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A Level H1 Chemistry Acids Bases Salts Quiz

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Questions

A-Level Chemistry H1 Quiz - Acids Bases Salts

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

Duration: 45 minutes
Total Marks: 40

Instructions:

Answer all questions.
Write your answers in the spaces provided.
Show all working for calculation questions.
Use the Data Booklet where appropriate.

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

1. Which statement correctly describes a weak acid?
[1]
A. It has a low concentration of hydrogen ions.
B. It is partially dissociated in aqueous solution.
C. It reacts slowly with metals.
D. It has a pH close to 7.

2. What is the conjugate base of the hydrogen carbonate ion, $HCO_3^-$ ?
[1]
A. $H_2CO_3$
B. $CO_3^{2-}$
C. $OH^-$
D. $H_3O^+$

3. Which oxide is amphoteric?
[1]
A. $Na_2O$
B. $MgO$
C. $Al_2O_3$
D. $SO_2$

4. Calculate the pH of a $0.050 \text{ mol dm}^{-3}$ solution of hydrochloric acid.
[1]
Answer: _______________

5. Define the term Brønsted-Lowry base.
[1]

6. Write the expression for the ionic product of water, $K_w$ .
[1]
$K_w =$ _________________________

7. State the approximate pH of a $0.1 \text{ mol dm}^{-3}$ solution of sodium hydroxide at 298 K.
[1]
Answer: _______________

8. Which indicator is most suitable for the titration of a weak acid with a strong base?
[1]
A. Methyl orange (pH range 3.1–4.4)
B. Bromophenol blue (pH range 3.0–4.6)
C. Phenolphthalein (pH range 8.3–10.0)
D. Litmus (pH range 5.0–8.0)

9. Explain why a solution of ammonium chloride, $NH_4Cl$ , is acidic.
[2]

10. The $K_a$ value for ethanoic acid is $1.7 \times 10^{-5} \text{ mol dm}^{-3}$ . What does a small $K_a$ value indicate about the strength of the acid?
[1]

Section B: Calculations & Equilibria (18 Marks)

11. Propanoic acid, $C_2H_5COOH$ , is a weak acid with $K_a = 1.3 \times 10^{-5} \text{ mol dm}^{-3}$ at 298 K.

(a) Write the equation for the dissociation of propanoic acid in water.
[1]

(b) Calculate the pH of a $0.10 \text{ mol dm}^{-3}$ solution of propanoic acid.
[3]

Answer space

Answer: pH = _______________

Answer space

Answer: _______________ %

12. A buffer solution is prepared by mixing $50.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ ethanoic acid ( $CH_3COOH$ ) with $50.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ sodium ethanoate ( $CH_3COONa$ ). The $K_a$ of ethanoic acid is $1.7 \times 10^{-5} \text{ mol dm}^{-3}$ .

(a) Calculate the pH of this buffer solution.
[3]

Answer space

Answer: pH = _______________

(b) Explain, with the aid of an equation, how this buffer solution resists a change in pH when a small amount of strong acid ( $H^+$ ) is added.
[3]

(c) Calculate the new pH if $1.0 \text{ cm}^3$ of $1.0 \text{ mol dm}^{-3}$ $HCl$ is added to the buffer solution in (a). Assume volumes are additive.
[4]

Answer space

Answer: pH = _______________

13. 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 $K_{sp}$ of $Mg(OH)_2$ .
[1]
$K_{sp} =$ _________________________

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

Answer space

Answer: Solubility = _______________ $\text{mol dm}^{-3}$

Section C: Structured Responses & Applications (12 Marks)

14. Titration curves provide important information about acid-base reactions.

(a) Sketch the pH curve for the titration of $25.0 \text{ cm}^3$ of $0.10 \text{ mol dm}^{-3}$ ethanoic acid with $0.10 \text{ mol dm}^{-3}$ sodium hydroxide. Label the equivalence point.
[3]

Answer space

(b) Explain why the pH at the equivalence point is greater than 7.
[2]

