A Level H2 Chemistry Acids Bases Salts Quiz

A-Level Chemistry H2 Quiz - Acids Bases Salts

Answer Key

Question 1

(b) $SO_4^{2-}$ [1]

Explanation: A conjugate base is formed when a Brønsted–Lowry acid donates a proton ($H^+$). $HSO_4^-$ loses one $H^+$ to become $SO_4^{2-}$. Option (a) is the conjugate acid (gaining a proton), (c) is the conjugate base of water, and (d) is unrelated.

Question 2

(b) $3.98 \times 10^{-11}$ mol dm$^{-3}$ [1]

Explanation:

• $pH = 3.40$, so $[H^+] = 10^{-3.40} = 3.98 \times 10^{-4}$ mol dm$^{-3}$
• At 25 °C: $K_w = [H^+][OH^-] = 1.00 \times 10^{-14}$
• $[OH^-] = \frac{K_w}{[H^+]} = \frac{1.00 \times 10^{-14}}{3.98 \times 10^{-4}} = 2.51 \times 10^{-11}$ mol dm$^{-3}$

Correction: The correct answer is (c) $2.51 \times 10^{-11}$ mol dm$^{-3}$.

Teaching note: Students must remember that $K_w = 1.00 \times 10^{-14}$ at 25 °C and use $[OH^-] = K_w / [H^+]$. A common error is to simply take $10^{-pH}$ and call it $[OH^-]$.

Question 3

(b) $NH_4Cl$ [1]

Explanation: $NH_4Cl$ is a salt of a weak base ($NH_3$) and a strong acid ($HCl$). The $NH_4^+$ ion hydrolyses in water: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$, producing $H_3O^+$ and making the solution acidic. $Na_2CO_3$ and $CH_3COONa$ are salts of strong bases and weak acids (basic solutions). $KNO_3$ is a salt of a strong acid and strong base (neutral).

Question 4

(b) 2.88 [1]

Explanation:

• For a weak acid: $[H^+] = \sqrt{K_a \times c} = \sqrt{1.74 \times 10^{-5} \times 0.100} = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3}$ mol dm$^{-3}$
• $pH = -\log(1.32 \times 10^{-3}) = 2.88$

Teaching note: The approximation $[H^+] = \sqrt{K_a \cdot c}$ assumes that dissociation is small, which is valid when $c \gg K_a$. Students should always check that the dissociation is less than 5% for this approximation to be valid.

Question 5

(c) Phenolphthalein (pH range 8.2–10.0) [1]

Explanation: The titration of a weak acid with a strong base produces a basic salt at the equivalence point (pH > 7). The equivalence point occurs in the basic region, so phenolphthalein, which changes colour in the pH range 8.2–10.0, is the most appropriate indicator. Methyl orange changes colour in the acidic range and would give a premature endpoint.

Question 6

Definition: A Brønsted–Lowry acid is a proton ($H^+$) donor. [1]

Example: Water ($H_2O$) — it can donate a proton to become $OH^-$ (acting as an acid) or accept a proton to become $H_3O^+$ (acting as a base). [1]

Teaching note: Other acceptable examples include $HCO_3^-$, $HSO_4^-$, $H_2PO_4^-$, and $HPO_4^{2-}$. These are amphiprotic species.

Question 7

$NaCl$ is formed from a strong acid ($HCl$) and a strong base ($NaOH$). Neither ion hydrolyses in water, so the solution remains neutral (pH = 7). [1]

$CH_3COONa$ is formed from a weak acid ($CH_3COOH$) and a strong base ($NaOH$). The ethanoate ion ($CH_3COO^-$) is the conjugate base of a weak acid and undergoes hydrolysis: [1]

$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ [1]

This produces $OH^-$ ions, making the solution basic (pH > 7). [1]

Question 8

$K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}$ [1]

Teaching note: The $K_a$ expression excludes pure liquids and solids. Water is omitted because it is the solvent. Students should remember to write the products (ions) in the numerator and the undissociated acid in the denominator.

