DSE Chemistry Diagnostic: Acids, Bases and Salts

Unit Test 1: Weak Acid pH Calculation

Question

Ethanoic acid ($CH_{3}COOH$) is a weak acid with $K_{a} = 1.8 \times 10^{-5}$ mol/dm$^{3}$ at 25$^{\circ}$C.

(a) Calculate the pH of a 0.10 mol/dm$^{3}$ solution of ethanoic acid. [4 marks]

(b) Calculate the pH of a 0.010 mol/dm$^{3}$ solution of ethanoic acid. [2 marks]

(c) A student claims that diluting a weak acid by a factor of 10 will increase the pH by exactly 1. Evaluate this claim by comparing your answers to (a) and (b). [2 marks]

Worked Solution

(a) $CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$

$K_{a} = \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}$

At equilibrium, let $[H^{+}] = x$:

$1.8 \times 10^{-5} = \frac{x \cdot x}{0.10 - x}$

Assuming $x \ll 0.10$ (i.e., dissociation is small):

$1.8 \times 10^{-5} \approx \frac{x^{2}}{0.10}$

$x^{2} = 1.8 \times 10^{-6}$

$x = 1.34 \times 10^{-3} \text{ mol/dm}^{3}$

Check: $x/0.10 = 1.34\% \lt 5\%$ — assumption valid.

$pH = -\log(1.34 \times 10^{-3}) = 2.87$

(b) $1.8 \times 10^{-5} = \frac{x^{2}}{0.010}$

$x^{2} = 1.8 \times 10^{-7}$

$x = 4.24 \times 10^{-4} \text{ mol/dm}^{3}$

$pH = -\log(4.24 \times 10^{-4}) = 3.37$

(c) The pH change: $3.37 - 2.87 = 0.50$.

The claim is incorrect for weak acids. Diluting by a factor of 10 increases the pH by less than 1. This is because dilution shifts the equilibrium to the right (Le Chatelier”s principle), causing a greater fraction of the acid to dissociate. The $[H^{+}]$ does not decrease by a full factor of 10.

(For a strong acid, the claim would be correct since $[H^{+}]$ directly decreases by a factor of 10, increasing pH by exactly 1.)

Unit Test 2: Buffer Solution

Question

A buffer solution is prepared by mixing 100 cm$^{3}$ of 0.20 mol/dm$^{3}$ ethanoic acid ($CH_{3}COOH$, $K_{a} = 1.8 \times 10^{-5}$) with 100 cm$^{3}$ of 0.10 mol/dm$^{3}$ sodium ethanoate ($CH_{3}COONa$).

(a) Calculate the pH of this buffer solution. [4 marks]

(b) Calculate the new pH after adding 5.0 cm$^{3}$ of 0.10 mol/dm$^{3}$ HCl to 50.0 cm$^{3}$ of the buffer. [4 marks]

(c) Explain why this buffer resists changes in pH when a small amount of strong acid is added. [2 marks]

Worked Solution

(a) After mixing, total volume = 200 cm$^{3}$.

$[CH_{3}COOH] = \frac{0.20 \times 100}{200} = 0.10 \text{ mol/dm}^{3}$

$[CH_{3}COO^{-}] = \frac{0.10 \times 100}{200} = 0.050 \text{ mol/dm}^{3}$

Using the Henderson-Hasselbalch equation:

$pH = pK_{a} + \log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$pK_{a} = -\log(1.8 \times 10^{-5}) = 4.74$

$pH = 4.74 + \log\frac{0.050}{0.10} = 4.74 + \log(0.5) = 4.74 - 0.30 = 4.44$

(b) In 50.0 cm$^{3}$ of buffer:

$n(CH_{3}COOH) = 0.10 \times 0.050 = 0.00500 \text{ mol}$

$n(CH_{3}COO^{-}) = 0.050 \times 0.050 = 0.00250 \text{ mol}$

Moles of $HCl$ added: $n(HCl) = 0.10 \times 0.0050 = 0.000500$ mol

The $H^{+}$ from HCl reacts with $CH_{3}COO^{-}$:

