A Level H2 Chemistry Practice Paper 3

TuitionGoWhere Practice Paper — Chemistry H2 A-Level

Answer Key: Acids, Bases & Salts

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Section A: Multiple Choice

1. B — $SO_4^{2-}$

Explanation: A conjugate base is formed when an acid donates a proton ($H^+$). $HSO_4^-$ loses one $H^+$ to become $SO_4^{2-}$. Option A ($H_2SO_4$) is the conjugate acid (gaining a proton), not the conjugate base. This tests understanding of conjugate acid-base pairs: acid → conjugate base + $H^+$.

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2. B — $4.0 \times 10^{-11}$ mol dm$^{-3}$

Explanation:

• $pH = 3.40$, so $[H_3O^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm$^{-3}$
• At 25 °C: $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$
• $[OH^-] = \frac{1.0 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11}$ mol dm$^{-3}$

Wait — let me recalculate: $10^{-3.40} = 3.981 \times 10^{-4}$. Then $[OH^-] = 1.0 \times 10^{-14} / 3.981 \times 10^{-4} = 2.51 \times 10^{-11}$ mol dm$^{-3}$.

Hmm, that gives $2.5 \times 10^{-11}$, which is option C. Let me re-examine.

Actually: $10^{-3.40} = 10^{-3} \times 10^{-0.40} = 1.0 \times 10^{-3} \times 0.3981 = 3.981 \times 10^{-4}$

$[OH^-] = 1.0 \times 10^{-14} / 3.981 \times 10^{-4} = 2.512 \times 10^{-11}$ mol dm$^{-3}$

This rounds to $2.5 \times 10^{-11}$ mol dm$^{-3}$, which is C.

Corrected Answer: C — $2.5 \times 10^{-11}$ mol dm$^{-3}$

Common mistake: Students may confuse $[H_3O^+]$ with $[OH^-]$ and select option A, or miscalculate the antilog.

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3. C — $NH_4Cl$

Explanation: $NH_4Cl$ is a salt formed from a weak base ($NH_3$) and a strong acid ($HCl$). The $NH_4^+$ ion undergoes hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$, producing $H_3O^+$ ions and making the solution acidic. $Na_2CO_3$ and $CH_3COONa$ produce basic solutions (salts of strong base + weak acid). $KNO_3$ is neutral (strong acid + strong base).

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4. A — 3.20

Explanation:

• For a weak acid: $[H^+] \approx \sqrt{K_a \times C} = \sqrt{4.0 \times 10^{-6} \times 0.10} = \sqrt{4.0 \times 10^{-7}} = 6.32 \times 10^{-4}$ mol dm$^{-3}$
• $pH = -\log(6.32 \times 10^{-4}) = 3.20$

Common mistake: Students may incorrectly calculate $\sqrt{4.0 \times 10^{-6}} = 2.0 \times 10^{-3}$ and get pH = 2.70 (option B), forgetting to multiply by the concentration first.

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5. D — The resulting solution is not a buffer because only $CH_3COONa$ is present.

Explanation: Moles of $CH_3COOH = 0.050 \times 0.20 = 0.010$ mol. Moles of $NaOH = 0.050 \times 0.20 = 0.010$ mol. The acid and base react in a 1:1 ratio, so all the $CH_3COOH$ is completely neutralised to form $CH_3COONa$. The resulting solution contains only $CH_3COONa$ (a salt), with no remaining weak acid to act as a buffer component. A buffer requires a weak acid and its conjugate base in comparable amounts.

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6. B — 7.0

Explanation: $NaOH$ is a strong base and $HCl$ is a strong acid. At the equivalence point, the salt formed is $NaCl$, which is neutral (from strong acid + strong base). The pH at the equivalence point is 7.0. This contrasts with weak acid–strong base titrations where the equivalence point pH > 7 due to hydrolysis of the conjugate base.