Question 9

$pH = 3.00$, so $[H^+] = 10^{-3.00} = 1.00 \times 10^{-3}$ mol dm$^{-3}$ [1]

For a weak monobasic acid HA: $K_a = \frac{[H^+]^2}{c - [H^+]}$

Since $[H^+] = 1.00 \times 10^{-3}$ and $c = 0.050$:

$K_a = \frac{(1.00 \times 10^{-3})^2}{0.050 - 1.00 \times 10^{-3}} = \frac{1.00 \times 10^{-6}}{0.049} = 2.04 \times 10^{-5} \text{ mol dm}^{-3}$

Using the approximation (since dissociation is small):

$K_a \approx \frac{(1.00 \times 10^{-3})^2}{0.050} = 2.0 \times 10^{-5} \text{ mol dm}^{-3}$ [1]

Marking: 1 mark for correct $[H^+]$, 1 mark for correct $K_a$ value.

Question 10

Assumption 1: The concentration of $H^+$ ions from the dissociation of water is negligible compared to that from the acid. [1]

Assumption 2: The degree of dissociation is small, so the equilibrium concentration of the undissociated acid is approximately equal to its initial concentration (i.e., $[HA]_{eq} \approx [HA]_{initial}$). [1]

Teaching note: These assumptions allow the simplification $K_a = \frac{[H^+]^2}{c}$ instead of $K_a = \frac{[H^+]^2}{c - [H^+]}$. The approximation is generally valid when the percentage dissociation is less than 5%.

Question 11

(a) Titration 3 (24.80 cm³) is anomalous. [1] It differs significantly from titres 1 and 2 (23.95 and 23.90 cm³), which are concordant (within 0.10 cm³ of each other). The anomalous result should be excluded because it is not concordant with the other reliable titres.

(b) Mean volume $= \frac{23.95 + 23.90}{2} = 23.925 = 23.93$ cm³ (to 2 d.p.) [1]

(c) $NaOH + HCl \rightarrow NaCl + H_2O$ [1]

Moles of $NaOH = 0.100 \times \frac{23.93}{1000} = 2.393 \times 10^{-3}$ mol

Since the ratio is 1:1, moles of $HCl = 2.393 \times 10^{-3}$ mol [1]

Concentration of $HCl = \frac{2.393 \times 10^{-3}}{25.0/1000} = 0.0957$ mol dm$^{-3}$ [1]

(d) Colourless to pink (or colourless to pale pink). [1]

Teaching note: Phenolphthalein is colourless in acidic solution and pink in basic solution. At the endpoint of an acid–base titration with NaOH, the solution turns from colourless to pink.

Question 12

(a) $K_a = \frac{[H^+][X^-]}{[HX]}$ [1]

(b) $[H^+] = 10^{-2.72} = 1.91 \times 10^{-3}$ mol dm$^{-3}$ [1]

(c) For a weak acid, $[H^+] = [X^-]$ at equilibrium.

$K_a = \frac{(1.91 \times 10^{-3})^2}{0.200 - 1.91 \times 10^{-3}} = \frac{3.65 \times 10^{-6}}{0.1981} = 1.84 \times 10^{-5} \text{ mol dm}^{-3}$

Using the approximation:

$K_a \approx \frac{(1.91 \times 10^{-3})^2}{0.200} = 1.82 \times 10^{-5} \text{ mol dm}^{-3}$

Answer: $K_a = 1.82 \times 10^{-5}$ mol dm$^{-3}$ (to 3 s.f.) [2]

(d) Percentage dissociation $= \frac{[H^+]}{c} \times 100\% = \frac{1.91 \times 10^{-3}}{0.200} \times 100\% = 0.955\%$ [1]

Question 13

(a) After mixing equal volumes, the concentration of each component is halved:

$[CH_3COOH] = [CH_3COO^-] = \frac{0.100}{2} = 0.050$ mol dm$^{-3}$

Using the Henderson–Hasselbalch equation:

$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

$pK_a = -\log(1.74 \times 10^{-5}) = 4.76$ [1]

Since $[CH_3COO^-] = [CH_3COOH]$, $\log(1) = 0$

$pH = 4.76 + 0 = 4.76$ [1]

(b) The buffer contains $CH_3COOH$ (the weak acid) and $CH_3COO^-$ (the conjugate base). When HCl is added, the $H^+$ ions from HCl react with the ethanoate ions: [1]

$CH_3COO^- + H^+ \rightarrow CH_3COOH$ [1]

This removes the added $H^+$ ions by converting them into undissociated ethanoic acid, so the pH remains approximately constant. [1] The equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$ shifts to the left, consuming the added $H^+$. [1]

(c) The pH would increase. [1] Adding more sodium ethanoate increases $[CH_3COO^-]$. From the Henderson–Hasselbalch equation, increasing the ratio $\frac{[CH_3COO^-]}{[CH_3COOH]}$ increases the $\log$ term, thus increasing the pH. [1]

Question 14

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

(b) Let the solubility of $Mg(OH)_2 = s$ mol dm$^{-3}$.

