A Level H2 Chemistry Acids Bases Salts Quiz

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A-Level Chemistry H2 Quiz - Acids Bases Salts

Name: __________________________
Class: __________________________
Date: __________________________
Score: _________ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

1. Answer all questions.
2. Write your answers in the spaces provided.
3. The use of a scientific calculator is allowed.
4. A Data Booklet is provided for reference.
5. Show all working for calculation questions.

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Section A: Multiple Choice & Basic Concepts (10 Marks)

1. Which of the following statements correctly describes the behavior of a buffer solution composed of ethanoic acid ($CH_3COOH$) and sodium ethanoate ($CH_3COONa$)?
A. The pH remains exactly 7.00 regardless of added acid or base.
B. It resists changes in pH by neutralizing added $H^+$ with $CH_3COO^-$ and added $OH^-$ with $CH_3COOH$.
C. It contains equal concentrations of $H^+$ and $OH^-$ ions.
D. The $pK_a$ of the acid changes when small amounts of strong acid are added.
[1]

2. Calculate the pH of a $0.050 \text{ mol dm}^{-3}$ solution of barium hydroxide, $Ba(OH)_2$, assuming complete dissociation.
A. 1.00
B. 1.30
C. 12.70
D. 13.00
[1]

3. Which indicator is most suitable for the titration of $25.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ ammonia ($NH_3$) with $0.10 \text{ mol dm}^{-3}$ hydrochloric acid ($HCl$)?
A. Phenolphthalein (pH range 8.3 – 10.0)
B. Thymolphthalein (pH range 9.3 – 10.5)
C. Methyl orange (pH range 3.1 – 4.4)
D. Bromothymol blue (pH range 6.0 – 7.6)
[1]

4. Explain why an aqueous solution of aluminum chloride, $AlCl_3$, is acidic.

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[2]

5. The $pK_a$ of methanoic acid ($HCOOH$) is 3.75. Calculate the $K_a$ value.
[1]

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Section B: Salt Hydrolysis & Buffer Calculations (10 Marks)

6. A solution is prepared by dissolving $0.82 \text{ g}$ of sodium ethanoate ($CH_3COONa$, $M_r = 82$) in water to make $500 \text{ cm}^3$ of solution. Calculate the concentration of the sodium ethanoate solution in $\text{mol dm}^{-3}$.
[2]

7. Given that the $pK_a$ of ethanoic acid is 4.76 and $K_w = 1.0 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$, calculate the pH of the sodium ethanoate solution prepared in Question 6.
[4]

8. A buffer solution contains $0.10 \text{ mol dm}^{-3}$ propanoic acid ($C_2H_5COOH$) and $0.20 \text{ mol dm}^{-3}$ sodium propanoate ($C_2H_5COONa$). The $K_a$ of propanoic acid is $1.3 \times 10^{-5} \text{ mol dm}^{-3}$. Calculate the initial pH of this buffer solution.
[2]

9. Calculate the new pH of the buffer solution described in Question 8 after adding $1.0 \text{ cm}^3$ of $1.0 \text{ mol dm}^{-3}$ $HCl$ to $100 \text{ cm}^3$ of the buffer. Assume the volume change is negligible.
[4]

10. Magnesium hydroxide, $Mg(OH)_2$, is sparingly soluble in water. Write the expression for the solubility product constant, $K_{sp}$, of $Mg(OH)_2$.
[1]

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Section C: Solubility & Titration Curves (10 Marks)

11. The $K_{sp}$ of $Mg(OH)_2$ is $1.8 \times 10^{-11} \text{ mol}^3 \text{ dm}^{-9}$ at $25^\circ C$. Calculate the solubility of $Mg(OH)_2$ in $\text{mol dm}^{-3}$.
[3]

12. Explain whether $Mg(OH)_2$ is more or less soluble in a solution of pH 2 compared to pure water.
[2]

13. A student titrates $25.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ ethanoic acid ($CH_3COOH$) with $0.100 \text{ mol dm}^{-3}$ sodium hydroxide ($NaOH$). Sketch the titration curve for this reaction. Label the equivalence point and the region where the solution acts as a buffer.
[3]

14. Explain why the pH at the equivalence point of the titration in Question 13 is greater than 7.

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[2]

15. Calculate the pH of the solution in Question 13 at the half-equivalence point. ($pK_a$ of $CH_3COOH = 4.76$).
[2]

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Section D: Advanced Titration Analysis (10 Marks)

16. In a separate experiment, $25.0 \text{ cm}^3$ of a weak monoprotic acid $HA$ is titrated with $0.100 \text{ mol dm}^{-3}$ $NaOH$. At the addition of $12.5 \text{ cm}^3$ of $NaOH$, the pH is 4.80. Determine the $pK_a$ of the acid $HA$.
[1]

17. For the titration in Question 16, the equivalence point is reached when $25.0 \text{ cm}^3$ of $NaOH$ has been added. Determine the initial concentration of the acid $HA$.
[2]

18. Suggest why the pH change near the equivalence point is less sharp for the weak acid-strong base titration in Question 16 compared to a strong acid-strong base titration.
[2]

19. Define the term "buffer solution" and state the two essential components required to form one.
[2]

20. Calculate the ratio of $[A^-]/[HA]$ required to prepare a buffer solution with a pH of 5.00, given that the $pK_a$ of the weak acid $HA$ is 4.76.
[3]

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End of Quiz

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A-Level Chemistry H2 Quiz - Acids Bases Salts (Answer Key)

1. B
Explanation: A buffer resists pH change. Added $H^+$ reacts with the conjugate base ($CH_3COO^-$) to form weak acid. Added $OH^-$ reacts with the weak acid ($CH_3COOH$) to form conjugate base and water.

