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

Proton transfer ► H 2 SO 4 + HNO 3  H 2 NO HSO 4 - ► acid base acid base ► (conjugate) (conjugate) ► Strong acid gives weak conjugate base ► Vice versa applies ► Acid loses proton  conjugate base ► Base gains a proton  conjugate acid ► Card – include example

Have a go ► Conjugate acid/base problems sheet ► Ext: What do you notice about water in these examples? There is a name for this – any suggestions? ► (Smith Older p 167 Q1)

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 – molymods CH 3 COOH ► Strong acid ► Dissociates completely in water producing H 3 O + as a result of good proton donor properties ► Card ► Weak acid ► Partial dissociation – poor proton donor ► Soluble bases produces an alkali - strong base/alkali - lots of OH - in solution

Strong and weak acids ► Use molymods to make CH 3 COOH and HCl representations

Typical reactions of acids ► Metals e.g. magnesium ► Carbonates e.g.CaCO 3 ► Bases e.g. MgO/CaO (metal oxides) ► The reacting species is H + (aq) = H 3 O + (aq) ► Try to work out suitable ionic equations – HCl used ► Cards ► Work out equations for nitric acid. ► Ext – CaCO 3 chips react with sulphuric rapidly initially, but soon the reaction slows to a virtual stop. Explain

Ionic equations – insert state symbols ► Mg + 2H + (aq)  Mg 2+ (aq) + H 2 (g) ► CaCO 3 + 2H +  Ca 2+ + CO 2 + H 2 O ► MgO + 2H + (aq)  Mg 2+ + H 2 O ► Issue notes ► Note – ions are represented individually when they are IN SOLUTION

Acid dissociation constants ► CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + ► Write an expression for Kc. ► Which concentration has little significance and why?

Acid dissociation constants K a K a= [CH 3 COO - ] [H 3 O + ] [H 3 O + ] [CH 3 COOH] Water concentration removed (incorporated into Kc) as it is large and therefore effectively constant

HA = H + + A - ► General relationship of the form (card) Ka=[A-] [H 3 O + ] [HA] So units always ?

Mol dm -3 ► Questions on K a ► Page 2 of notes

Water as an equilibrium system – need this for pH of alkalis ► 2H 2 O(l) = H 3 O + (aq) + OH-(aq) ► Water concentration large and constant ► Kw-the ionic product of water: ► Kw = [H 3 O + ] [OH-] = mol 2 dm -6 ► Read “ionic product of water” p2 – answer any questions.

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

The pH scale – what is it? ► p=-log 10 pH = -log 10 [H + ] = log [H + ] NOTE Ka does the same trick i.e. pKa = -log 10 [Ka] = log [Ka]

pH calculations – strong acids ► The concentration of H + must be determined first, it is then inserted into the relationship for pH ► For strong monoprotic acids ► [H+] = concentration of acid examples ► For strong diprotic acids e.g. sulphuric ► [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 mol dm -3 HCl has pH = 3 ► ► Need to go both ways!! – “play and learn”

Try these ► Q’s p 3-4 ► Extension – why do complications arise with solutions such as 3.50 mol dm -3 H 3 PO 4 ? ► What data would you require in order to calculate accurate pH values?

pH calculations – strong alkalis ► Need K w – this allows [H + ] to be calculated from [OH - ]: ► [H + ] x [OH - ] = mol 2 dm 25 0 C ► Kw = constant at constant temperature ► Calculate [OH - ] from info given e.g. ► Use Kw to calculate [H+] see above ► Feed into pH equation to determine pH of solution.

pH calculations – strong alkalis ► Example ► 2.0g NaOH in 5 litres of solution. ► = 0.40g dm -3 ► = 0.4/40 mol dm -3 ► =0.01 mol dm -3 = mol dm -3 ► [H + ] x [OH - ] = ► So [H + ] = / = ► pH = 12

pH of weak acids in simple solution i.e.just dissolved in water ► [H + ] = [CH 3 COO - ] ► So, as Ka=[CH 3 COO - ] [H 3 O + ] [CH 3 COOH] Ka= [H 3 O + ] 2 [CH3COOH] Take the square root of both sides to give [H + ] [H 3 O + ] 2 = Ka[CH 3 COOH]

