A Level H1 Chemistry Acids Bases Salts Quiz

A-Level Chemistry H1 Quiz - Acids Bases Salts

Answer Key & Teaching Notes

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Section A: Multiple Choice

1. Answer: A

Explanation: A Brønsted–Lowry acid is defined as a proton ($H^+$) donor. Option C describes an Arrhenius acid, which is a related but narrower definition. Option B describes a Brønsted–Lowry base, and Option D describes an alkaline substance. At A-Level, the Brønsted–Lowry definition is the one most commonly tested.

Common mistake: Students confuse Arrhenius and Brønsted–Lowry definitions. While all Brønsted–Lowry acids produce $H^+$ in water, the defining feature is proton donation, not merely producing $H^+$ ions.

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2. Answer: C

Explanation: For a strong monoprotic acid, $[H^+] = 0.05 \text{ mol dm}^{-3}$ (complete dissociation). Therefore:

$\text{pH} = -\log_{10}[H^+] = -\log_{10}(0.05) = -\log_{10}(5 \times 10^{-2}) = 2 - \log_{10}5 = 2 - 0.70 = 1.30$

Common mistake: Students may incorrectly calculate $\log_{10}(0.05)$ as 0.3 (confusing $\log_{10}$ with $\ln$), or forget that strong acids fully dissociate so $[H^+]$ equals the acid concentration.

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3. Answer: C

Explanation: Sodium ethanoate is the salt of a weak acid (ethanoic acid) and a strong base (sodium hydroxide). The ethanoate ion ($CH_3COO^-$) undergoes hydrolysis in water:

$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$

This produces $OH^-$ ions, making the solution basic (pH > 7).

• $NH_4Cl$: salt of weak base + strong acid → acidic (pH < 7)
• $NaCl$: salt of strong acid + strong base → neutral (pH = 7)
• $KNO_3$: salt of strong acid + strong base → neutral (pH = 7)

Teaching note: The rule is: "the stronger wins." If the parent acid is weaker than the parent base is strong, the solution is basic, and vice versa.

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4. Answer: C

Explanation: A buffer solution consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). Here, $CH_3COOH$ is the weak acid and $CH_3COO^-$ (from $CH_3COONa$) is the conjugate base.

• A is incorrect: the pH of an ethanoate buffer is approximately 4.76 (equal to $pK_a$), not 7.
• B is incorrect: buffers resist pH change when either acid OR base is added.
• D is incorrect: the pH changes slightly but not dramatically. The buffer minimises the change; it does not eliminate it entirely.

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5. Answer: A

Explanation: In pure water, $[H^+] = [OH^-]$, so:

$K_w = [H^+][OH^-] = [H^+]^2$

$[H^+] = \sqrt{K_w} = \sqrt{5.5 \times 10^{-14}} = 2.345 \times 10^{-7} \text{ mol dm}^{-3}$

$\text{pH} = -\log_{10}(2.345 \times 10^{-7}) = 7 - \log_{10}(2.345) = 7 - 0.37 = 6.63$

Teaching note: Pure water is always neutral ($[H^+] = [OH^-]$), but the pH is only 7.00 at $25^\circ\text{C}$ where $K_w = 1.0 \times 10^{-14}$. At higher temperatures, $K_w$ increases, so pH of pure water decreases below 7 — but it is still neutral.

Common mistake: Students think pH 7 always means neutral. Neutrality means $[H^+] = [OH^-]$, which depends on temperature through $K_w$.

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Section B: Structured Questions

6. (a) [2 marks]

A weak acid is an acid that partially dissociates (ionises) in aqueous solution, establishing an equilibrium between the undissociated acid and its ions.

Mark allocation:

• 1 mark for "partially dissociates" or "partially ionises"
• 1 mark for reference to equilibrium / not fully dissociated

Common mistake: Writing "dilute acid" instead of "weak acid." Dilute refers to concentration; weak refers to the degree of dissociation. A concentrated weak acid is still weak.

(b) [2 marks]

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

Mark allocation:

• 1 mark for "resists pH change" / "minimises pH change"
• 1 mark for "on addition of small amounts of acid or base" AND reference to weak acid-conjugate base (or weak base-conjugate acid)

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7. (a) [1 mark]

$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

Note: Water is omitted from the expression because it is the solvent and its concentration is essentially constant.