15. Carbonic acid, $H_2CO_3$ , plays a key role in maintaining the pH of blood.

(a) Write the equation for the first dissociation of carbonic acid.
[1]

(b) In the blood, the concentration of $HCO_3^-$ is approximately 20 times that of $H_2CO_3$ . Given that $pK_{a1}$ for carbonic acid is 6.4, calculate the pH of blood using the Henderson-Hasselbalch equation or $K_a$ expression.
[3]

Answer space

Answer: pH = _______________

16. An unknown monoprotic acid, HA, has a concentration of $0.010 \text{ mol dm}^{-3}$ and a pH of 3.0.

(a) Determine whether HA is a strong or weak acid. Justify your answer with a calculation.
[2]

Answer space

(b) Calculate the $K_a$ value for HA.
[2]

Answer space

Answer: $K_a =$ _______________ $\text{mol dm}^{-3}$

17. A student titrates $25.0 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ ammonia ( $NH_3$ ) with $0.100 \text{ mol dm}^{-3}$ hydrochloric acid ( $HCl$ ).

(a) Write the ionic equation for the reaction between ammonia and hydrochloric acid.
[1]

(b) Suggest a suitable indicator for this titration and explain your choice.
[2]

18. Barium sulfate, $BaSO_4$ , is used in medical imaging. Its $K_{sp}$ is $1.1 \times 10^{-10} \text{ mol}^2 \text{ dm}^{-6}$ at 298 K.

(a) Calculate the solubility of $BaSO_4$ in pure water in $\text{mol dm}^{-3}$ .
[2]

Answer space

Answer: Solubility = _______________ $\text{mol dm}^{-3}$

(b) Explain why $BaSO_4$ is less soluble in a solution of sodium sulfate ( $Na_2SO_4$ ) than in pure water.
[2]

19. Methanoic acid ( $HCOOH$ ) has a $K_a$ of $1.8 \times 10^{-4} \text{ mol dm}^{-3}$ .

(a) Calculate the pH of a $0.050 \text{ mol dm}^{-3}$ solution of methanoic acid.
[3]

Answer space

Answer: pH = _______________

(b) State one assumption made in your calculation in (a).
[1]

20. The pH of a saturated solution of calcium hydroxide, $Ca(OH)_2$ , is 12.4 at 298 K.

(a) Calculate the concentration of hydroxide ions, $[OH^-]$ , in this solution.
[2]

Answer space

Answer: $[OH^-] =$ _______________ $\text{mol dm}^{-3}$

(b) Calculate the solubility product, $K_{sp}$ , for $Ca(OH)_2$ at this temperature.
[2]

Answer space

Answer: $K_{sp} =$ _______________ $\text{mol}^3 \text{ dm}^{-9}$

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Answers

A-Level Chemistry H1 Quiz - Acids Bases Salts (Answer Key)

Total Marks: 40

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

1. B
[1]
Reasoning: A weak acid is defined by its partial dissociation in water. Concentration (A) is independent of strength. Reaction rate (C) is kinetic, not equilibrium. pH (D) depends on concentration.

2. B
[1]
Reasoning: Conjugate base is formed by removing $H^+$ . $HCO_3^- \rightarrow CO_3^{2-} + H^+$ .

3. C
[1]
Reasoning: $Al_2O_3$ reacts with both acids and bases. $Na_2O$ and $MgO$ are basic; $SO_2$ is acidic.

4. 1.30
[1]
Reasoning: HCl is a strong acid, so $[H^+] = 0.050$ . $pH = -\log(0.050) = 1.30$ .

5. A proton ( $H^+$ ) acceptor.
[1]

6. $[H^+][OH^-]$
[1]

7. 13
[1]
Reasoning: $[OH^-] = 0.1$ . $pOH = 1$ . $pH = 14 - 1 = 13$ .

8. C
[1]
Reasoning: The equivalence point for weak acid-strong base is alkaline (pH > 7). Phenolphthalein changes color in this range.

