Acids Bases BuffersAcids Bases Buffers► Number of reaction types? Write down as

many as you can.► History – what do you currently know about

pH/acids/bases/buffers?► Universal indicator?

Acids Bases BuffersAcids Bases Buffers►Bronsted-Lowry Definitions - Bronsted-Lowry Definitions - cards cards ►ACID = ACID = ►Proton (HProton (H++) donor ) donor ►BASE = BASE = ►Proton (HProton (H++) acceptor) acceptor►Cards! Use them!!Cards! Use them!!

Proton transferProton transfer

►HH22SOSO4 4 + HNO+ HNO33 H H22NONO33++ + HSO + HSO44

--

►acid base acid baseacid base acid base► (conjugate) (conjugate)(conjugate) (conjugate)►Strong acid gives weak conjugate baseStrong acid gives weak conjugate base►Vice versa appliesVice versa applies►Acid loses proton Acid loses proton conjugate base conjugate base►Base gains a proton Base gains a proton conjugate acid conjugate acid►Card – include exampleCard – include example

Have a goHave a go

►Conjugate acid/base problems sheetConjugate acid/base problems sheet►Ext: What do you notice about water in Ext: What do you notice about water in

these examples? There is a name for these examples? There is a name for this – any suggestions?this – any suggestions?

►(Smith Older p 167 Q1)(Smith Older p 167 Q1)

Weak acids - Weak acids -

►Weak acids – equilibrium systems► A + B = C + D►Kc = ?►Modified eqm constant►Main effects – add/remove

components►Temp - possible

Strong vs weak acids – Strong vs weak acids – molymods CHmolymods CH33COOHCOOH

►Strong acid Strong acid ► Dissociates completely in water producing Dissociates completely in water producing

HH33OO++ as a result of good proton donor as a result of good proton donor propertiesproperties

►Card Card ►Weak acidWeak acid►Partial dissociation – poor proton donorPartial dissociation – poor proton donor►Soluble bases produces an alkali - strong Soluble bases produces an alkali - strong

base/alkali - lots of OHbase/alkali - lots of OH- - in solution in solution

Strong and weak acidsStrong and weak acids

►Use molymods to make CHUse molymods to make CH33COOH and COOH and HCl representationsHCl representations

Typical reactions of acidsTypical reactions of acids

► Metals e.g. magnesiumMetals e.g. magnesium► Carbonates e.g.CaCOCarbonates e.g.CaCO33

► Bases e.g. MgO/CaO (metal oxides) Bases e.g. MgO/CaO (metal oxides) ► The reacting species is HThe reacting species is H++(aq) = H(aq) = H33OO++(aq)(aq)

► Try to work out suitable ionic equations – HCl usedTry to work out suitable ionic equations – HCl used► CardsCards► Work out equations for nitric acid. Work out equations for nitric acid. ► Ext – CaCOExt – CaCO33 chips react with sulphuric rapidly chips react with sulphuric rapidly

initially, but soon the reaction slows to a virtual initially, but soon the reaction slows to a virtual stop. Explainstop. Explain

Ionic equations – insert state Ionic equations – insert state symbolssymbols

►Mg + 2HMg + 2H++ (aq) (aq) Mg Mg2+2+ (aq) (aq) + H+ H22 (g) (g)

►CaCOCaCO33 + 2H + 2H++ Ca Ca2+ 2+ + CO+ CO22 + H + H22OO

►MgO + 2HMgO + 2H+ + (aq)(aq) Mg Mg2+ 2+ + H+ H22OO

► Issue notesIssue notes►Note – ions are represented Note – ions are represented

individually when they are IN individually when they are IN SOLUTIONSOLUTION

Acid dissociation constantsAcid dissociation constants

► CHCH33COOH + HCOOH + H22O O CH CH33COOCOO-- + H + H33OO++

► Write an expression for Kc. Write an expression for Kc. ► Which concentration has little Which concentration has little

significance and why?significance and why?