(b) [3 marks]

For a weak acid, assuming $x$ is small compared to initial concentration:

$K_a = \frac{[H^+]^2}{[CH_3COOH]_{\text{initial}}}$

$[H^+]^2 = K_a \times [CH_3COOH] = 1.74 \times 10^{-5} \times 0.20 = 3.48 \times 10^{-6}$

$[H^+] = \sqrt{3.48 \times 10^{-6}} = 1.865 \times 10^{-3} \text{ mol dm}^{-3}$

$\text{pH} = -\log_{10}(1.865 \times 10^{-3}) = 2.73$

Mark allocation:

• 1 mark for correct $K_a$ expression or substitution
• 1 mark for correct $[H^+]$ calculation
• 1 mark for correct pH to 2 d.p.

Common mistake: Forgetting to take the square root, or using $[H^+] = K_a \times [HA]$ instead of $[H^+] = \sqrt{K_a \times [HA]}$. The approximation $[H^+] = \sqrt{K_a \cdot C}$ is valid when the degree of dissociation is small (< 5%).

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8. (a) [1 mark]

$HCl + NaOH \rightarrow NaCl + H_2O$

or in ionic form: $H^+ + OH^- \rightarrow H_2O$

(b) [2 marks]

$n(HCl) = 0.100 \times \frac{25.0}{1000} = 2.50 \times 10^{-3} \text{ mol}$

Since the mole ratio is 1:1:

$n(NaOH) = 2.50 \times 10^{-3} \text{ mol}$

$V(NaOH) = \frac{2.50 \times 10^{-3}}{0.100} = 2.50 \times 10^{-2} \text{ dm}^3 = 25.0 \text{ cm}^3$

Mark allocation:

• 1 mark for moles of HCl calculated correctly
• 1 mark for correct volume of NaOH

(c) [2 marks]

The pH at the equivalence point is 7.

Explanation: $HCl$ is a strong acid and $NaOH$ is a strong base. The salt formed ($NaCl$) is derived from a strong acid and strong base, so it does not hydrolyse. The solution is neutral at pH 7.

Mark allocation:

• 1 mark for stating pH = 7
• 1 mark for correct explanation (strong acid + strong base → neutral salt)

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9. (a) [2 marks]

Total volume after mixing $= 50.0 + 50.0 = 100.0 \text{ cm}^3$

$[CH_3COOH] = \frac{0.200 \times 50.0}{100.0} = 0.100 \text{ mol dm}^{-3}$

$[CH_3COONa] = \frac{0.100 \times 50.0}{100.0} = 0.050 \text{ mol dm}^{-3}$

Mark allocation:

• 1 mark for each correct concentration

(b) [3 marks]

Using the Henderson–Hasselbalch equation:

$\text{pH} = pK_a + \log_{10}\frac{[\text{salt}]}{[\text{acid}]}$

$pK_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76$

$\text{pH} = 4.76 + \log_{10}\left(\frac{0.050}{0.100}\right) = 4.76 + \log_{10}(0.50) = 4.76 - 0.30 = 4.46$

Alternative method using $K_a$ expression:

$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

$1.74 \times 10^{-5} = \frac{0.050 \times [H^+]}{0.100}$

$[H^+] = \frac{1.74 \times 10^{-5} \times 0.100}{0.050} = 3.48 \times 10^{-5} \text{ mol dm}^{-3}$

$\text{pH} = -\log_{10}(3.48 \times 10^{-5}) = 4.46$

Mark allocation:

• 1 mark for correct $pK_a$ or $K_a$ expression
• 1 mark for correct substitution
• 1 mark for correct final pH

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10. [3 marks]

When a small amount of $HCl$ is added to a buffer containing $CH_3COOH$ and $CH_3COO^-$:

The added $H^+$ ions react with the conjugate base ($CH_3COO^-$):

$CH_3COO^- + H^+ \rightarrow CH_3COOH$

This shifts the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ to the left (Le Chatelier's principle). The added $H^+$ is consumed by the $CH_3COO^-$ ions, converting them into $CH_3COOH$. Since the ratio $\frac{[CH_3COO^-]}{[CH_3COOH]}$ changes only slightly, the pH remains approximately constant.

