Involumetricanalysis, just sufficientvolumeofa solutionofknownconcentrationis allowed to completely react with the substance being analyzed in a sample. Theprocess of determining the ‘just sufficient’ volume is called titration. From themeasured volume and known concentration of the solution used, we can calculatethe concentration (or amount) of the substance inquestion.

3 A standard solution is a solutionwith accurately knownconcentration.

4 Thepointinanacid-alkalititrationatwhichtheacidandalkalijustreactcompletelywith eachother is called the equivalencepoint.

5 An acid-alkali indicator is used to indicate the end point of an acid-alkalititration.

Type of acid-alkali titration Suitable indicator

Strong acid-strong alkali methyl orangeorphenolphthalein

Weak acid-strong alkali phenolphthalein

Strong acid-weak alkali methyl orange

Weak acid-weak alkali —

6 Specialapparatusused involumetricanalysis includeelectronicbalance,volumetricflask, pipette, pipette filler andburette.

7 The following diagram summarizes the steps for calculations in volumetricanalysis.

knownconcentrationand volume

of A

number of moles of

substance in question

mass,concentrationor volume of substance in

question

numberof moles

of A

usecoefficients

in the balanced

equation to find the

mole ratio


8 Thefollowingdiagramsummarizesthestepstocalculatetheamountorconcentrationof the substancebeing analyzed in a sample.

number of moles of acid / alkali originally added to the sample

number of moles of acid / alkali left over

after reaction with the substance being analyzed in the

sample (information obtained from back

titration)

number of moles of acid / alkali reacts with the substance

being analyzed in the sample

number of moles of the substance being

analyzed in the sample

amount or concentration of the

substance being analyzed in the

sample

=–

use coefficientsin the balanced equation to find the mole ratio

ExamtipsExamtipsExamtipsExamtips ♦ Students should be able to give the meaning of the term ‘primarystandard’ in titrimetric analysis.

♦ A primary standard of a substance can be prepared by dissolving aknownmassofthesubstance inasolventandmakingupthesolutionto a known volume.

♦ As sodium hydroxide absorbs water / CO2 readily in air, we CANNOTprepareastandardsodiumhydroxidesolutionjustbydissolvingaknownmassofthesolid indistilledwaterandmakinguptoaknownvolumeof solution.

♦ Questions often ask about the correct procedure for washing eachglasswarebefore titration.

– Washapipetteandaburettefirstwithdistilledwaterandthenwiththe solution it is going to deliver.

– Wash a conical flask and a volumetric flask with distilled wateronly.

♦ Consider an aqueous solution of ethanoic acidwith a pH valueof 4.

Adding

– solid calcium carbonate; and

– dilute aqueous ammonia

would increase thepH valueof the acid.

♦ Thefirst titration isatrialandthedatashouldNOTbe included inthecalculation.

♦ ExaminationquestionsoftenaskstudentstosketchthepHchangeforthetitration of aweak acid (e.g. ethanoic acid)with dilute NaOH(aq).

20.0 30.0 40.010.00

4

6

8

10

12

14

2

0

pH

Volume of NaOH(aq) added (cm3)

phenolphthaleinchanges colour within this pH range

equivalencepoint

♦ Students should be able to choose a suitable indicator based on thepH change during a titration.

♦ The equivalent point of a neutralization reaction can also be detectedbymeasuring either temperature or electrical conductivity.

– The following graph shows the temperature change during thetitration betweenNaOH(aq) and HCl(aq).

X

YZ

Volume of acid added (cm3)

Temperature (°C)equivalence point of titration; temperature is the highest

equivalencepoint


Example

Thedrugtabletsshownbelowcanrelievestomachache.Itcontainsaluminiumhydroxide,Al(OH)3, as theonly active ingredient.

FORMULA: Each tablet contains Aluminium Hydroxide Dried Gel 200 mg, Magnesium Hydroxide 200mg, Dimethylpolysiloxane 20mg. DOSAGE: 2 tablets to be well chewed 30 minutes after meals, at bedtime, when symptoms occur or as directed by physician. Do not take more than 24 tablets in a 24 hour period. CAUTION: It is dangerous to exceed the stated does. Keep this drug out of the reach of children. Do not take this product if you are taking antibiotic drug containing Tetracycline.

a) Explain how the drug tablets can relieve stomach ache with the help of anequation. (2marks)

b)Explainwhy thedrug tablets shouldbe chewedbefore swallowing. (1mark)

c) Astudentperformedthefollowingexperimenttodeterminetheamountofaluminiumhydroxide contained inonedrug tablet.