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7. A — $1.3 \times 10^{-3}$ mol dm$^{-3}$

Explanation:

• $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$
• $K_{sp} = [Pb^{2+}][I^-]^2$
• If solubility = $s$, then $[Pb^{2+}] = s$ and $[I^-] = 2s$
• $K_{sp} = s \times (2s)^2 = 4s^3 = 8.5 \times 10^{-9}$
• $s^3 = 2.125 \times 10^{-9}$
• $s = \sqrt[3]{2.125 \times 10^{-9}} = 1.29 \times 10^{-3} \approx 1.3 \times 10^{-3}$ mol dm$^{-3}$

Common mistake: Forgetting the stoichiometric coefficient of $I^-$ and writing $K_{sp} = s \times s^2 = s^3$ instead of $4s^3$.

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8. C — Phenolphthalein

Explanation: In a weak acid–strong base titration, the equivalence point occurs at pH > 7 (basic) because the conjugate base of the weak acid hydrolyses. The pH change around the equivalence point falls in the range of approximately 7–10. Phenolphthalein has a pH range of 8.2–10.0, which falls within this steep pH change region. Methyl orange (3.1–4.4) would change colour too early, well before the equivalence point.

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9. B — $CO_3^{2-}$ ions react with water to produce $OH^-$.

Explanation: $Na_2CO_3$ is a salt of a strong base ($NaOH$) and a weak acid ($H_2CO_3$). The $CO_3^{2-}$ ion is the conjugate base of $HCO_3^-$ and undergoes hydrolysis: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$. This produces $OH^-$ ions, making the solution alkaline. $Na^+$ does not hydrolyse (it is the conjugate acid of a strong base).

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10. A — $2.0 \times 10^{-5}$ mol dm$^{-3}$

Explanation:

• $pH = 3.00$, so $[H^+] = 10^{-3.00} = 1.0 \times 10^{-3}$ mol dm$^{-3}$
• For $HX \rightleftharpoons H^+ + X^-$: $[H^+] = [X^-] = 1.0 \times 10^{-3}$ mol dm$^{-3}$
• Remaining $[HX] = 0.050 - 0.001 = 0.049 \approx 0.050$ mol dm$^{-3}$ (approximation valid since dissociation is small)
• $K_a = \frac{[H^+][X^-]}{[HX]} = \frac{(1.0 \times 10^{-3})^2}{0.050} = \frac{1.0 \times 10^{-6}}{0.050} = 2.0 \times 10^{-5}$ mol dm$^{-3}$

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11. C — The pH remains almost unchanged.

Explanation: This is a buffer solution containing a weak acid ($CH_3COOH$) and its conjugate base ($CH_3COO^-$). When a small amount of $HCl$ is added, the $H^+$ ions are consumed by the conjugate base: $CH_3COO^- + H^+ \rightarrow CH_3COOH$. This shifts the equilibrium but the ratio $[CH_3COO^-]/[CH_3COOH]$ changes only slightly, so the pH remains almost unchanged. This is the defining property of a buffer.

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12. A — $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$ mol$^2$ dm$^{-6}$

Explanation: The ionic product of water is defined as $K_w = [H_3O^+][OH^-]$. At 25 °C, $K_w = 1.0 \times 10^{-14}$ mol$^2$ dm$^{-6}$. Option B has the wrong value. Option C incorrectly uses a ratio. Option D incorrectly uses a sum.

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13. C — 40.0 cm³

Explanation:

• $H_2SO_4$ is diprotic: $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$
• Moles of $H_2SO_4 = 0.0200 \times 0.10 = 0.0020$ mol
• Moles of $NaOH$ needed $= 2 \times 0.0020 = 0.0040$ mol
• Volume of $NaOH = \frac{0.0040}{0.10} = 0.040$ dm$^3 = 40.0$ cm³

Common mistake: Forgetting that $H_2SO_4$ provides 2 moles of $H^+$ per mole of acid and using a 1:1 ratio, giving 20.0 cm³ (option B).