Mark allocation:

• 1 mark for stating that added $H^+$ reacts with the conjugate base ($CH_3COO^-$)
• 1 mark for reference to equilibrium shifting left / Le Chatelier's principle
• 1 mark for explaining that the ratio changes minimally, so pH is approximately constant

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11. (a) [1 mark]

From the titration curve, the pH at the equivalence point is approximately 8.7 (accept 8.5–9.0).

(b) [2 marks]

The pH at the equivalence point is greater than 7 because the salt formed is the salt of a weak acid and a strong base (e.g., sodium ethanoate). The conjugate base of the weak acid undergoes hydrolysis:

$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$

This produces $OH^-$ ions, making the solution slightly alkaline.

Mark allocation:

• 1 mark for identifying the salt as being from a weak acid + strong base
• 1 mark for explaining hydrolysis producing $OH^-$ ions

(c) [2 marks]

The half-equivalence point occurs at 12.5 cm³ (half the volume at the equivalence point).

Significance: At the half-equivalence point, exactly half the weak acid has been neutralised, so $[\text{acid}] = [\text{conjugate base}]$. From the Henderson–Hasselbalch equation:

$\text{pH} = pK_a + \log_{10}(1) = pK_a$

Therefore, pH = $pK_a$ at this point, allowing the $K_a$ of the weak acid to be determined directly from the titration curve.

Mark allocation:

• 1 mark for correct volume (12.5 cm³)
• 1 mark for stating that pH = $pK_a$ at this point

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12. (a) [1 mark]

$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$

(b) [4 marks]

$K_b = \frac{[OH^-]^2}{[NH_3]_{\text{initial}}}$

$[OH^-]^2 = K_b \times [NH_3] = 1.8 \times 10^{-5} \times 0.10 = 1.8 \times 10^{-6}$

$[OH^-] = \sqrt{1.8 \times 10^{-6}} = 1.342 \times 10^{-3} \text{ mol dm}^{-3}$

$\text{pOH} = -\log_{10}(1.342 \times 10^{-3}) = 2.87$

$\text{pH} = 14.00 - 2.87 = 11.13$

Alternative (direct pH calculation):

$[H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{1.342 \times 10^{-3}} = 7.45 \times 10^{-12}$

$\text{pH} = -\log_{10}(7.45 \times 10^{-12}) = 11.13$

Mark allocation:

• 1 mark for correct $K_b$ expression / substitution
• 1 mark for correct $[OH^-]$
• 1 mark for correct pOH
• 1 mark for correct final pH to 2 d.p.

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13. [4 marks]

Apparatus: Burette, pipette ($25.0 \text{ cm}^3$), pipette filler, conical flask, white tile, standard $NaOH$ solution, $HCl$ solution of unknown concentration, indicator (e.g., phenolphthalein or methyl orange).

Procedure:

1. Use a pipette to measure $25.0 \text{ cm}^3$ of the $HCl$ solution into a conical flask.
2. Add a few drops of indicator.
3. Fill the burette with the standard $NaOH$ solution and record the initial reading.
4. Add $NaOH$ from the burette to the conical flask, swirling continuously.
5. As the end point approaches, add $NaOH$ dropwise until the indicator changes colour permanently (e.g., phenolphthalein turns from colourless to pale pink).
6. Record the final burette reading and calculate the volume of $NaOH$ used.
7. Repeat the titration to obtain concordant results (within $0.10 \text{ cm}^3$).

Calculation: Use $n = \frac{cV}{1000}$ and the stoichiometric ratio to find the concentration of $HCl$.

Mark allocation:

• 1 mark for correct apparatus (burette, pipette, conical flask, indicator)
• 1 mark for correct procedure (pipetting acid, titrating with base)
• 1 mark for identifying the end point via indicator colour change
• 1 mark for mentioning repetition / concordant results

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14. (a) [1 mark]

$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$

(b) [2 marks]

The carbonate ion ($CO_3^{2-}$) is the conjugate base of the weak acid $HCO_3^-$ (which is itself the conjugate base of the weak acid $H_2CO_3$). Since $CO_3^{2-}$ is a relatively strong conjugate base (from a weak acid), it reacts with water (hydrolysis) to produce $OH^-$ ions. The $Na^+$ ion does not hydrolyse because it comes from the strong base $NaOH$. The net production of $OH^-$ ions makes the solution basic.