Step 1 Adrug tabletwasdissolved in50.0cm3of1.00moldm–3hydrochloric acidto forma solution.

Step 2 Thesolutionwasplacedinavolumetricflaskandthendilutedto250.0cm3withdistilledwater.

Step 3 25.0cm3 of the diluted solution were titrated with 0.190moldm–3 sodiumhydroxidesolutionusingasuitableindicator.18.1cm3ofsodiumhydroxidesolutionwereneeded to reach the endpoint.

i) Drawa labelleddiagram to show the set-upused in the titration. (3marks)

ii)Suggest a suitable indicator for the titration in Step 3, and state the expectedcolour change at the endpoint. (2marks)

iii)Calculate the number of moles of excess hydrochloric acid in 25.0cm3 of thediluted solution from thedataobtained in the titration. (2marks)

iv)Hence calculate themassof aluminiumhydroxide inonedrug tablet. (3 marks)

(Relative atomicmasses:H=1.0,C=12.0,Al = 27.0)

Answer

a) Aluminium hydroxide in the tablet neutralizes the excess hydrochloric acid in thestomach. (1)

Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l) (1)

b)Chewingbreaksdownthetabletsintosmallerpieces.Thisincreasesthesurfaceareaof the tablets and thus increases the reaction rate (brings faster relief of pain). (1)

c) i)

(3)white tile

0.190 mol dm–3

sodium hydroxide solution

burette

solution of drug tablet + indicator

conical flask

(1mark for correct set-up;2marks for correct labels; 0mark if the set-up isnotworkable)

ii)Methyl orange: from red to yellow / Phenolphthalein: from colourless to red orpink (2)

iii) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 0.190moldm–3

18.1cm3

NumberofmolesofNaOH in18.1cm3 solution=0.190moldm–3 x 18.11000

dm3

= 0.00344mol (1)

Accordingtotheequation,1moleofHClrequires1moleofNaOHforcompleteneutralization.

i.e. numberofmolesof excessHCl in25.0cm3diluted solution = 0.00344mol (1)

– Thefollowinggraphshowstheelectricalconductivitychangeduringthe titration betweenH2SO4(aq) andBa(OH)2(aq).

Volume of H2SO4(aq) added to Ba(OH)2(aq) (cm3)low

high

Con

duct

ivity

As sulphuric acid is added, it removes both the barium ions (byprecipitation) andhydroxide ions (by neutralization)

Ba(OH)2(aq)+H2SO4(aq) BaSO4(s) + 2H2O(l)

At the equivalence point, all the barium ions and hydroxide ionshavebeenremoved.Hencetheelectricalconductivityofthereactionmixture falls almost to zero.

Topic 430 Acids and Bases

➤Questions often ask students to suggest a suitable indicator for a certaintitration and state the colour change.

➤Do NOT omit the dilution factor in the calculations. Otherwise, a wronganswerwouldbeobtained.

RemarksRemarks*

iv)NumberofmolesofHCl added in Step 1 = 1.00moldm–3 x 50.01000

dm3

= 0.0500mol (1)

NumberofmolesofHCl left over after reactionwithdrug tablet in Step 1

= 0.00344mol x 250.0cm3

25.0cm3

= 0.0344mol

NumberofmolesofHCl reactedwithAl(OH)3 indrug tablet = (0.0500 – 0.0344)mol = 0.0156mol (1)

Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

Accordingtotheequation,1moleofAl(OH)3requires3molesofHClforcompleteneutralization.

i.e. numberofmolesofAl(OH)3 indrug tablet= 0.01563

mol

= 0.00520mol

MassofAl(OH)3 indrug tablet = 0.00520mol x 78.0gmol–1

= 0.406g (1)

\ one drug tablet contains 0.406gof aluminiumhydroxide.

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