$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$

$[Mg^{2+}] = s$, $[OH^-] = 2s$

$K_{sp} = s \times (2s)^2 = 4s^3 = 5.61 \times 10^{-12}$ [1]

$s^3 = \frac{5.61 \times 10^{-12}}{4} = 1.4025 \times 10^{-12}$

$s = \sqrt[3]{1.4025 \times 10^{-12}} = 1.12 \times 10^{-4}$ mol dm$^{-3}$ [1]

(c) $[OH^-] = 2s = 2 \times 1.12 \times 10^{-4} = 2.24 \times 10^{-4}$ mol dm$^{-1}$

$pOH = -\log(2.24 \times 10^{-4}) = 3.65$ [1]

$pH = 14.00 - 3.65 = 10.35$ [1]

Question 15

(a) Using the Henderson–Hasselbalch equation:

$pH = pK_a + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$

$5.00 = 4.76 + \log\frac{[CH_3COO^-]}{[CH_3COOH]}$ [1]

$\log\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.24$

$\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{0.24} = 1.74$ [1]

(b) Let $[CH_3COOH] = x$, then $[CH_3COO^-] = 1.74x$

$x + 1.74x = 0.200$

$2.74x = 0.200$

$x = 0.0730$ mol dm$^{-1}$ [1]

$[CH_3COOH] = 0.073$ mol dm$^{-3}$, $[CH_3COO^-] = 0.127$ mol dm$^{-3}$ [1]

(c) The buffer has a limited reservoir of $CH_3COO^-$ ions. If a large amount of strong acid is added, all the $CH_3COO^-$ will be consumed, and the buffer capacity is exceeded. After this point, any additional $H^+$ will cause a sharp decrease in pH. [1]

Question 16

(a) Curve III corresponds to the diprotic strong acid. [1] It has two equivalence points (at 12.5 cm³ and 25.0 cm³), which is characteristic of a diprotic acid where the two protons are neutralised in two stages. The first half-equivalence occurs at 12.5 cm³ (half of 25.0 cm³), and the second at 25.0 cm³. [1]

(b) Curve II represents a weak acid titrated with a strong base. At the equivalence point, the salt formed is the sodium salt of the weak acid (e.g., $CH_3COONa$). The conjugate base of the weak acid hydrolyses in water: [1]

$A^- + H_2O \rightleftharpoons HA + OH^-$

This produces $OH^-$ ions, making the solution basic, so the pH at the equivalence point is greater than 7. [1]

(c) $NaOH + HCl \rightarrow NaCl + H_2O$

Moles of $NaOH = 0.100 \times \frac{24.80}{1000} = 2.48 \times 10^{-3}$ mol [1]

Moles of $HCl = 2.48 \times 10^{-3}$ mol (1:1 ratio)

Concentration of $HCl = \frac{2.48 \times 10^{-3}}{25.0/1000} = 0.0992$ mol dm$^{-3}$ [1]

Question 17

(a) $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$ [1]

(b) The limewater turns milky / cloudy / white precipitate forms. [1]

(c) The red litmus paper turns blue. [1] Calcium oxide reacts with water to form calcium hydroxide: $CaO + H_2O \rightarrow Ca(OH)_2$. Calcium hydroxide is a base, so the solution is alkaline and turns red litmus blue. [1]

(d) $Ca(OH)_2$ is fully dissociated: $Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^-$

$[OH^-] = 2 \times 0.020 = 0.040$ mol dm$^{-3}$ [1]

$pOH = -\log(0.040) = 1.40$

$pH = 14.00 - 1.40 = 12.60$ [1]

Question 18

(a) Increasing acid strength: $CH_3COOH < ClCH_2COOH < FCH_2COOH$ [1]

The larger the $K_a$ value, the stronger the acid (greater degree of dissociation). $K_a$ values: $CH_3COOH$ ($1.74 \times 10^{-5}$) < $ClCH_2COOH$ ($1.38 \times 10^{-3}$) < $FCH_2COOH$ ($2.57 \times 10^{-3}$). [1]

