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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?
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Acids Bases Buffers ► Bronsted-Lowry Definitions - cards ► ACID = ► Proton (H + ) donor ► BASE = ► Proton (H + ) acceptor ► Cards! Use them!!
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Proton transfer ► H 2 SO 4 + HNO 3 H 2 NO 3 + + 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
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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)
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Weak acids - ► ► Weak acids – equilibrium systems ► ► A + B = C + D ► ► Kc = ? ► ► Modified eqm constant ► ► Main effects – add/remove components ► ► Temp - possible
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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
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Strong and weak acids ► Use molymods to make CH 3 COOH and HCl representations
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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
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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
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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?
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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
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HA = H + + A - ► General relationship of the form (card) Ka=[A-] [H 3 O + ] [HA] So units always ?
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Mol dm -3 ► Questions on K a ► Page 2 of notes
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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-] = 10 -14 mol 2 dm -6 ► Read “ionic product of water” p2 – answer any questions.
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pH ► When did you first meet pH? ► Why is it important? ► What actually is it?? ► Universal indicator?
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The pH scale – what is it? ► p=-log 10 pH = -log 10 [H + ] = log 10 1. [H + ] NOTE Ka does the same trick i.e. pKa = -log 10 [Ka] = log 10 1. [Ka]
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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
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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”
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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?
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pH calculations – strong alkalis ► Need K w – this allows [H + ] to be calculated from [OH - ]: ► [H + ] x [OH - ] = 10 -14 mol 2 dm -6 @ 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.
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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 = 10 -2 mol dm -3 ► [H + ] x [OH - ] = 10 -14 ► So [H + ] = 10 -14 / 10 -2 = 10 -12 ► pH = 12
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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]
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Try these - pH calculations 2 ► Q 3 (a) and (b) [pKa]? ► If pKa =4.5 then Ka=? ► 3.16 x 10 -5 mol dm -3 ► Use known Q/A to learn calculator tricks
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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
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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
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Ph change during a titration equivalence point = end point ► What is an end point in a titration?
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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.
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Ph change during a titration equivalence point = end point
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See “Task” page 5 of notes ► An exercise in reading! ► Ask if unsure. ► Refer to expt already carried out – see if it agrees.
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pH Curves for various combinations
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Ka can be determined by the pH at the half neutralisation point. Expt
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The ideal indicator has the equivalence point in the middle of its range
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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
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See “indicators [1]” presentation ►
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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
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Buffer Solutions ► Definition ► A buffer solution is one which minimises the change in pH on addition of small amounts of acid or alkali
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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
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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
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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!
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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
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Notes q’s – answers p9 ► Q1 0.001 or 10 -3 mol dm -3 ► Q2 3.16 x 10 -5 mol dm -3 ► Q3 2.9g ► Q4 0.439g
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Hints for notes Q’s ► Q1 log 1/10 -3 = ► Q1 log 1/10 -4.25 = ► Q 3 ethanoic acid contains 7.5g dm -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 [CH 3 COO - ] in mol dm -3 ► ► X 82 gives mass per litre ► ► Divide by 2 gives mass in 500 cm 3
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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
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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
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