Mark allocation:

• 1 mark for identifying that $CO_3^{2-}$ is the conjugate base of a weak acid
• 1 mark for explaining that hydrolysis produces $OH^-$ ions, making the solution basic

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15. (a) [2 marks]

$HCl$ is a strong acid and dissociates completely, contributing $0.10 \text{ mol dm}^{-3}$ of $H^+$ ions. Ethanoic acid is a weak acid with $K_a = 1.74 \times 10^{-5}$, so it contributes only a very small amount of $H^+$ (approximately $\sqrt{1.74 \times 10^{-5} \times 0.10} \approx 1.3 \times 10^{-3} \text{ mol dm}^{-3}$). The $H^+$ from $HCl$ is approximately 77 times greater than that from ethanoic acid. Additionally, the high $[H^+]$ from $HCl$ suppresses the dissociation of ethanoic acid (common ion effect), making its contribution even smaller.

Mark allocation:

• 1 mark for identifying $HCl$ as strong (complete dissociation) and $CH_3COOH$ as weak
• 1 mark for explaining the common ion effect / that $H^+$ from $HCl$ suppresses $CH_3COOH$ dissociation

(b) [2 marks]

Since the contribution from ethanoic acid is negligible:

$[H^+] \approx 0.10 \text{ mol dm}^{-3} \text{ (from HCl)}$

$\text{pH} = -\log_{10}(0.10) = 1.00$

Mark allocation:

• 1 mark for stating $[H^+] = 0.10 \text{ mol dm}^{-3}$
• 1 mark for pH = 1.00

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Section C: Application & Data Interpretation

16. (a) [2 marks]

Buffer B will show the smaller pH change.

Explanation: Buffer B has a higher total concentration of buffer components ($0.20 \text{ mol dm}^{-3}$ each of acid and salt) compared to Buffer A ($0.10 \text{ mol dm}^{-3}$ each). A buffer with higher concentrations of both components has a greater buffer capacity — it can neutralise more added acid or base before the pH changes significantly.

Mark allocation:

• 1 mark for identifying Buffer B
• 1 mark for explaining buffer capacity in terms of higher concentrations

(b) [5 marks]

Before addition of HCl:

$[CH_3COOH] = [CH_3COONa] = 0.10 \text{ mol dm}^{-3}$

$\text{pH} = pK_a + \log_{10}\left(\frac{0.10}{0.10}\right) = 4.76 + \log_{10}(1) = 4.76$

After addition of HCl:

Moles of HCl added $= 0.10 \times \frac{5.0}{1000} = 5.0 \times 10^{-4} \text{ mol}$

Initial moles in $100 \text{ cm}^3$:

• $n(CH_3COOH) = 0.10 \times \frac{100}{1000} = 0.010 \text{ mol}$
• $n(CH_3COO^-) = 0.10 \times \frac{100}{1000} = 0.010 \text{ mol}$

The added $H^+$ reacts with $CH_3COO^-$:

• New $n(CH_3COO^-) = 0.010 - 5.0 \times 10^{-4} = 0.0095 \text{ mol}$
• New $n(CH_3COOH) = 0.010 + 5.0 \times 10^{-4} = 0.0105 \text{ mol}$

Since total volume is the same for both species (approximately $105 \text{cm}^3$, but the ratio is unchanged):

$\text{pH} = 4.76 + \log_{10}\left(\frac{0.0095}{0.0105}\right) = 4.76 + \log_{10}(0.9048) = 4.76 - 0.043 = 4.72$

Change in pH $= 4.76 - 4.72 = 0.04$ (very small change, demonstrating buffer action)

Mark allocation:

• 1 mark for correct initial pH
• 1 mark for calculating moles of HCl added
• 1 mark for correct new moles of acid and salt
• 1 mark for correct final pH
• 1 mark for correct calculation showing small pH change