(b) Fluorine is more electronegative than chlorine. [1] The $F$ atom exerts a stronger electron-withdrawing inductive effect through the carbon chain, which stabilises the conjugate base ($FCH_2COO^-$) more effectively than chlorine stabilises $ClCH_2COO^-$. [1] The greater the stabilisation of the conjugate base, the more the equilibrium favours dissociation, and the stronger the acid. [1]

(c) $K_a = 1.38 \times 10^{-3}$, $c = 0.100$ mol dm$^{-3}$

$[H^+] = \sqrt{K_a \times c} = \sqrt{1.38 \times 10^{-3} \times 0.100} = \sqrt{1.38 \times 10^{-4}} = 1.17 \times 10^{-2}$ mol dm$^{-3}$ [1]

$pH = -\log(1.17 \times 10^{-2}) = 1.93$ [1]

Note: The approximation is borderline here since dissociation is ~11.7%. For a more exact answer, solve the quadratic: $K_a = \frac{x^2}{0.100 - x}$, giving $x = 1.06 \times 10^{-2}$, $pH = 1.97$. Either method accepted.

(d) Fluoroacetic acid ($FCH_2COOH$) would have the highest conductivity. [1] It has the largest $K_a$ value, meaning it dissociates to the greatest extent in solution, producing the highest concentration of ions ($H^+$ and $FCH_2COO^-$), which results in the highest electrical conductivity. [1]

Question 19

(a) $CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)$ [1]

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$ [1]

(b) Moles of $NaOH = 0.500 \times \frac{28.40}{1000} = 1.42 \times 10^{-2}$ mol [1]

(c) From the 1:1 ratio, moles of excess $HCl = 1.42 \times 10^{-2}$ mol [1]

(d) Total moles of $HCl$ added $= 1.00 \times \frac{50.0}{1000} = 5.00 \times 10^{-2}$ mol

Moles of $HCl$ reacted with $CaCO_3 = 5.00 \times 10^{-2} - 1.42 \times 10^{-2} = 3.58 \times 10^{-2}$ mol [1]

(e) From the equation: $CaCO_3 : HCl = 1 : 2$

Moles of $CaCO_3 = \frac{3.58 \times 10^{-2}}{2} = 1.79 \times 10^{-2}$ mol [1]

Mass of $CaCO_3 = 1.79 \times 10^{-2} \times 100.1 = 1.79$ g [1]

Percentage of $CaCO_3 = \frac{1.79}{1.50} \times 100\% = 119\%$

Note: This result exceeds 100%, which suggests either the data is idealised or there is an error in the question values. In an exam context, students should report the calculated value. For a realistic answer, the numbers would need adjustment. Accept the calculated value: 119% (or note the inconsistency). [1]

Revised marking note: If the question is used in practice, adjust the NaOH titre to a more realistic value (e.g., 18.40 cm³) to give a reasonable percentage. For this answer key, the method is correct and the calculation follows through.

Question 20

(a) FA1: $Al^{3+}$ [1] — White precipitate with NaOH that dissolves in excess (forming $[Al(OH)_4]^-$) and no gas with warming confirms $Al^{3+}$. [1]

FA2: $Zn^{2+}$ [1] — White precipitate with NaOH that dissolves in excess (forming $[Zn(OH)_4]^{2-}$) and no gas with warming confirms $Zn^{2+}$. [1]

FA3: $NH_4^+$ [1] — No precipitate with NaOH, but warming produces a gas (ammonia) that turns damp red litmus blue, confirming $NH_4^+$. [1]

Note: FA1 and FA2 cannot be distinguished by these tests alone. Additional tests (e.g., with $NH_3$ solution) would be needed. For the purpose of this question, either assignment of $Al^{3+}$ to FA1 and $Zn^{2+}$ to FA2 (or vice versa) is acceptable if the reasoning is consistent. A common exam approach is to add a distinguishing test. Here, accept either assignment with correct reasoning.

(b) $Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)$ [1]

(c) $Zn(OH)_2(s) + 2OH^-(aq) \rightarrow [Zn(OH)_4]^{2-}(aq)$ [1]

Note: Alternatively, the overall equation: $Zn^{2+}(aq) + 4OH^-(aq) \rightarrow [Zn(OH)_4]^{2-}(aq)$

(d) $NH_4^+(aq) + OH^-(aq) \rightarrow NH_3(g) + H_2O(l)$ [1]

Total: 50 marks