Structured Questions
Answer the questions below in the spaces provided. Check your answers with the mark schemes and examiner tips.
Question 1: Standard pH Calculations
9 marksCalculate the pH of each of the following solutions at 298 K, giving all answers to 2 decimal places. (\(K_w = 1.00 \times 10^{-14}\text{ mol}^2\text{ dm}^{-6}\) at 298 K).
(a) Calculate the pH of a 0.035 mol dm⁻³ solution of nitric acid (HNO₃). [2]
(b) Calculate the pH of a 0.025 mol dm⁻³ solution of potassium hydroxide (KOH). [3]
(c) Propanoic acid (CH₃CH₂COOH) is a weak acid with \(K_a = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\) at 298 K. Calculate the pH of a 0.120 mol dm⁻³ solution of propanoic acid, stating any assumptions made. [4]
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(a)
- HNO₃ is a strong acid, so [H⁺] = 0.035 mol dm⁻³ [1]
- \(\text{pH} = -\log_{10}(0.035) = \mathbf{1.46}\) [1]
(b)
- KOH is a strong base, so [OH⁻] = 0.025 mol dm⁻³ [1]
- \([\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.00 \times 10^{-14}}{0.025} = 4.00 \times 10^{-13}\text{ mol dm}^{-3}\) [1]
- \(\text{pH} = -\log_{10}(4.00 \times 10^{-13}) = \mathbf{12.40}\) [1]
(c)
- \([\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.35 \times 10^{-5} \times 0.120}\) [1]
- \([\text{H}^+] = \sqrt{1.62 \times 10^{-6}} = 1.27 \times 10^{-3}\text{ mol dm}^{-3}\) [1]
- \(\text{pH} = -\log_{10}(1.27 \times 10^{-3}) = \mathbf{2.90}\) (accept 2.89) [1]
- Assumptions (any one from):
• Assume [H⁺] = [A⁻] at equilibrium (i.e. negligible H⁺ contribution from the autoionisation of water)
• Assume [HA] at equilibrium equals the initial [HA] (i.e. dissociation of the weak acid is negligible) [1]
Question 2: pH Curves & Indicators
6 marksA titration is carried out by adding 0.100 mol dm⁻³ sodium hydroxide solution to 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid solution.
(a) Outline the key characteristics of the pH titration curve for this reaction. Specify the initial pH, the equivalence point pH, and the pH range of the vertical section. [3]
(b) Explain how you would select a suitable indicator for this titration using the pH curve. [3]
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(a)
- Initial pH should start in the weak acid range (typically pH 2.5 to 3.0) [1]
- The equivalence point pH is basic / greater than 7 (typically pH 8.5 to 9.0) because the salt formed (sodium ethanoate) hydrolyses slightly [1]
- The vertical section of the pH curve lies entirely in the basic region, typically spanning from pH 7.0 to 10.0 [1]
(b)
- The chosen indicator must change colour completely within the steep / vertical section of the pH titration curve [1]
- The pH range of the indicator's transition / colour change must fall entirely within this vertical section [1]
- Phenolphthalein (pH range 8.3 to 10.0) is suitable because its transition range matches the vertical section of a weak acid-strong base titration curve [1]
Question 3: Buffer Solution Calculations & Action
8 marksA buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid (CH₃COOH) with 50.0 cm³ of 0.150 mol dm⁻³ sodium ethanoate (CH₃COONa). For ethanoic acid, \(K_a = 1.76 \times 10^{-5}\text{ mol dm}^{-3}\) at 298 K.
(a) Calculate the pH of this buffer solution at 298 K. [4]
(b) Explain, with the help of a chemical equation, how this buffer solution minimises pH change when a small volume of hydrochloric acid is added. [4]
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(a)
- Moles of \(\text{CH}_3\text{COOH} = 0.200 \times \frac{50.0}{1000} = \mathbf{0.0100\text{ mol}}\) [1]
- Moles of CH₃COO⁻ (from salt) \(= 0.150 \times \frac{50.0}{1000} = \mathbf{0.00750\text{ mol}}\) [1]
- \([\text{H}^+] = K_a \times \frac{\text{Moles of HA}}{\text{Moles of A}^-} = 1.76 \times 10^{-5} \times \frac{0.0100}{0.00750} = 2.35 \times 10^{-5}\text{ mol dm}^{-3}\) [1]
- \(\text{pH} = -\log_{10}(2.35 \times 10^{-5}) = \mathbf{4.63}\) [1]
(b)
- Hydrochloric acid dissociates to release H⁺ ions into the solution [1]
- The conjugate base (ethanoate ions, CH₃COO⁻) in the buffer reacts with the added H⁺ ions to form ethanoic acid:
CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq) [2] (1 mark for correct equation; 1 mark for state symbols/explanation) - Because the added H⁺ ions are converted into the weak acid, the concentration of hydrogen ions remains almost constant, minimising the change in pH [1]