A Level H2 Chemistry Practice Paper 3

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Chemistry H2
Level: A-Level
Paper: Practice Paper (Version 3 of 5)
Topic Focus: Acids, Bases and Salts
Duration: 1 hour 30 minutes
Total Marks: 60

Name: __________________________
Class: __________________________
Date: __________________________

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Instructions to Candidates

• Write your name, class, and date in the spaces provided.
• Answer all questions.
• The use of an approved scientific calculator is expected.
• A Data Booklet is provided for reference.
• You may lose marks if you do not show your working or if you do not use appropriate units.
• At the end of the examination, fasten all your work securely together.
• The number of marks is given in brackets [ ] at the end of each question or part question.

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Section A: Structured Questions

Answer all questions in this section.

1 Ethanoic acid, $CH_3COOH$, is a weak acid with a $K_a$ value of $1.74 \times 10^{-5} \text{ mol dm}^{-3}$ at 298 K.

(a) Define the term pH. [1]

(b) Calculate the pH of a $0.100 \text{ mol dm}^{-3}$ solution of ethanoic acid. State any assumptions made in your calculation. [3]

(c) A buffer solution is prepared by mixing $50.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ ethanoic acid with $50.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ sodium ethanoate. (i) Calculate the pH of this buffer solution. [2] (ii) Explain, with the aid of an equation, how this buffer solution resists changes in pH when a small amount of dilute hydrochloric acid is added. [2]

2 Propanoic acid ($C_2H_5COOH$) reacts with methanol ($CH_3OH$) in the presence of an acid catalyst to form an ester.

(a) Write the equation for this reaction, using structural formulae. [2]

(b) The equilibrium constant, $K_c$, for this reaction is 4.0 at a specific temperature. If 1.0 mol of propanoic acid and 1.0 mol of methanol are mixed and allowed to reach equilibrium, calculate the amount (in mol) of the ester present at equilibrium. [4]

3 The solubility product, $K_{sp}$, of magnesium hydroxide, $Mg(OH)_2$, is $1.8 \times 10^{-11} \text{ mol}^3 \text{ dm}^{-9}$ at 298 K.

(a) Write the expression for the solubility product of $Mg(OH)_2$. [1]

(b) Calculate the solubility of $Mg(OH)_2$ in pure water in $\text{mol dm}^{-3}$. [2]

(c) Explain why $Mg(OH)_2$ is more soluble in a solution of dilute hydrochloric acid than in pure water. [2]

4 An unknown monoprotic weak acid, HA, has a concentration of $0.050 \text{ mol dm}^{-3}$. The pH of this solution is measured to be 2.88 at 298 K.

(a) Calculate the concentration of hydrogen ions, $[H^+]$, in the solution. [1]

(b) Calculate the acid dissociation constant, $K_a$, for HA. [3]

(c) Suggest a suitable indicator for the titration of this weak acid with a strong base, sodium hydroxide. Explain your choice by referring to the pH at the equivalence point. [2]

5 Consider the following species: $NH_3$, $NH_4^+$, $H_2O$, and $OH^-$.

(a) Identify the conjugate acid-base pairs from the reaction below: $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$ [2]

(b) Ammonia acts as a base in the reaction above. Define a Brønsted-Lowry base. [1]

(c) Explain why ammonia is a weaker base than methylamine, $CH_3NH_2$. [2]

6 A student performs a titration of $25.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ sodium hydroxide with $0.100 \text{ mol dm}^{-3}$ hydrochloric acid.