9. $NH_4^+$ is the conjugate acid of a weak base ( $NH_3$ ). It hydrolyses in water:
[2]
$NH_4^+ (aq) + H_2O (l) \rightleftharpoons NH_3 (aq) + H_3O^+ (aq)$
The production of $H_3O^+$ makes the solution acidic.
[1 for equation/hydrolysis concept, 1 for linking to acidity]

10. The acid is weak / partially dissociated.
[1]

Section B: Calculations & Equilibria (18 Marks)

11.
(a) $C_2H_5COOH (aq) \rightleftharpoons C_2H_5COO^- (aq) + H^+ (aq)$
[1] (Must have reversible arrow and state symbols)

(b) pH = 2.94
[3]
Working:
$K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{[H^+]^2}{[HA]_{initial}}$ (Assumption: dissociation is small)
$[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.3 \times 10^{-5} \times 0.10}$
$[H^+] = \sqrt{1.3 \times 10^{-6}} = 1.14 \times 10^{-3} \text{ mol dm}^{-3}$
$pH = -\log(1.14 \times 10^{-3}) = 2.94$
[1 for formula, 1 for calculation of $[H^+]$ , 1 for pH]

(c) 1.14%
[2]
Working:
$\% \text{ dissociation} = \frac{[H^+]}{[HA]_{initial}} \times 100$
$= \frac{1.14 \times 10^{-3}}{0.10} \times 100 = 1.14\%$
[1 for substitution, 1 for answer]

12.
(a) pH = 4.77
[3]
Working:
Since volumes and concentrations are equal, $[CH_3COOH] = [CH_3COO^-]$ .
$pH = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right)$
$pK_a = -\log(1.7 \times 10^{-5}) = 4.77$
$\log(1) = 0$
$pH = 4.77$
[1 for $pK_a$ , 1 for ratio logic, 1 for answer]

(b) Explanation of Buffer Action:
[3]
The added $H^+$ ions react with the ethanoate ions ( $CH_3COO^-$ ) from the salt:
$CH_3COO^- (aq) + H^+ (aq) \rightarrow CH_3COOH (aq)$
This removes most of the added $H^+$ , keeping the pH relatively constant.
[1 for equation, 1 for identifying reacting species, 1 for explaining removal of $H^+$ ]

(c) pH = 4.59
[4]
Working:
Initial moles:
$n(\text{acid}) = 0.050 \times 0.10 = 0.0050 \text{ mol}$
$n(\text{salt}) = 0.050 \times 0.10 = 0.0050 \text{ mol}$
Moles $H^+$ added: $0.0010 \times 1.0 = 0.0010 \text{ mol}$

New moles:
$n(\text{acid}) = 0.0050 + 0.0010 = 0.0060 \text{ mol}$
$n(\text{salt}) = 0.0050 - 0.0010 = 0.0040 \text{ mol}$

New pH:
$pH = 4.77 + \log\left(\frac{0.0040}{0.0060}\right)$
$pH = 4.77 + \log(0.667)$
$pH = 4.77 - 0.176 = 4.59$
[1 for initial moles, 1 for new moles, 1 for substitution, 1 for final answer]

13.
(a) $K_{sp} = [Mg^{2+}][OH^-]^2$
[1]

(b) Solubility = $1.65 \times 10^{-4} \text{ mol dm}^{-3}$
[3]
Working:
Let solubility be $s$ .
$[Mg^{2+}] = s$ , $[OH^-] = 2s$
$K_{sp} = (s)(2s)^2 = 4s^3$
$1.8 \times 10^{-11} = 4s^3$
$s^3 = \frac{1.8 \times 10^{-11}}{4} = 4.5 \times 10^{-12}$
$s = \sqrt[3]{4.5 \times 10^{-12}} = 1.65 \times 10^{-4}$
[1 for expression in terms of s, 1 for algebra, 1 for answer]

Section C: Structured Responses & Applications (12 Marks)

14.
(a) Sketch:
[3]