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14. B — 10.52

Explanation:

• $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$
• $K_{sp} = [Mg^{2+}][OH^-]^2 = 4s^3 = 1.8 \times 10^{-11}$
• $s^3 = 4.5 \times 10^{-12}$
• $s = 1.65 \times 10^{-4}$ mol dm$^{-3}$
• $[OH^-] = 2s = 3.30 \times 10^{-4}$ mol dm$^{-3}$
• $pOH = -\log(3.30 \times 10^{-4}) = 3.48$
• $pH = 14.00 - 3.48 = 10.52$

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15. C — $Na_2CO_3$

Explanation: The extent of hydrolysis depends on the strength of the parent acid and base. $NaCl$ is neutral (strong acid + strong base, no hydrolysis). $NH_4Cl$ undergoes cationic hydrolysis (weak base cation). $Na_2CO_3$ undergoes anionic hydrolysis with $CO_3^{2-}$, which is the conjugate base of the weak acid $HCO_3^-$ — since $H_2CO_3$ is weak, $CO_3^{2-}$ is relatively strongly basic and undergoes extensive hydrolysis. $NH_4CH_3COO$ has both ions hydrolysing but the effects partially cancel ($K_a \approx K_b$ for the parent acid and base). $CO_3^{2-}$ is the strongest base among the anions listed because it is the conjugate base of a weak acid ($HCO_3^-$) which itself comes from a weak acid ($H_2CO_3$).

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Section B: Structured Questions

16. Buffer Solutions and pH Calculations [8 marks]

(a) [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 consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).

Marking:

• [1] for mentioning resistance to pH change
• [1] for identifying the components (weak acid + conjugate base, or weak base + conjugate acid)

(b)(i) [3 marks]

Step 1: Calculate moles of each component after mixing.

• Moles of $HCOOH = 0.100 \times 0.50 = 0.050$ mol
• Moles of $HCOONa = 0.100 \times 0.30 = 0.030$ mol
• Total volume $= 100 + 100 = 200$ cm³ $= 0.200$ dm³

Step 2: Calculate concentrations in the mixture.

• $[HCOOH] = \frac{0.050}{0.200} = 0.25$ mol dm$^{-3}$
• $[HCOO^-] = \frac{0.030}{0.200} = 0.15$ mol dm$^{-3}$

Step 3: Apply the Henderson-Hasselbalch equation.

$pH = pK_a + \log\frac{[A^-]}{[HA]}$

$pK_a = -\log(1.8 \times 10^{-4}) = 3.74$

$pH = 3.74 + \log\frac{0.15}{0.25} = 3.74 + \log(0.60) = 3.74 + (-0.22) = 3.52$

Marking:

• [1] for correct moles/concentrations
• [1] for correct $pK_a$ and correct substitution into Henderson-Hasselbalch
• [1] for correct final answer (pH = 3.52)

(b)(ii) [3 marks]

When a small amount of dilute $HCl$ is added, the $H^+$ ions from $HCl$ react with the methanoate ions ($HCOO^-$) in the buffer:

$HCOO^-(aq) + H^+(aq) \rightarrow HCOOH(aq)$

This removes the added $H^+$ ions by converting them into un-ionised $HCOOH$. The equilibrium:

$HCOOH(aq) \rightleftharpoons H^+(aq) + HCOO^-(aq)$

shifts to the left (Le Chatelier's principle) as $HCOO^-$ is consumed. Since the buffer contains relatively large reservoirs of both $HCOOH$ and $HCOO^-$, the ratio $[HCOO^-]/[HCOOH]$ changes only slightly, and hence the pH remains nearly constant.

Marking:

• [1] for stating that added $H^+$ reacts with $HCOO^-$
• [1] for the equation showing the reaction
• [1] for explaining that the ratio changes little / reference to Le Chatelier's principle

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17. Solubility Product and Precipitation [8 marks]

(a) [1 mark]

$K_{sp} = [Ba^{2+}][SO_4^{2-}]$

Marking:

• [1] for correct expression (no state symbols needed, no solids included)

(b) [2 marks]

$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$

Let solubility $= s$ mol dm$^{-3}$

$[Ba^{2+}] = s$ and $[SO_4^{2-}] = s$

$K_{sp} = s \times s = s^2 = 1.1 \times 10^{-10}$

$s = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \approx 1.0 \times 10^{-5} \text{ mol dm}^{-3}$

Marking:

• [1] for correct setup ($s^2 = K_{sp}$)
• [1] for correct answer ($1.0 \times 10^{-5}$ or $1.05 \times 10^{-5}$ mol dm$^{-3}$)

(c) [3 marks]

In 0.10 mol dm$^{-3}$ $Na_2SO_4$, $[SO_4^{2-}] \approx 0.10$ mol dm$^{-3}$ (from the dissolved salt, assuming the contribution from dissolved $BaSO_4$ is negligible).