(a) Sketch the titration curve for this strong acid-strong base titration. Label the axes and indicate the approximate pH at the equivalence point. [3]

(b) Why is the pH at the equivalence point exactly 7.0 at 298 K? [1]

(c) If the student accidentally adds $0.10 \text{ cm}^3$ excess acid after the equivalence point, estimate the new pH of the solution. (Assume the total volume is approximately $50.0 \text{ cm}^3$). [3]

7 Aluminum chloride, $AlCl_3$, is dissolved in water.

(a) The resulting solution is acidic. Write an equation to explain this observation, showing the formation of the hexaaquaaluminum(III) ion and its subsequent reaction with water. [2]

(b) What would be observed if sodium carbonate solution is added to the aqueous aluminum chloride solution? Explain the observations chemically. [3]

8 The ionic product of water, $K_w$, is $1.0 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$ at 298 K.

(a) Define $K_w$. [1]

(b) The dissociation of water is an endothermic process. Predict and explain how the pH of pure water changes as the temperature increases from 298 K to 350 K. [3]

9 Benzoic acid ($C_6H_5COOH$) is a weak acid used as a food preservative.

(a) Write the expression for $K_a$ of benzoic acid. [1]

(b) A solution contains $0.020 \text{ mol dm}^{-3}$ benzoic acid and $0.040 \text{ mol dm}^{-3}$ sodium benzoate. Given $K_a = 6.3 \times 10^{-5} \text{ mol dm}^{-3}$, calculate the pH of the solution. [2]

(c) Explain why benzoic acid is more effective as a preservative in acidic foods (pH < 4) than in neutral foods. [2]

10 Sulfuric acid, $H_2SO_4$, is a strong diprotic acid.

(a) Write the two dissociation steps for sulfuric acid in water. [2]

(b) Explain why the first dissociation constant ($K_{a1}$) is much larger than the second dissociation constant ($K_{a2}$). [2]

(c) Calculate the pH of $0.010 \text{ mol dm}^{-3}$ $H_2SO_4$, assuming complete dissociation of both protons. [2]

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Section B: Data-Based and Application Questions

Answer all questions in this section.

11 The table below shows the pH values of $0.10 \text{ mol dm}^{-3}$ solutions of four different acids at 298 K.

Acid	Formula	pH
A	$HCl$	1.0
B	$CH_3COOH$	2.9
C	$HCOOH$	2.4
D	$ClCH_2COOH$	1.9

(a) Arrange the acids in order of increasing strength. [1]

(b) Explain the difference in acidity between Acid B (ethanoic acid) and Acid D (chloroethanoic acid) in terms of their molecular structure. [3]

(c) Calculate the $K_a$ value for Acid C (methanoic acid). [3]

12 Calcium hydroxide, $Ca(OH)_2$, is slightly soluble in water. A saturated solution of calcium hydroxide is known as limewater.

(a) Write the equilibrium equation for the dissolution of calcium hydroxide. [1]

(b) In an experiment, $25.0 \text{ cm}^3$ of saturated limewater was titrated against $0.050 \text{ mol dm}^{-3}$ hydrochloric acid. The average titre was $12.5 \text{ cm}^3$. (i) Calculate the concentration of hydroxide ions, $[OH^-]$, in the limewater. [2] (ii) Calculate the solubility product, $K_{sp}$, of calcium hydroxide. [3]

(c) If solid calcium chloride is added to the saturated limewater, what effect will this have on the solubility of calcium hydroxide? Explain your answer using Le Chatelier’s principle. [2]

13 Amino acids contain both an amino group ($-NH_2$) and a carboxyl group ($-COOH$). Glycine ($H_2NCH_2COOH$) is the simplest amino acid.

(a) Draw the structure of glycine in its zwitterionic form. [1]

(b) Explain what is meant by the term isoelectric point. [1]

(c) Glycine has two $pK_a$ values: $pK_{a1} = 2.34$ (for $-COOH$) and $pK_{a2} = 9.60$ (for $-NH_3^+$). (i) Calculate the isoelectric point of glycine. [1] (ii) Sketch the titration curve for the addition of NaOH to a solution of glycine hydrochloride ($^+H_3NCH_2COOH$). Label the regions corresponding to the two buffering actions. [3]

14 Indicators are weak acids or bases that change color depending on the pH of the solution. Let HIn represent the acidic form and In⁻ represent the basic form of an indicator.