Try these - pH calculations 2 ► Q 3 (a) and (b) [pKa]? ► If pKa =4.5 then Ka=? ► 3.16 x mol dm -3 ► Use known Q/A to learn calculator tricks

pH Calculations 1 - answers ► Q1 a: 0.7, b:1.0 c:6.15 d:12.7 e:13.5 ► Q2 a: 3.2x10 -4 mol dm -3 b: 0.05M c:1.26M d:0.01M H 2 SO 4 ► Q3 a: 1.3 b: 0.05 c: 12 ► Q4 a: 1.23

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

Ph change during a titration equivalence point = end point ► What is an end point in a titration?

What is an end point? ► The point when reactants have been mixed in exactly the proportions given in the equation ► In an acid base titration this corresponds to a point half way up/down the steep portion of the graph.

Ph change during a titration equivalence point = end point

See “Task” page 5 of notes ► An exercise in reading! ► Ask if unsure. ► Refer to expt already carried out – see if it agrees.

pH Curves for various combinations

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

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

INDICATORS Indicators are weak acids which have a different colour to their conjugate base HIn H + + In - colour 1 colour 2 low pH: equilibrium pushed left = colour 1 high pH: equilibrium pushed right = colour 2

See “indicators [1]” presentation ►

Enthalpy of neutralisation ► AS Energetics: ► ΔE = m x s x Δt ► Convert to 1 mole – i.e. Divide by the number of moles used for above values

Buffer Solutions ► Definition ► A buffer solution is one which minimises the change in pH on addition of small amounts of acid or alkali

Buffer solutions are prepared by taking the salt of a weak acid (or base) and dissolving it in the acid (or base) itself. ► e.g. CH 3 COOH +H 2 O  CH 3 COO - + H 3 O + ► In the buffer the concentrations of CH 3 COOH and CH 3 COO - are BOTH high. ► So if H + is added the following occurs ► H + + CH 3 COO -  CH 3 COOH ► Similarly if OH - ions are added ► OH - + CH 3 COOH  CH 3 COO - + H 2 O i.e. much of the added OH- is removed

Calculations involving buffer 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 - ] [H 3 O + ] ► ► [HA] ► ► This is the basis for calculations

Ka= [A - ] [H 3 O + ] [HA] ► Concentrations of ► Concentrations of [A - ] and [H 3 O + ] are ► ► NOT the same ► ► So these quantities are calculated separately ► ► Deduce numbers – feed them in!

Example ► 5.0g of sodium ethanoate is added to 500cm 3 of 0.2 mol dm -3 ethanoic acid which is then made up to 1 litre with distilled water. ► Calculate the pH of the resulting solution given Ka = 1.8 x10 -5 mol dm -3 (4.53) ► See Q’s page 8 of notes ► pH calculations 3 ► Textbook Q’s

Notes q’s – answers p9 ► Q or mol dm -3 ► Q x mol dm -3 ► Q3 2.9g ► Q g

Hints for notes Q’s ► Q1 log 1/10 -3 = ► Q1 log 1/ = ► Q 3 ethanoic acid contains 7.5g dm -3 ► ► pH 4.5 so [H + ] = ► ► = 3.2 x mol dm -3 ► ► Rearrange Ka expression and insert values ► ► To give [CH 3 COO - ] in mol dm -3 ► ► X 82 gives mass per litre ► ► Divide by 2 gives mass in 500 cm 3

pH calcs 3 Answers ► 1 (a) 3.35 (b) 4.73 ► 2 (a) new pH = 3.38 so change = 0.03 ► 2 (b) 13 to 11.9 ► 3 (a) 5.14 (b) 57.8g

Applications of Buffer Solutions ► Body fluids e.g. blood – a change of pH value of 0.1 can cause loss of consciousness ► Cosmetic products ► Detergents