$K_{sp} = [Ba^{2+}][SO_4^{2-}]$

$1.1 \times 10^{-10} = [Ba^{2+}] \times 0.10$

$[Ba^{2+}] = \frac{1.1 \times 10^{-10}}{0.10} = 1.1 \times 10^{-9} \text{ mol dm}^{-3}$

The solubility is $1.1 \times 10^{-9}$ mol dm$^{-3}$, which is much lower than in pure water. This is due to the common ion effect: the presence of $SO_4^{2-}$ from $Na_2SO_4$ shifts the equilibrium $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$ to the left (Le Chatelier's principle), suppressing the dissolution of $BaSO_4$.

Marking:

• [1] for correct calculation
• [1] for identifying the common ion effect
• [1] for explaining the shift in equilibrium

(d) [2 marks]

After mixing equal volumes, the concentrations are halved:

• $[Ba^{2+}] = \frac{0.0010}{2} = 5.0 \times 10^{-4}$ mol dm$^{-3}$
• $[SO_4^{2-}] = \frac{0.0010}{2} = 5.0 \times 10^{-4}$ mol dm$^{-3}$

Ionic product $= [Ba^{2+}][SO_4^{2-}] = (5.0 \times 10^{-4})^2 = 2.5 \times 10^{-7}$

Since ionic product ($2.5 \times 10^{-7}$) $> K_{sp}$ ($1.1 \times 10^{-10}$), a precipitate of $BaSO_4$ will form.

Marking:

• [1] for correct calculation of ionic product
• [1] for correct comparison with $K_{sp}$ and conclusion

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18. Acid-Base Titrations and Indicators [8 marks]

(a) [3 marks]

Titration	Rough	1	2	3
Final reading / cm³	24.80	24.35	24.30	24.40
Initial reading / cm³	0.00	0.00	0.00	0.00
Titre / cm³	24.80	24.35	24.30	24.40

Titrations 2 and 3 are concordant (within 0.10 cm³ of each other: $|24.35 - 24.30| = 0.05$ cm³ ✓; $|24.35 - 24.40| = 0.05$ cm³ ✓).

All three accurate titrations (1, 2, 3) are within 0.10 cm³ of each other, so all three can be used.

$\text{Mean titre} = \frac{24.35 + 24.30 + 24.40}{3} = \frac{73.05}{3} = 24.35 \text{ cm}^3$

Marking:

• [1] for correct table format with all readings
• [1] for identifying concordant titres and excluding the rough titration
• [1] for correct mean titre calculation (24.35 cm³)

(b) [2 marks]

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Moles of $NaOH = 0.02435 \times 0.100 = 2.435 \times 10^{-3}$ mol

Moles of $CH_3COOH = 2.435 \times 10^{-3}$ mol (1:1 ratio)

$[CH_3COOH] = \frac{2.435 \times 10^{-3}}{0.0250} = 0.0974 \text{ mol dm}^{-3}$

Marking:

• [1] for correct moles calculation
• [1] for correct concentration (0.0974 mol dm$^{-3}$)

(c) [3 marks]

This is a weak acid ($CH_3COOH$) – strong base ($NaOH$) titration. At the equivalence point, the solution contains $CH_3COONa$, which hydrolyses to give a solution with pH > 7 (approximately pH 8.7–9.0). The steep portion of the titration curve around the equivalence point spans approximately pH 7 to pH 10.

Phenolphthalein changes colour in the pH range 8.2–10.0, which falls within the steep pH change region around the equivalence point. Therefore, the colour change from colourless to pink will occur sharply at the equivalence point, giving an accurate result.

