Chemistry • Year 12 • Module 6 • Lesson 9
pH of Weak Acids & Bases: Ka, Kb & ICE Tables
Apply ICE table methods to real data, interpret a pH vs concentration graph, and reason through cause-and-effect in weak acid and base systems.
1. Interpret a pH vs concentration graph for acetic acid
The graph below shows the calculated pH of acetic acid (CH3COOH, Ka = 1.8 × 10−5) solutions as concentration increases from 0.001 to 1.0 mol/L at 25°C. pH values were calculated using the exact ICE table method. The Australian Wine Research Institute (AWRI) notes that wine typically contains 6–8 g/L acetic acid (M = 60.05 g/mol), giving a concentration of approximately 0.10–0.13 mol/L. 10 marks
1.1 Describe the trend shown in the graph: how does pH change as concentration increases from 0.001 to 1.0 mol/L? 2 marks
1.2 Use the graph to estimate the pH of 0.010 mol/L CH3COOH. Show how you read the value. 1 mark
1.3 The graph is plotted on a logarithmic concentration axis. Use the formula pH ≈ ½(pKa − log[HA]) to explain why the graph is approximately linear on this scale. 2 marks
1.4 AWRI research indicates that wine in the range 0.10–0.13 mol/L acetic acid has pH ≈ 2.87–2.84. A winemaker claims that doubling the acetic acid concentration from 0.10 to 0.20 mol/L will halve the pH. Evaluate this claim. 2 marks
1.5 Predict what would happen to the pH of the wine solution if a small amount of sodium acetate (CH3COONa) were added. Justify using the equilibrium expression for Ka. 3 marks
2. Cause-and-effect chain, HF assumption failure
Hydrofluoric acid (HF, Ka = 6.8 × 10−4) is encountered in aluminium processing in Australia. A student applies the simplifying assumption to 0.020 mol/L HF and calculates [H+] = √(6.8 × 10−4 × 0.020) = 3.69 × 10−3 mol/L. Complete the cause-and-effect chain below by filling in the empty boxes. 5 marks
2.5 How many pH units does the error from using the approximation correspond to? (Calculate pH from both [H+] values.) 2 marks
3. Interpret weak base Kb data, ammonia and methylamine
The table below compares two industrially relevant weak bases. Ammonia (NH3) is central to Australian fertiliser manufacture; methylamine (CH3NH2) is used in pharmaceutical synthesis. 8 marks
| Base | Kb | c (mol/L) | Kb/c | Assumption valid? | [OH−] = √(Kb×c) | pOH | pH |
|---|---|---|---|---|---|---|---|
| NH3 | 1.8 × 10−5 | 0.100 | |||||
| NH3 | 1.8 × 10−5 | 0.010 | |||||
| CH3NH2 | 4.4 × 10−4 | 0.100 |
3.1 Complete all blank cells in the table above. Show full working for at least one row below. 6 marks
3.2 Compare the pH of 0.100 mol/L NH3 with 0.010 mol/L NH3. Explain why diluting the ammonia solution does not decrease its pH by a factor of 10. 2 marks
4. Predict and justify, HCN in the cyanide leach process
Newcrest Mining uses a cyanide leach process that involves dissolving cyanide salts at pH > 10.5 to keep cyanide as CN−(aq) rather than HCN(aq). HCN is a toxic weak acid with Ka = 6.2 × 10−10. A process engineer accidentally lowers the leach solution pH from 10.5 to 9.0 by over-adding sulfuric acid. 4 marks
4.1 Predict what happens to the proportion of cyanide present as HCN(aq) when pH drops from 10.5 to 9.0. Justify your prediction using the Ka expression for HCN: Ka = [H+][CN−] / [HCN]. 2 marks
4.2 At pH 9.0, [H+] = 1.0 × 10−9 mol/L. Use the Ka expression to calculate the ratio [HCN] : [CN−] at pH 9.0. Is HCN the dominant form? 2 marks
Q1, Graph interpretation
1.1 As concentration increases from 0.001 to 1.0 mol/L, pH decreases (the solution becomes more acidic). The relationship is approximately linear on the log-concentration axis, with each 10-fold increase in concentration lowering pH by approximately 0.5 units. 1 mark for “pH decreases as concentration increases”; 1 mark for quantitative description (approximately 0.5 pH units per decade or similar).