(a) Derive the relationship between pH, $pK_{In}$, and the ratio $\frac{[In^-]}{[HIn]}$. [2]

(b) Methyl orange has a $pK_{In}$ of 3.7. Its acidic form is red and its basic form is yellow. (i) What is the color of methyl orange at pH 2.0? [1] (ii) Explain why methyl orange is unsuitable for the titration of ethanoic acid with sodium hydroxide. [2]

15 The diagram below represents the distribution of species for carbonic acid ($H_2CO_3$) in aqueous solution as a function of pH. (Note: Imagine a graph where $H_2CO_3$ dominates at low pH, $HCO_3^-$ peaks around pH 6-7, and $CO_3^{2-}$ dominates at high pH. $pK_{a1} \approx 6.4$, $pK_{a2} \approx 10.3$.)

(a) Identify the predominant species at pH 4.0, pH 8.0, and pH 12.0. [3]

(b) At what pH is $[H_2CO_3] = [HCO_3^-]$? [1]

(c) Explain how the $H_2CO_3 / HCO_3^-$ system acts as a buffer in human blood. [2]

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Section C: Long Structured Questions

Answer all questions in this section.

16 This question concerns the preparation and analysis of a buffer solution.

(a) Describe how you would prepare $1.0 \text{ dm}^3$ of a buffer solution with pH 5.0 using $0.10 \text{ mol dm}^{-3}$ ethanoic acid ($K_a = 1.74 \times 10^{-5}$) and solid sodium ethanoate ($M_r = 82.0$). Show all calculations. [5]

(b) To $100 \text{ cm}^3$ of this buffer, $10.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ NaOH is added. Calculate the new pH of the solution. [4]

(c) Compare the change in pH calculated in (b) with the change in pH if the same amount of NaOH were added to $100 \text{ cm}^3$ of pure water (initially pH 7.0). Comment on the efficiency of the buffer. [3]

17 Solubility equilibria are important in qualitative analysis.

(a) Explain the concept of fractional precipitation. [2]

(b) A solution contains $0.010 \text{ mol dm}^{-3}$ $Cl^-$ and $0.010 \text{ mol dm}^{-3}$ $CrO_4^{2-}$ ions. Silver nitrate solution is added dropwise. Given: $K_{sp}(AgCl) = 1.8 \times 10^{-10} \text{ mol}^2 \text{ dm}^{-6}$ $K_{sp}(Ag_2CrO_4) = 1.1 \times 10^{-12} \text{ mol}^3 \text{ dm}^{-9}$

(i) Calculate the concentration of $Ag^+$ required to initiate precipitation of AgCl. [2]
(ii) Calculate the concentration of $Ag^+$ required to initiate precipitation of $Ag_2CrO_4$. [2]
(iii) Which salt precipitates first? Explain your answer. [2]

(c) In the Mohr method for determining chloride concentration, potassium chromate is used as an indicator. Explain the chemical basis for the endpoint detection in this titration. [3]

18 The strength of organic acids is influenced by substituent effects.

(a) Compare the acidity of ethanoic acid ($CH_3COOH$), chloroethanoic acid ($ClCH_2COOH$), and dichloroethanoic acid ($Cl_2CHCOOH$). Explain the trend. [4]

(b) Phenol ($C_6H_5OH$) is a weaker acid than ethanoic acid but stronger than ethanol ($C_2H_5OH$). (i) Explain why phenol is more acidic than ethanol. [3] (ii) Explain why phenol is less acidic than ethanoic acid. [3]

19 Hydrolysis of salts affects the pH of aqueous solutions.

(a) Predict whether the aqueous solutions of the following salts are acidic, alkaline, or neutral. Explain your reasoning for each. (i) Sodium ethanoate ($CH_3COONa$) [2] (ii) Ammonium chloride ($NH_4Cl$) [2] (iii) Sodium chloride ($NaCl$) [1]

(b) Calculate the pH of a $0.10 \text{ mol dm}^{-3}$ solution of ammonium chloride. ($K_b$ for $NH_3 = 1.8 \times 10^{-5} \text{ mol dm}^{-3}$). [4]

20 Practical skills in acid-base chemistry.

(a) Describe the procedure for calibrating a pH meter before use. [2]