Acid dissociation constantsAcid dissociation constants

KKaa == [CH[CH33COOCOO--]] [H[H

33OO++]]   

[CH[CH33COOH]COOH]

Water concentration removed (incorporated into Kc) as it is large and therefore effectively constant

HA = HHA = H++ + A + A--

►General relationship of the form (card)General relationship of the form (card)

Ka=[A-] [H3O+] 

[HA]

So units always ?

Mol dmMol dm-3-3

►Questions on KQuestions on Kaa

►Page 2 of notesPage 2 of notes

Water as an equilibrium Water as an equilibrium system –need this for pH of system –need this for pH of

alkalisalkalis

►2H2H22O(l) = HO(l) = H33OO++(aq) + OH-(aq)(aq) + OH-(aq)

►Water concentration large and Water concentration large and constantconstant

►Kw-the Kw-the ionic product of water:ionic product of water:►Kw = [HKw = [H33OO++] [OH-] = 10] [OH-] = 10-14-14 mol mol22 dm dm-6-6

►Read “ionic product of water” p2 – Read “ionic product of water” p2 – answer any questions.answer any questions.

pHpH

►When did you first meet pH?When did you first meet pH?►Why is it important?Why is it important?►What actually is it??What actually is it??►Universal indicator?Universal indicator?

The pH scale – what is it?The pH scale – what is it?

►p=-logp=-log1010

pH = -log10 [H+] = log10 1 . [H+]NOTE Ka does the same tricki.e. pKa = -log10 [Ka] = log10 1 . [Ka]

pH calculations – strong acidspH calculations – strong acids

► The concentration of The concentration of HH+ + must be must be determineddetermined first, it is then inserted into the first, it is then inserted into the relationship for pHrelationship for pH

► For strong monoprotic acids For strong monoprotic acids ► [H+] = concentration of acid examples[H+] = concentration of acid examples► For strong diprotic acids e.g. sulphuricFor strong diprotic acids e.g. sulphuric► [H+] = 2 x concentration of acid[H+] = 2 x concentration of acid

Using your calculator:

► Key in the concentration of H+

►Use reciprocal key - x-1 or 1/x►Then log the answer.►Different calculators work in different ways.►Exam trick - do sum with known answer►E.g. 0.001 mol dm-3 HCl has pH = 3►Need to go both ways!! – “play and learn”

Try theseTry these

►Q’s p 3-4Q’s p 3-4►Extension – why do complications arise Extension – why do complications arise

with solutions such as 3.50 mol dmwith solutions such as 3.50 mol dm-3-3 HH33POPO44??

►What data would you require in order What data would you require in order to calculate accurate pH values?to calculate accurate pH values?

pH calculations – strong pH calculations – strong alkalisalkalis

► Need KNeed Kww – this allows [H – this allows [H++] to be calculated from ] to be calculated from [OH[OH--]:]:

► [H[H++] x [OH] x [OH--] = 10] = 10-14 -14 mol mol22 dm dm-6 -6 @ 25@ 2500CC

► Kw = constant at constant temperatureKw = constant at constant temperature► Calculate [OHCalculate [OH--] from info given e.g. ] from info given e.g. ► Use Kw to calculate [H+] see aboveUse Kw to calculate [H+] see above► Feed into pH equation to determine pH of Feed into pH equation to determine pH of

solution.solution.

pH calculations – strong pH calculations – strong alkalisalkalis

►ExampleExample►2.0g NaOH in 5 litres of solution.2.0g NaOH in 5 litres of solution.►= 0.40g dm= 0.40g dm-3-3 ►= 0.4/40 mol dm= 0.4/40 mol dm-3-3

►=0.01 mol dm=0.01 mol dm-3 -3 = 10= 10-2-2 mol dm mol dm-3 -3

►[H[H++] x [OH] x [OH--] = 10] = 10-14-14

►So [HSo [H++] = 10] = 10-14-14/ 10/ 10-2 -2 = 10= 10-12-12

►pH = 12pH = 12

pH of weak acids in simple pH of weak acids in simple solution i.e.just dissolved in solution i.e.just dissolved in

waterwater►[H[H++] = [CH] = [CH33COOCOO--]]