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17. (a) [1 mark]

Decreasing acid strength: Methanoic acid > Ethanoic acid > Carbonic acid

(or $HCOOH > CH_3COOH > H_2CO_3$)

(b) [3 marks]

$[H^+]^2 = K_a \times [HCOOH] = 1.6 \times 10^{-4} \times 0.050 = 8.0 \times 10^{-6}$

$[H^+] = \sqrt{8.0 \times 10^{-6}} = 2.828 \times 10^{-3} \text{ mol dm}^{-3}$

$\text{pH} = -\log_{10}(2.828 \times 10^{-3}) = 2.55$

Mark allocation:

• 1 mark for correct substitution into $K_a$ expression
• 1 mark for correct $[H^+]$
• 1 mark for correct pH to 2 d.p.

(c) [1 mark]

The larger the $K_a$ value, the greater the degree of dissociation of the acid in water, and hence the stronger the acid. $K_a$ is a quantitative measure of acid strength: a higher $K_a$ means the equilibrium lies further to the right (more dissociation), producing more $H^+$ ions.

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18. (a) [2 marks]

$\text{pOH} = 14.00 - 13.0 = 1.0$

$[OH^-] = 10^{-\text{pOH}} = 10^{-1.0} = 0.10 \text{ mol dm}^{-3}$

Mark allocation:

• 1 mark for correct pOH
• 1 mark for correct $[OH^-]$

(b) [4 marks]

$n(NaOH) = 0.10 \times \frac{35.0}{1000} = 3.50 \times 10^{-3} \text{ mol}$

The balanced equation is:

$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$

Mole ratio: $n(H_2SO_4) : n(NaOH) = 1 : 2$

$n(H_2SO_4) = \frac{3.50 \times 10^{-3}}{2} = 1.75 \times 10^{-3} \text{ mol}$

$[H_2SO_4] = \frac{1.75 \times 10^{-3}}{25.0/1000} = \frac{1.75 \times 10^{-3}}{0.0250} = 0.070 \text{ mol dm}^{-3}$

Mark allocation:

• 1 mark for moles of NaOH
• 1 mark for correct stoichiometric ratio (1:2)
• 1 mark for moles of $H_2SO_4$
• 1 mark for correct concentration

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19. (a) [1 mark]

$n(CH_3COONa) = \frac{8.20}{82.0} = 0.100 \text{ mol}$

(b) [3 marks]

$[CH_3COONa] = \frac{0.100}{0.250} = 0.400 \text{ mol dm}^{-3}$

$[CH_3COOH] = 0.100 \text{ mol dm}^{-3}$

Using the Henderson–Hasselbalch equation:

$\text{pH} = pK_a + \log_{10}\left(\frac{[\text{salt}]}{[\text{acid}]}\right) = 4.76 + \log_{10}\left(\frac{0.400}{0.100}\right)$

$\text{pH} = 4.76 + \log_{10}(4.00) = 4.76 + 0.60 = 5.36$

Mark allocation:

• 1 mark for correct concentration of salt
• 1 mark for correct substitution into Henderson–Hasselbalch equation
• 1 mark for correct final pH

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20. (a) [2 marks]

Methyl orange changes colour over the pH range approximately 3.1–4.4 (red to yellow). For a weak acid–strong base titration, the equivalence point occurs at a pH greater than 7 (typically pH 8–9) because the salt formed is basic. Methyl orange would change colour well before the equivalence point is reached, leading to a significant titration error (underestimation of the volume of base needed).

Mark allocation:

• 1 mark for stating that the equivalence point pH > 7 for weak acid–strong base
• 1 mark for explaining that methyl orange changes colour in acidic range, causing premature end point

(b) [2 marks]

Phenolphthalein is a more suitable indicator.

Explanation: Phenolphthalein changes colour over the pH range approximately 8.2–10.0 (colourless to pink). This range falls within the steep portion of the titration curve for a weak acid–strong base titration, which occurs around pH 7–10. The colour change therefore coincides with the equivalence point, giving an accurate result.

Mark allocation:

• 1 mark for suggesting phenolphthalein
• 1 mark for explaining that its pH range matches the equivalence point region

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End of Answer Key

Total Marks: 50