(b) A student titrates a weak acid with a strong base using a pH meter. The resulting curve shows a gradual increase in pH, followed by a steep vertical section, and then a leveling off. (i) How can the $pK_a$ of the weak acid be determined from this curve? [2] (ii) Why is a pH meter preferred over a visual indicator for the titration of a very weak acid ($pK_a > 9$) with a strong base? [2]

(c) Suggest two sources of error in a titration experiment and how they can be minimized. [4]

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End of Paper

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

Answer Key and Marking Scheme (Version 3)

Topic: Acids, Bases and Salts
Total Marks: 60

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Section A: Structured Questions

1 (a) $pH = -\log_{10}[H^+]$ [1] (b) Assumption: $[H^+] = [A^-]$ and $[HA]_{eq} \approx [HA]_{initial}$ (since $K_a$ is small). [1] $K_a = \frac{[H^+]^2}{[HA]}$ $[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.74 \times 10^{-5} \times 0.100}$ [1] $[H^+] = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$ $pH = -\log(1.32 \times 10^{-3}) = 2.88$ [1] (c) (i) Since volumes and concentrations are equal, $[acid] = [salt]$. $pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right) = pK_a + \log(1) = pK_a$ [1] $pK_a = -\log(1.74 \times 10^{-5}) = 4.76$ [1] (ii) $CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$ [1] The added $H^+$ ions react with the conjugate base ($CH_3COO^-$) to form weak acid, minimizing the change in $[H^+]$. [1]

2 (a) $C_2H_5COOH + CH_3OH \rightleftharpoons C_2H_5COOCH_3 + H_2O$ [1 for reactants/products, 1 for equilibrium sign/conditions] (b) Let $x$ be the moles of ester formed. Initial: Acid=1.0, Alcohol=1.0, Ester=0, Water=0 Eq: Acid=$1.0-x$, Alcohol=$1.0-x$, Ester=$x$, Water=$x$ $K_c = \frac{[Ester][Water]}{[Acid][Alcohol]} = \frac{(x/V)(x/V)}{((1-x)/V)((1-x)/V)} = \frac{x^2}{(1-x)^2}$ [1] $\sqrt{4.0} = \frac{x}{1-x} \Rightarrow 2 = \frac{x}{1-x}$ [1] $2(1-x) = x \Rightarrow 2 - 2x = x \Rightarrow 3x = 2$ [1] $x = 0.67 \text{ mol}$ [1]

3 (a) $K_{sp} = [Mg^{2+}][OH^-]^2$ [1] (b) Let $s$ be solubility in $\text{mol dm}^{-3}$. $[Mg^{2+}] = s$, $[OH^-] = 2s$ $K_{sp} = (s)(2s)^2 = 4s^3$ [1] $1.8 \times 10^{-11} = 4s^3$ $s^3 = 4.5 \times 10^{-12}$ $s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$ [1] (c) $H^+$ ions from HCl react with $OH^-$ ions to form water: $H^+ + OH^- \rightarrow H_2O$. [1] This decreases $[OH^-]$, shifting the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ to the right (Le Chatelier), increasing solubility. [1]

4 (a) $[H^+] = 10^{-pH} = 10^{-2.88} = 1.32 \times 10^{-3} \text{ mol dm}^{-3}$ [1] (b) $K_a = \frac{[H^+][A^-]}{[HA]}$ Assume $[H^+] = [A^-]$ and $[HA]_{eq} \approx 0.050$ $K_a = \frac{(1.32 \times 10^{-3})^2}{0.050}$ [2] $K_a = 3.48 \times 10^{-5} \text{ mol dm}^{-3}$ [1] (c) Phenolphthalein. [1] The equivalence point for weak acid-strong base titration is alkaline (pH 8-9). Phenolphthalein changes color in the range 8.3-10.0, which falls within the vertical section of the titration curve. [1]