Marking:

• [1] for identifying this as a weak acid–strong base titration with equivalence point pH > 7
• [1] for stating the pH range of phenolphthalein (8.2–10.0)
• [1] for explaining that this range falls within the steep region of the curve

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19. Salt Hydrolysis and pH [6 marks]

(a) [1 mark]

Salt hydrolysis is the reaction of the ions of a salt with water to produce $H^+$ or $OH^-$ ions, resulting in a solution that is not neutral (i.e., pH ≠ 7).

Marking:

• [1] for a clear definition involving reaction of salt ions with water producing $H^+$ or $OH^-$

(b)(i) $AlCl_3$ — Acidic [2 marks]

$AlCl_3$ is a salt of a strong acid ($HCl$) and a weak base ($Al(OH)_3$). The $Al^{3+}$ ion undergoes hydrolysis:

$Al^{3+}(aq) + 3H_2O(l) \rightleftharpoons Al(OH)_3(s) + 3H^+(aq)$

Or more precisely (stepwise):

$[Al(H_2O)_6]^{3+}(aq) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}(aq) + H^+(aq)$

The highly charged $Al^{3+}$ ion polarises the coordinated water molecules, making it easier for a proton to be released. This produces $H^+$ ions, making the solution acidic.

Marking:

• [1] for correct prediction (acidic)
• [1] for correct hydrolysis equation

(b)(ii) $K_2CO_3$ — Basic [2 marks]

$K_2CO_3$ is a salt of a strong base ($KOH$) and a weak acid ($H_2CO_3$). The $CO_3^{2-}$ ion undergoes hydrolysis:

$CO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HCO_3^-(aq) + OH^-(aq)$

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

Marking:

• [1] for correct prediction (basic)
• [1] for correct hydrolysis equation

(b)(iii) $NaNO_3$ — Neutral [1 mark]

$NaNO_3$ is a salt of a strong acid ($HNO_3$) and a strong base ($NaOH$). Neither $Na^+$ nor $NO_3^-$ undergoes hydrolysis, so the solution is neutral (pH = 7).

Marking:

• [1] for correct prediction (neutral) with valid reasoning

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Section C: Data Interpretation and Application

20. Polyprotic Acids and Titration Curves [15 marks]

(a) [3 marks]

The first equivalence point occurs when all $H_3PO_4$ has been converted to $H_2PO_4^-$:

$H_3PO_4 + NaOH \rightarrow NaH_2PO_4 + H_2O$

At this point, the solution contains $NaH_2PO_4$ (the amphoteric species $H_2PO_4^-$). The pH is determined by the amphoteric behaviour of $H_2PO_4^-$, which can act as both an acid and a base. For an amphoteric species, the pH is approximately:

$pH \approx \frac{pK_{a1} + pK_{a2}}{2}$

$pK_{a1} = -\log(7.5 \times 10^{-3}) = 2.12$

$pK_{a2} = -\log(6.2 \times 10^{-8}) = 7.21$

$pH \approx \frac{2.12 + 7.21}{2} = \frac{9.33}{2} = 4.67 \approx 4.7$

The pH is not 7 because $H_2PO_4^-$ is not a neutral salt — it is an amphoteric ion whose pH depends on the average of $pK_{a1}$ and $pK_{a2}$.

Marking:

• [1] for identifying the species present at the first equivalence point ($H_2PO_4^-$)
• [1] for recognising it is amphoteric
• [1] for the calculation using $(pK_{a1} + pK_{a2})/2$

(b) [3 marks]

The third ionisation constant $K_{a3} = 4.8 \times 10^{-13}$ is extremely small, meaning $HPO_4^{2-}$ is an extremely weak acid. The third equivalence point would require:

$HPO_4^{2-} + NaOH \rightarrow Na_3PO_4 + H_2O$

At the third equivalence point, the solution would contain $PO_4^{3-}$, which is a relatively strong conjugate base (since $K_{a3}$ is very small, $K_b$ for $PO_4^{3-}$ is large). The pH at the third equivalence point would be very high (>12), and the pH change would be very gradual because the very weak third dissociation provides very little buffering capacity. Additionally, at such high pH, the effect of atmospheric $CO_2$ dissolution becomes significant, making the endpoint indistinct. The very small $K_{a3}$ means the third buffer region has very low capacity, and the steep pH rise is not sharp enough to detect with an indicator.