►So, as So, as Ka = [CH3COO-] [H3O+] [CH3COOH] Ka = [H3O+]2

[CH3COOH]

Take the square root of both sides to give [H+]

[H3O+]2 = Ka[CH3COOH]

Try these - pH calculations 2Try these - pH calculations 2

►Q 3 (a) and (b) [pKa]?Q 3 (a) and (b) [pKa]?► If pKa =4.5 then Ka=?If pKa =4.5 then Ka=?►3.16 x 103.16 x 10-5 -5 mol dmmol dm-3-3

►Use known Q/A to learn calculator Use known Q/A to learn calculator trickstricks

pH Calculations 1 - answerspH Calculations 1 - answers

►Q1 a: 0.7, b:1.0 c:6.15 d:12.7 e:13.5Q1 a: 0.7, b:1.0 c:6.15 d:12.7 e:13.5►Q2 a: 3.2x10Q2 a: 3.2x10-4-4mol dmmol dm-3 -3 b: 0.05M b: 0.05M

c:1.26M d:0.01M Hc:1.26M d:0.01M H22SOSO44

►Q3 a: 1.3 b: 0.05 c: 12Q3 a: 1.3 b: 0.05 c: 12►Q4 a: 1.23Q4 a: 1.23

pH Calculations 2 - AnswerspH Calculations 2 - Answers

►Calcs 2Calcs 2►Q1 a:1.40 b:0.52 c:14.1 d:12Q1 a:1.40 b:0.52 c:14.1 d:12►Q2 a:1.58x10Q2 a:1.58x10-4-4 b:0.158 b:0.158►Q3 a:2.75 b:5.13Q3 a:2.75 b:5.13►Q4 Methanoic>butanoic>HCNQ4 Methanoic>butanoic>HCN►Q5 a: 1.80 b: 13 c: 13.72Q5 a: 1.80 b: 13 c: 13.72

Ph change during a titrationPh change during a titrationequivalence point = end equivalence point = end

pointpoint►What is an end point in a titration?What is an end point in a titration?

What is an end point?What is an end point?

►The point when reactants have been The point when reactants have been mixed in exactly the proportions given mixed in exactly the proportions given in the equationin the equation

► In an acid base titration this In an acid base titration this corresponds to a point half way corresponds to a point half way up/down the steep portion of the up/down the steep portion of the graph.graph.

Ph change during a titrationPh change during a titrationequivalence point = end equivalence point = end

pointpoint

See “Task” page 5 of notesSee “Task” page 5 of notes

►An exercise in reading!An exercise in reading!►Ask if unsure.Ask if unsure.►Refer to expt already carried out – see Refer to expt already carried out – see

if it agrees.if it agrees.

pH Curves for various pH Curves for various combinationscombinations

Ka can be determined by the pH Ka can be determined by the pH at the half neutralisation point. at the half neutralisation point.

ExptExpt

The ideal indicator has the The ideal indicator has the equivalence point in the middle of its equivalence point in the middle of its

rangerange

INDICATORSINDICATORS

Indicators are weak acids which have a Indicators are weak acids which have a different colour to their conjugate basedifferent colour to their conjugate base

HInHIn H H++ + + InIn--

colour 1colour 1 colour 2colour 2

low pH: equilibrium pushed left = low pH: equilibrium pushed left = colour 1colour 1

high pH: equilibrium pushed right = high pH: equilibrium pushed right = colour colour 22

See “indicators [1]” See “indicators [1]” presentationpresentation

►

Enthalpy of neutralisationEnthalpy of neutralisation

►AS Energetics: AS Energetics:

►ΔΔE = m x s x E = m x s x ΔΔtt

►Convert to 1 mole – i.e. Divide by the Convert to 1 mole – i.e. Divide by the number of moles used for above number of moles used for above valuesvalues

Buffer SolutionsBuffer Solutions

►DefinitionDefinition►A buffer solution is one which A buffer solution is one which

minimises the changeminimises the change in pH on in pH on addition of small amounts of acid or addition of small amounts of acid or alkalialkali

Buffer solutions are prepared by taking the Buffer solutions are prepared by taking the salt of a weak acid (or base) and dissolving it salt of a weak acid (or base) and dissolving it

in the acid (or base) itself.in the acid (or base) itself.