5 (a) Pair 1: $NH_3$ (base) / $NH_4^+$ (acid) [1] Pair 2: $H_2O$ (acid) / $OH^-$ (base) [1] (b) A proton ($H^+$) acceptor. [1] (c) The methyl group in methylamine is electron-releasing (positive inductive effect, +I). [1] This increases the electron density on the nitrogen atom, making the lone pair more available for donation to a proton, thus making it a stronger base than ammonia. [1]

6 (a)

• Y-axis: pH (0-14), X-axis: Volume of acid added. [1]
• Start pH ~13, End pH ~1. [1]
• Vertical section centered at pH 7. [1] (b) Salt formed is NaCl, which is neutral. Neither $Na^+$ nor $Cl^-$ hydrolyzes. [1] $[H^+] = [OH^-] = 10^{-7} \text{ mol dm}^{-3}$. (c) Moles excess $H^+ = 0.10 \text{ cm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.01 \text{ mmol} = 1.0 \times 10^{-5} \text{ mol}$. [1] Total volume $\approx 50.0 \text{ cm}^3 = 0.050 \text{ dm}^3$. $[H^+] = \frac{1.0 \times 10^{-5}}{0.050} = 2.0 \times 10^{-4} \text{ mol dm}^{-3}$. [1] $pH = -\log(2.0 \times 10^{-4}) = 3.70$. [1]

7 (a) $Al^{3+}(aq) + 6H_2O(l) \rightarrow [Al(H_2O)_6]^{3+}(aq)$ [1] $[Al(H_2O)_6]^{3+}(aq) + H_2O(l) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H_3O^+(aq)$ [1] (b) Effervescence (bubbles of gas) and a white precipitate. [1] $Al^{3+}$ is acidic; $CO_3^{2-}$ is basic. They undergo mutual hydrolysis. $2Al^{3+} + 3CO_3^{2-} + 3H_2O \rightarrow 2Al(OH)_3(s) + 3CO_2(g)$. [2]

8 (a) $K_w = [H^+][OH^-]$ [1] (b) Since dissociation is endothermic, increasing T shifts equilibrium to the right. [1] $[H^+]$ and $[OH^-]$ both increase. [1] Since $pH = -\log[H^+]$, pH decreases (becomes < 7). Note: Water remains neutral as $[H^+]=[OH^-]$. [1]

9 (a) $K_a = \frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}$ [1] (b) $pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right)$ $pK_a = -\log(6.3 \times 10^{-5}) = 4.20$ $pH = 4.20 + \log\left(\frac{0.040}{0.020}\right) = 4.20 + \log(2) = 4.20 + 0.30 = 4.50$ [2] (c) The active preservative species is the undissociated benzoic acid molecule, which can penetrate bacterial cell membranes. [1] At low pH, equilibrium shifts towards the undissociated acid ($C_6H_5COOH$). At neutral pH, it exists mainly as the benzoate ion, which cannot penetrate cells effectively. [1]

10 (a)

1. $H_2SO_4 \rightarrow H^+ + HSO_4^-$ [1]
2. $HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$ [1] (b) Removing a proton from a neutral molecule ($H_2SO_4$) is easier than removing a positive proton from a negatively charged ion ($HSO_4^-$) due to electrostatic attraction. [2] (c) $[H^+]$ from 1st dissociation = 0.010 M. $[H^+]$ from 2nd dissociation $\approx$ 0.010 M (assuming strong/complete for simplicity in this context, though technically $K_{a2}$ is weak, usually A-Level questions specify "assume complete" or give $K_{a2}$. If complete: Total $[H^+] = 0.020$ M). $pH = -\log(0.020) = 1.70$. [2] (Note: If treating 2nd step as weak, calculation is more complex, but "assume complete" is standard for this mark allocation unless $K_{a2}$ is provided.)