2. D
Calculation:
$[OH^-] = 2 \times [Ba(OH)_2] = 2 \times 0.050 = 0.10 \text{ mol dm}^{-3}$.
$pOH = -\log(0.10) = 1.00$.
$pH = 14.00 - 1.00 = 13.00$.

3. C
Explanation: This is a Weak Base ($NH_3$) + Strong Acid ($HCl$) titration. The equivalence point is acidic (pH < 7, typically around 5). Methyl orange (range 3.1–4.4) is suitable. Phenolphthalein changes color in the basic region, which is before the equivalence point in this titration.

4.
$Al^{3+}$ is a small, highly charged cation. It polarizes the O-H bonds in the water molecules of its hydration shell $[Al(H_2O)_6]^{3+}$. This weakens the O-H bond, allowing a proton ($H^+$) to be released to the surrounding water molecules, forming $H_3O^+$.
Equation: $[Al(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H_3O^+$.

5.
$K_a = 10^{-pK_a} = 10^{-3.75} = 1.78 \times 10^{-4} \text{ mol dm}^{-3}$.

6.
Moles of $CH_3COONa = \frac{0.82}{82} = 0.010 \text{ mol}$.
Concentration = $\frac{0.010 \text{ mol}}{0.500 \text{ dm}^3} = 0.020 \text{ mol dm}^{-3}$.

7.
Salt hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-4.76}} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.75 \times 10^{-10}$.
$[OH^-] = \sqrt{K_b \times [Salt]} = \sqrt{5.75 \times 10^{-10} \times 0.020} = \sqrt{1.15 \times 10^{-11}} = 3.39 \times 10^{-6} \text{ mol dm}^{-3}$.
$pOH = -\log(3.39 \times 10^{-6}) = 5.47$.
$pH = 14.00 - 5.47 = 8.53$.

8.
Using Henderson-Hasselbalch:
$pH = pK_a + \log\left(\frac{[Salt]}{[Acid]}\right)$.
$pK_a = -\log(1.3 \times 10^{-5}) = 4.89$.
$pH = 4.89 + \log\left(\frac{0.20}{0.10}\right) = 4.89 + 0.30 = 5.19$.

9.
Moles of $H^+$ added = $1.0 \times 10^{-3} \text{ dm}^3 \times 1.0 \text{ mol dm}^{-3} = 0.001 \text{ mol}$.
Initial moles in $100 \text{ cm}^3$:
Acid = $0.10 \times 0.1 = 0.010 \text{ mol}$.
Salt = $0.20 \times 0.1 = 0.020 \text{ mol}$.
Reaction: $H^+ + C_2H_5COO^- \rightarrow C_2H_5COOH$.
New moles Acid = $0.010 + 0.001 = 0.011 \text{ mol}$.
New moles Salt = $0.020 - 0.001 = 0.019 \text{ mol}$.
New $pH = 4.89 + \log\left(\frac{0.019}{0.011}\right) = 4.89 + 0.24 = 5.13$.

10.
$K_{sp} = [Mg^{2+}][OH^-]^2$.

11.
Let solubility be $s \text{ mol dm}^{-3}$.
$[Mg^{2+}] = s$, $[OH^-] = 2s$.
$K_{sp} = (s)(2s)^2 = 4s^3$.
$1.8 \times 10^{-11} = 4s^3$.
$s^3 = 4.5 \times 10^{-12}$.
$s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4} \text{ mol dm}^{-3}$.

12.
More soluble. In pH 2, $[H^+]$ is high. $H^+$ reacts with $OH^-$ ions from the equilibrium $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$ to form water. This decreases $[OH^-]$, shifting the equilibrium position to the right (Le Chatelier’s Principle) to dissolve more solid.

13.
Sketch:

• Start pH approx 3 (weak acid).
• Gradual rise (buffer region).
• Vertical section at equivalence point (25 cm³ NaOH).
• Equivalence point pH > 7 (approx 8-9).
• Final pH approaches 13 (excess strong base).
• Label "Buffer Region" around 12.5 cm³.
• Label "Equivalence Point" at 25 cm³.

14.
At equivalence, all $CH_3COOH$ is converted to $CH_3COO^-$. The ethanoate ion hydrolyzes: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$. The production of $OH^-$ ions makes the solution alkaline (pH > 7).

15.
At half-equivalence, $[CH_3COOH] = [CH_3COO^-]$.
$pH = pK_a + \log(1) = pK_a$.
$pH = 4.76$.

16.
At half-equivalence volume ($12.5 \text{ cm}^3$ is half of $25.0 \text{ cm}^3$), $pH = pK_a$.
Therefore, $pK_a = 4.80$.

17.
At equivalence, moles acid = moles base.
Moles NaOH = $0.025 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.0025 \text{ mol}$.
Moles HA = $0.0025 \text{ mol}$.
Concentration HA = $\frac{0.0025 \text{ mol}}{0.025 \text{ dm}^3} = 0.100 \text{ mol dm}^{-3}$.

18.
In a weak acid-strong base titration, a buffer solution exists before the equivalence point. This buffer resists changes in pH. Additionally, the salt formed hydrolyzes, and the equilibrium is not as complete/sharp as the neutralization of $H^+$ and $OH^-$ in strong-strong titrations, resulting in a smaller change in pH per drop of titrant near the equivalence point.

19.
A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).

20.
Using Henderson-Hasselbalch:
$pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)$
$5.00 = 4.76 + \log\left(\frac{[A^-]}{[HA]}\right)$
$0.24 = \log\left(\frac{[A^-]}{[HA]}\right)$
$\frac{[A^-]}{[HA]} = 10^{0.24} = 1.74$.