1.2 Reading the graph at x = 0.010 mol/L (second gridline), pH ≈ 3.4. Accept 3.3–3.5. 1 mark.
1.3 On a log scale, the x-axis plots log[HA]. The approximation pH ≈ ½(pKa − log[HA]) = ½pKa + ½log(1/[HA]) shows that pH is a linear function of log[HA] with slope −½. Since the x-axis is already log[HA], the plot of pH vs log[HA] is a straight line with gradient −0.5. 1 mark for identifying linear relationship with slope −½; 1 mark for connecting to the formula correctly.
1.4 The claim is incorrect. pH is not proportional to concentration; it is proportional to the logarithm of [H+], which is itself the square root of concentration (approximately). Doubling c from 0.10 to 0.20 mol/L gives pH ≈ ½(4.74 − log 0.20) = ½(4.74 + 0.699) = 2.72, compared to 2.87 at 0.10 mol/L, a decrease of only 0.15 pH units, not halving the pH. 1 mark for correctly rejecting the claim; 1 mark for quantitative or conceptual justification.
1.5 pH would increase (solution becomes less acidic / more neutral). Adding CH3COO− increases [A−] in the Ka expression Ka = [H+][A−] / [HA]. For Ka to remain constant (Le Chatelier), [H+] must decrease, so pH increases. This is the common ion effect, and the result is a buffer solution. 1 mark for “pH increases”; 1 mark for Ka / common ion reasoning; 1 mark for naming or describing the buffer effect.
Q2, HF assumption failure chain
Effect 1: α = (3.69 × 10−3 / 0.020) × 100% = 18.45% which is greater than 5%.
Effect 2: The simplifying assumption is invalid; the student should use the quadratic formula instead.
Effect 3: x = (−6.8 × 10−4 + √((6.8 × 10−4)2 + 4 × 6.8 × 10−4 × 0.020)) / 2. Discriminant = 4.624 × 10−7 + 5.44 × 10−5 = 5.486 × 10−5. √ = 7.407 × 10−3. x = (−6.8 × 10−4 + 7.407 × 10−3) / 2 = 6.727 × 10−3 / 2 = 3.36 × 10−3 mol/L.
2.5 Approximate pH = −log(3.69 × 10−3) = 2.43; Correct pH = −log(3.36 × 10−3) = 2.47. Difference = 0.04 pH units. 1 mark for each pH calculated correctly.
Q3, Weak base Kb data
Table answers:
- NH3, c = 0.100: Kb/c = 1.8 × 10−4 (valid); [OH−] = √(1.8 × 10−6) = 1.342 × 10−3; pOH = 2.87; pH = 11.13.
- NH3, c = 0.010: Kb/c = 1.8 × 10−3 (valid, just); [OH−] = √(1.8 × 10−7) = 4.243 × 10−4; pOH = 3.37; pH = 10.63.
- CH3NH2, c = 0.100: Kb/c = 4.4 × 10−3 > 0.0025 (invalid, note); using approximation: [OH−] = √(4.4 × 10−5) = 6.633 × 10−3; α = 6.63% > 5%; approximation invalid, but the question may accept approximation for this level. Exact quadratic: x = (−4.4 × 10−4 + √((4.4 × 10−4)2 + 4 × 4.4 × 10−4 × 0.100))/2 = 6.41 × 10−3; pOH = 2.19; pH = 11.81.
3.2 Diluting NH3 tenfold from 0.100 to 0.010 mol/L decreases pH from 11.13 to 10.63, a change of only 0.5 pH units, not 1.0. This is because NH3 is a weak base: as concentration decreases, a larger fraction ionises (α increases), partially offsetting the dilution. For a strong base, tenfold dilution would reduce [OH−] by exactly 10-fold (1.0 pH unit). The equilibrium shifts to maintain the Kb relationship, dampening the pH change. 1 mark for correct pH comparison; 1 mark for “weak base ionises more on dilution” / equilibrium shift explanation.
Q4, HCN in cyanide leach process
4.1 When pH drops from 10.5 to 9.0, [H+] increases (from 10−10.5 to 10−9). In Ka = [H+][CN−] / [HCN], an increase in [H+] with Ka fixed means [HCN] / [CN−] increases, the equilibrium shifts left, so more cyanide is present as the toxic HCN form rather than CN−. 1 mark for direction (more HCN); 1 mark for Ka reasoning.
4.2 Rearranging: [HCN] / [CN−] = [H+] / Ka = (1.0 × 10−9) / (6.2 × 10−10) = 1.61. So [HCN] : [CN−] ≈ 1.61 : 1. Yes, at pH 9.0, HCN is the dominant form (approximately 62% of total cyanide). This represents a serious safety hazard as HCN is volatile and toxic. 1 mark for ratio calculation; 1 mark for “HCN is dominant” conclusion.