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Section B: Data-Based and Application Questions

11 (a) A < D < C < B (Increasing strength means lower pH for same conc, so order of strength: B < C < D < A. Question asks increasing strength: B, C, D, A). [1] Correction: Lowest pH is strongest. A(1.0) > D(1.9) > C(2.4) > B(2.9). Order of increasing strength: B, C, D, A. (b) Chlorine is electronegative and exerts a negative inductive effect (-I). [1] This withdraws electron density from the carboxyl group, weakening the O-H bond and stabilizing the conjugate base ($ClCH_2COO^-$) by dispersing the negative charge. [2] (c) $[H^+] = 10^{-2.4} = 3.98 \times 10^{-3} \text{ mol dm}^{-3}$. $K_a = \frac{[H^+]^2}{[HA]} = \frac{(3.98 \times 10^{-3})^2}{0.10}$ [2] $K_a = 1.58 \times 10^{-4} \text{ mol dm}^{-3}$. [1]

12 (a) $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$ [1] (b) (i) Moles $H^+ = 0.050 \times \frac{12.5}{1000} = 6.25 \times 10^{-4} \text{ mol}$. Ratio $H^+ : OH^-$ is 1:1. Moles $OH^- = 6.25 \times 10^{-4} \text{ mol}$. $[OH^-] = \frac{6.25 \times 10^{-4}}{0.025} = 0.025 \text{ mol dm}^{-3}$. [2] (ii) $[Ca^{2+}] = \frac{1}{2}[OH^-] = 0.0125 \text{ mol dm}^{-3}$. $K_{sp} = [Ca^{2+}][OH^-]^2 = (0.0125)(0.025)^2$ [2] $K_{sp} = 7.81 \times 10^{-6} \text{ mol}^3 \text{ dm}^{-9}$. [1] (c) Solubility decreases. [1] Adding $CaCl_2$ increases $[Ca^{2+}]$. By Le Chatelier’s principle, the equilibrium shifts to the left to remove excess $Ca^{2+}$, causing precipitation of $Ca(OH)_2$. [1]

13 (a) $^+H_3N-CH_2-COO^-$ [1] (b) The pH at which the amino acid exists primarily as a zwitterion and has no net electrical charge. [1] (c) (i) $pI = \frac{pK_{a1} + pK_{a2}}{2} = \frac{2.34 + 9.60}{2} = 5.97$. [1] (ii) Curve starts at low pH (~1). Two buffer regions (flat parts) centered at pH 2.34 and 9.60. Two equivalence points (vertical sections). [3]

14 (a) $K_{In} = \frac{[H^+][In^-]}{[HIn]}$ $\frac{[In^-]}{[HIn]} = \frac{K_{In}}{[H^+]}$ $\log\left(\frac{[In^-]}{[HIn]}\right) = \log K_{In} - \log[H^+] = -pK_{In} + pH$ $pH = pK_{In} + \log\left(\frac{[In^-]}{[HIn]}\right)$ [2] (b) (i) Red. (pH < pKa, acid form dominates). [1] (ii) The equivalence point for weak acid-strong base is ~pH 8-9. Methyl orange changes color at pH 3.1-4.4. The color change would occur long before the equivalence point, leading to a large titration error. [2]

15 (a) pH 4.0: $H_2CO_3$ [1] pH 8.0: $HCO_3^-$ [1] pH 12.0: $CO_3^{2-}$ [1] (b) pH = $pK_{a1} = 6.4$. [1] (c) $H^+(aq) + HCO_3^-(aq) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons H_2O(l) + CO_2(g)$. Added acid is removed by $HCO_3^-$. Added base is removed by $H_2CO_3$. This maintains blood pH around 7.4. [2]

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Section C: Long Structured Questions