CH_{3}COO^{-} + H^{+} \rightarrow CH_{3}COOH}

New moles:

$n(CH_{3}COO^{-}) = 0.00250 - 0.000500 = 0.00200 \text{ mol}$

$n(CH_{3}COOH) = 0.00500 + 0.000500 = 0.00550 \text{ mol}$

New total volume = $50.0 + 5.0 = 55.0$ cm$^{3}$:

$[CH_{3}COOH] = \frac{0.00550}{0.0550} = 0.100 \text{ mol/dm}^{3}$

$[CH_{3}COO^{-}] = \frac{0.00200}{0.0550} = 0.0364 \text{ mol/dm}^{3}$

$pH = 4.74 + \log\frac{0.0364}{0.100} = 4.74 + \log(0.364) = 4.74 - 0.439 = 4.30$

(c) The buffer contains a weak acid ($CH_{3}COOH$) and its conjugate base ($CH_{3}COO^{-}$). When a strong acid ($H^{+}$) is added, the $H^{+}$ ions react with $CH_{3}COO^{-}$ to form $CH_{3}COOH$Consuming most of the added $H^{+}$ and preventing a significant drop in pH.

Unit Test 3: Salt Hydrolysis

Question

(a) Predict whether an aqueous solution of ammonium chloride ($NH_{4}Cl$) is acidic, alkaline, or neutral. Explain your answer using the concept of hydrolysis. [3 marks]

(b) Predict whether an aqueous solution of sodium ethanoate ($CH_{3}COONa$) is acidic, alkaline, or neutral. Explain. [3 marks]

(c) Predict whether an aqueous solution of sodium chloride ($NaCl$) is acidic, alkaline, or neutral. Explain. [1 mark]

Worked Solution

(a) $NH_{4}Cl$ solution is acidic ($pH \lt 7$).

$NH_{4}Cl$ dissociates completely in water: $NH_{4}Cl \rightarrow NH_{4}^{+} + Cl^{-}$.

The ammonium ion ($NH_{4}^{+}$) is the conjugate acid of the weak base ammonia ($NH_{3}$). It undergoes hydrolysis:

$NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{3} + H_{3}O^{+}$

This reaction releases $H_{3}O^{+}$ ions, making the solution acidic. The chloride ion ($Cl^{-}$) is the conjugate base of a strong acid ($HCl$) and does not hydrolyse.

(b) $CH_{3}COONa$ solution is alkaline ($pH \gt 7$).

$CH_{3}COONa$ dissociates completely: $CH_{3}COONa \rightarrow CH_{3}COO^{-} + Na^{+}$.

The ethanoate ion ($CH_{3}COO^{-}$) is the conjugate base of the weak acid ethanoic acid ($CH_{3}COOH$). It undergoes hydrolysis:

$CH_{3}COO^{-} + H_{2}O \rightleftharpoons CH_{3}COOH + OH^{-}$

This reaction produces $OH^{-}$ ions, making the solution alkaline. The sodium ion ($Na^{+}$) does not hydrolyse.

(c) $NaCl$ solution is neutral ($pH = 7$).

Both $Na^{+}$ (conjugate acid of the strong base $NaOH$) and $Cl^{-}$ (conjugate base of the strong acid $HCl$) do not undergo hydrolysis. Neither ion affects the $pH$ of the solution.

Integration Test 1: Titration Curve + Indicator Choice

Question

25.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ ammonia solution ($NH_{3}$, $K_{b} = 1.8 \times 10^{-5}$) is titrated with 0.100 mol/dm$^{3}$ hydrochloric acid.