►e.g. CHe.g. CH33COOH +HCOOH +H22O O CH CH33COOCOO-- + H + H33OO++

► In the buffer the concentrations of In the buffer the concentrations of CHCH33COOH and CHCOOH and CH33COOCOO- - are BOTH high.are BOTH high.

►So if HSo if H++ is added the following occurs is added the following occurs►HH+ + + CH + CH33COOCOO-- CH CH33COOH COOH

►Similarly if OHSimilarly if OH-- ions are added ions are added ►OHOH-- + CH + CH33COOH COOH CH CH33COOCOO-- + H + H22O O

i.e. much of the added OH- is removedi.e. much of the added OH- is removed

Calculations involving buffer Calculations involving buffer solutions:solutions:

►For HA H+(aq) + A-(aq)► Adding [A-] to the system will shift the

equilibrium to the► left►Hence Ka will►Not change►Ka= [A-] [H3O+]  ► [HA]►This is the basis for calculations

Ka= [A-] [H3O+]  [HA]

►Concentrations of Concentrations of [A-] and [H3O+] are

►NOT the same►So these quantities are

calculated separately►Deduce numbers – feed them

in!

ExampleExample

►5.0g of sodium ethanoate is added to 5.0g of sodium ethanoate is added to 500cm500cm33 of 0.2 mol dm of 0.2 mol dm-3-3 ethanoic acid ethanoic acid which is then made up to 1 litre with which is then made up to 1 litre with distilled water.distilled water.

►Calculate the pH of the resulting solution Calculate the pH of the resulting solution given Ka = 1.8 x10given Ka = 1.8 x10-5-5 mol dm mol dm-3-3 (4.53) (4.53)

►See Q’s page 8 of notesSee Q’s page 8 of notes►pH calculations 3 pH calculations 3 ►Textbook Q’sTextbook Q’s

Notes q’s – answers p9Notes q’s – answers p9

►Q1 0.001 or 10Q1 0.001 or 10-3-3 mol dm mol dm-3-3

►Q2 3.16 x 10Q2 3.16 x 10-5-5 mol dm mol dm-3-3

►Q3 2.9gQ3 2.9g►Q4 0.439gQ4 0.439g

Hints for notes Q’sHints for notes Q’s

► Q1 log 1/10Q1 log 1/10-3-3 = = ► Q1 log 1/10Q1 log 1/10-4.25-4.25 = =► Q 3 ethanoic acidQ 3 ethanoic acid contains 7.5g contains 7.5g dmdm-3-3 ► pH 4.5 so [H+] = 10-4.5 ► = 3.2 x 10-5 mol dm-3

► Rearrange Ka expression and insert values► To give [CH3COO-] in mol dm-3

► X 82 gives mass per litre► Divide by 2 gives mass in 500 cm3

pH calcs 3 AnswerspH calcs 3 Answers

►1 (a) 3.35 (b) 4.731 (a) 3.35 (b) 4.73►2 (a) new pH = 3.38 so change = 0.032 (a) new pH = 3.38 so change = 0.03►2 (b) 13 to 11.92 (b) 13 to 11.9►3 (a) 5.14 (b) 57.8g3 (a) 5.14 (b) 57.8g

Applications of Buffer Applications of Buffer SolutionsSolutions

►Body fluids e.g. blood – a change of pH Body fluids e.g. blood – a change of pH value of 0.1 can cause loss of value of 0.1 can cause loss of consciousness consciousness

►Cosmetic productsCosmetic products►DetergentsDetergents