16 (a) $pH = pK_a + \log\left(\frac{[salt]}{[acid]}\right)$ $5.0 = 4.76 + \log\left(\frac{[salt]}{0.10}\right)$ $0.24 = \log\left(\frac{[salt]}{0.10}\right)$ $\frac{[salt]}{0.10} = 10^{0.24} = 1.74$ $[salt] = 0.174 \text{ mol dm}^{-3}$. [2] For $1.0 \text{ dm}^3$, moles salt = 0.174 mol. Mass = $0.174 \times 82.0 = 14.27 \text{ g}$. [2] Procedure: Dissolve 14.3 g of sodium ethanoate in some $0.10 \text{ M}$ ethanoic acid, then make up to $1.0 \text{ dm}^3$ with the same acid. [1] (b) Initial moles: Acid = $0.10 \times 0.1 = 0.01$ mol. Salt = $0.174 \times 0.1 = 0.0174$ mol. Moles NaOH added = $0.010 \times 0.010 = 0.0001$ mol. Reaction: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$. New moles Acid = $0.01 - 0.0001 = 0.0099$ mol. New moles Salt = $0.0174 + 0.0001 = 0.0175$ mol. $pH = 4.76 + \log\left(\frac{0.0175}{0.0099}\right) = 4.76 + 0.25 = 5.01$. [4] (c) In water: $[OH^-] = \frac{0.0001}{0.11} \approx 9 \times 10^{-4}$. pOH ~3. pH ~11. Change from 7 to 11 is 4 units. [2] Buffer changed by only 0.01 pH units. Buffer is highly effective. [1]

17 (a) Precipitation of ions from a solution by careful addition of a precipitating agent, where salts precipitate in order of their solubility products. [2] (b) (i) $[Ag^+] = \frac{K_{sp}(AgCl)}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.010} = 1.8 \times 10^{-8} \text{ mol dm}^{-3}$. [2] (ii) $[Ag^+]^2 = \frac{K_{sp}(Ag_2CrO_4)}{[CrO_4^{2-}]} = \frac{1.1 \times 10^{-12}}{0.010} = 1.1 \times 10^{-10}$. $[Ag^+] = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \text{ mol dm}^{-3}$. [2] (iii) AgCl precipitates first because it requires a lower concentration of $Ag^+$ ($1.8 \times 10^{-8}$ vs $1.05 \times 10^{-5}$). [2] (c) $Ag^+$ reacts with $Cl^-$ first to form white AgCl. [1] Once all $Cl^-$ is consumed, excess $Ag^+$ reacts with $CrO_4^{2-}$ to form brick-red $Ag_2CrO_4$ precipitate. [1] The appearance of the red color indicates the endpoint. [1]

18 (a) Order of acidity: Ethanoic < Chloroethanoic < Dichloroethanoic. [1] Cl is electronegative (-I effect). [1] More Cl atoms withdraw more electron density, stabilizing the carboxylate anion more effectively. [1] This makes the O-H bond more polar and easier to break, increasing $K_a$. [1] (b) (i) Phenoxide ion ($C_6H_5O^-$) is stabilized by resonance delocalization of the negative charge into the benzene ring. Ethoxide ion has no such stabilization. [3] (ii) Ethanoate ion ($CH_3COO^-$) has resonance delocalization over two electronegative oxygen atoms, which is more effective than delocalization into the carbon ring of phenoxide. Also, O-H bond in carboxylic acids is more polar. [3]

19 (a) (i) Alkaline. $CH_3COO^-$ hydrolyzes: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$. [2] (ii) Acidic. $NH_4^+$ hydrolyzes: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$. [2] (iii) Neutral. Derived from strong acid and strong base; no hydrolysis. [1] (b) $K_a(NH_4^+) = \frac{K_w}{K_b(NH_3)} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$. [1] $[H^+] = \sqrt{K_a \times [Salt]} = \sqrt{5.56 \times 10^{-10} \times 0.10}$ [2] $[H^+] = 7.46 \times 10^{-6}$. $pH = -\log(7.46 \times 10^{-6}) = 5.13$. [1]

20 (a) Rinse electrode with distilled water. [1] Immerse in buffer solutions of known pH (e.g., 4.0 and 7.0) and adjust calibration settings. [1] (b) (i) $pK_a$ is the pH at the half-equivalence point (where volume of base added is half that required for equivalence). [2] (ii) The pH change at the equivalence point for very weak acids is gradual, not vertical. Visual indicators do not show a sharp color change, making detection difficult. A pH meter detects the inflection point accurately. [2] (c)

1. Parallax error in reading burette. Minimize by reading at eye level. [2]
2. Air bubbles in burette jet. Minimize by flushing jet before starting. [2] (Other valid errors: Wet conical flask, incorrect indicator choice, etc.)