Marking:

• [1] for noting that $K_{a3}$ is extremely small
• [1] for explaining that the third equivalence point pH would be very high / indistinct
• [1] for explaining the lack of a sharp pH change / buffer capacity issue

(c) [2 marks]

At the first half-equivalence point, exactly half of the $H_3PO_4$ has been neutralised:

$[H_3PO_4] = [H_2PO_4^-]$

Using the Henderson-Hasselbalch equation:

$pH = pK_{a1} + \log\frac{[H_2PO_4^-]}{H_3PO_4]}$

Since $[H_2PO_4^-] = [H_3PO_4]$, the ratio $= 1$, and $\log(1) = 0$.

Therefore: $pH = pK_{a1} + 0 = pK_{a1}$

Marking:

• [1] for stating that $[H_3PO_4] = [H_2PO_4^-]$ at the half-equivalence point
• [1] for applying Henderson-Hasselbalch and showing $\log(1) = 0$

(d) [2 marks]

At the second equivalence point, all $H_2PO_4^-$ has been converted to $HPO_4^{2-}$. The pH at this point is approximately:

$pH \approx \frac{pK_{a2} + pK_{a3}}{2} = \frac{7.21 + 12.32}{2} = \frac{19.53}{2} = 9.77$

Phenolphthalein changes colour in the range pH 8.2–10.0. Since the equivalence point pH (~9.77) falls within this range, and the steep portion of the titration curve around the second equivalence point also falls within this range, phenolphthalein is a suitable indicator for the second equivalence point.

Marking:

• [1] for calculating or estimating the pH at the second equivalence point
• [1] for concluding that phenolphthalein is suitable because its range overlaps with the equivalence point pH

(e) [3 marks]

At the second equivalence point, all the original $H_3PO_4$ has been converted to $HPO_4^{2-}$.

Moles of $H_3PO_4$ initially $= 0.0250 \times 0.100 = 2.50 \times 10^{-3}$ mol

From the stoichiometry:

• First equivalence: $H_3PO_4 + NaOH \rightarrow NaH_2PO_4 + H_2O$ (1:1)
• Second equivalence: $NaH_2PO_4 + NaOH \rightarrow Na_2HPO_4 + H_2O$ (1:1)

At the second equivalence point, moles of $HPO_4^{2-} = 2.50 \times 10^{-3}$ mol

Total volume $= 25.0 + 50.0 = 75.0$ cm³ $= 0.0750$ dm³

$[HPO_4^{2-}] = \frac{2.50 \times 10^{-3}}{0.0750} = 0.0333 \text{ mol dm}^{-3}$

Marking:

• [1] for correct moles of $HPO_4^{2-}$
• [1] for correct total volume
• [1] for correct concentration (0.0333 mol dm$^{-3}$)

(f) [2 marks]

$K_{a2} = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$

For the solution containing 0.10 mol dm$^{-3}$ $NaH_2PO_4$ and 0.10 mol dm$^{-3}$ $Na_2HPO_4$:

Using the Henderson-Hasselbalch equation:

$pH = pK_{a2} + \log\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}$

$pK_{a2} = -\log(6.2 \times 10^{-8}) = 7.21$

Since $[HPO_4^{2-}] = [H_2PO_4^-] = 0.10$ mol dm$^{-3}$:

$pH = 7.21 + \log\frac{0.10}{0.10} = 7.21 + \log(1) = 7.21 + 0 = 7.21$

Marking:

• [1] for correct $K_{a2}$ expression and $pK_{a2}$ value
• [1] for correct pH calculation (7.21)

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Mark Summary:

Section	Marks
A: Q1–15 (MCQ)	15
B: Q16	8
B: Q17	8
B: Q18	8
B: Q19	6
C: Q20	15
Total	60