Starts at pH ~3 (weak acid).
Gradual rise, then steep vertical section around pH 7-9.
Equivalence point marked at volume $25.0 \text{ cm}^3$ .
Ends at pH ~13 (strong base).
[1 for start pH, 1 for shape/equivalence position, 1 for end pH]

(b) Explanation:
[2]
At the equivalence point, the solution contains sodium ethanoate.
The ethanoate ion hydrolyses: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ .
The production of $OH^-$ ions makes the solution alkaline (pH > 7).
[1 for salt hydrolysis equation/concept, 1 for linking $OH^-$ to pH]

15.
(a) $H_2CO_3 (aq) \rightleftharpoons H^+ (aq) + HCO_3^- (aq)$
[1]

(b) pH = 7.7
[3]
Working:
$pH = pK_a + \log\left(\frac{[HCO_3^-]}{[H_2CO_3]}\right)$
$pH = 6.4 + \log(20)$
$pH = 6.4 + 1.30 = 7.70$
[1 for formula, 1 for substitution, 1 for answer]

16.
(a) Weak Acid.
[2]
If strong, $[H^+]$ would be $0.010 \text{ mol dm}^{-3}$ , giving pH 2.0.
Since pH is 3.0, $[H^+] = 0.001 \text{ mol dm}^{-3}$ , which is less than the initial concentration.
[1 for comparison/calculation, 1 for conclusion]

(b) $K_a = 1.1 \times 10^{-4} \text{ mol dm}^{-3}$
[2]
Working:
$[H^+] = 10^{-3} = 0.001$
$[HA]_{eq} = 0.010 - 0.001 = 0.009$ .
$K_a = \frac{[H^+]^2}{[HA]_{eq}} = \frac{(0.001)^2}{0.009} = 1.11 \times 10^{-4}$ .
[1 for substitution, 1 for answer]

17.
(a) $NH_3 (aq) + H^+ (aq) \rightarrow NH_4^+ (aq)$
[1]

(b) Methyl orange.
[2]
The titration involves a weak base and strong acid, so the equivalence point is acidic (pH < 7).
Methyl orange changes color in the acidic range (3.1–4.4), matching the steep part of the curve.
[1 for indicator, 1 for reasoning]

18.
(a) Solubility = $1.05 \times 10^{-5} \text{ mol dm}^{-3}$
[2]
Working:
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = s^2$
$s = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5}$
[1 for formula, 1 for answer]

(b) Common Ion Effect.
[2]
$Na_2SO_4$ provides $SO_4^{2-}$ ions.
According to Le Chatelier’s principle, increasing $[SO_4^{2-}]$ shifts the equilibrium $BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq)$ to the left, reducing solubility.
[1 for identifying common ion, 1 for equilibrium shift explanation]

19.
(a) pH = 2.52
[3]
Working:
$[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.8 \times 10^{-4} \times 0.050}$
$[H^+] = \sqrt{9.0 \times 10^{-6}} = 3.0 \times 10^{-3}$
$pH = -\log(3.0 \times 10^{-3}) = 2.52$
[1 for formula, 1 for calculation, 1 for pH]

(b) The degree of dissociation is small / $[HA]_{initial} \approx [HA]_{equilibrium}$ .
[1]

20.
(a) $[OH^-] = 2.51 \times 10^{-2} \text{ mol dm}^{-3}$
[2]
Working:
$pH = 12.4 \Rightarrow pOH = 14 - 12.4 = 1.6$
$[OH^-] = 10^{-1.6} = 2.51 \times 10^{-2}$
[1 for pOH, 1 for concentration]

(b) $K_{sp} = 7.9 \times 10^{-6} \text{ mol}^3 \text{ dm}^{-9}$
[2]
Working:
$Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-$
$[Ca^{2+}] = \frac{1}{2}[OH^-] = 1.255 \times 10^{-2}$
$K_{sp} = [Ca^{2+}][OH^-]^2 = (1.255 \times 10^{-2})(2.51 \times 10^{-2})^2$
$K_{sp} \approx 7.9 \times 10^{-6}$
[1 for stoichiometry/concentration, 1 for final Ksp]