(a) Calculate the pH of the ammonia solution before any acid is added. [3 marks]

(b) Calculate the pH at the equivalence point. [3 marks]

(c) State and explain the most suitable indicator for this titration. [2 marks]

Worked Solution

(a) $NH_{3} + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$

$K_{b} = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.8 \times 10^{-5}$

Let $[OH^{-}] = x$:

$1.8 \times 10^{-5} = \frac{x^{2}}{0.100}$

$x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ mol/dm}^{3}$

$pOH = -\log(1.34 \times 10^{-3}) = 2.87$

$pH = 14 - 2.87 = 11.13$

(b) At the equivalence point, all $NH_{3}$ has been converted to $NH_{4}^{+}$.

Moles of $NH_{3}$ = $0.100 \times 0.0250 = 0.00250$ mol

Volume of HCl needed = $25.0$ cm$^{3}$ (equimolar), total volume = $50.0$ cm$^{3}$.

$[NH_{4}^{+}] = \frac{0.00250}{0.0500} = 0.0500 \text{ mol/dm}^{3}$

$NH_{4}^{+}$ hydrolyses: $NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{3} + H_{3}O^{+}$

$K_{a} = \frac{K_{w}}{K_{b}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

$5.56 \times 10^{-10} = \frac{x^{2}}{0.0500}$

$x = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}$

$pH = -\log(5.27 \times 10^{-6}) = 5.28$

(c) The equivalence point pH is 5.28, which is acidic. The most suitable indicator is one whose colour change range includes pH 5.28. Methyl orange (pH range 3.1—4.4) is too low. Bromocresol green (pH range 3.8—5.4) would be suitable. Methyl red (pH range 4.4—6.2) is also a good choice. Phenolphthalein (pH range 8.3—10.0) would NOT be suitable as the colour change occurs well above the equivalence point pH.

Integration Test 2: pH Mixing and Neutralisation

Question

(a) Calculate the pH when 10.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ NaOH is added to 40.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ HCl. [3 marks]

(b) Calculate the pH when 30.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ NaOH is added to 40.0 cm$^{3}$ of 0.100 mol/dm$^{3}$ HCl. [3 marks]

(c) Explain why the pH changes much more dramatically between the two scenarios in (a) and (b) than between adding 10.0 cm$^{3}$ and 20.0 cm$^{3}$ of NaOH. [2 marks]

Worked Solution

(a) Moles of $HCl$: $0.100 \times 0.0400 = 0.00400$ mol

Moles of NaOH: $0.100 \times 0.0100 = 0.00100$ mol

$HCl$ is in excess by: $0.00400 - 0.00100 = 0.00300$ mol

Total volume = $40.0 + 10.0 = 50.0$ cm$^{3}$

$[H^{+}] = \frac{0.00300}{0.0500} = 0.0600 \text{ mol/dm}^{3}$

$pH = -\log(0.0600) = 1.22$

(b) Moles of $HCl$: $0.00400$ mol

Moles of NaOH: $0.100 \times 0.0300 = 0.00300$ mol

$HCl$ is in excess by: $0.00400 - 0.00300 = 0.00100$ mol

Total volume = $40.0 + 30.0 = 70.0$ cm$^{3}$

$[H^{+}] = \frac{0.00100}{0.0700} = 0.01429 \text{ mol/dm}^{3}$

$pH = -\log(0.01429) = 1.85$

(c) The pH changes from 1.22 to 1.85 (a change of 0.63) when NaOH added increases from 10.0 to 30.0 cm$^{3}$. The change per 10 cm$^{3}$ is relatively small because the solution still contains a large excess of $H^{+}$ (a strong acid). The pH is in the region where the logarithmic scale compresses large changes in $[H^{+}]$ into small changes in pH.

However, near the equivalence point (at 40.0 cm$^{3}$ of NaOH), adding even a small amount of NaOH causes a dramatic pH change because the $[H^{+}]$ approaches very small values where the logarithmic relationship amplifies the change. This is the characteristic steep region of a strong acid-strong base titration curve.

Integration Test 3: Buffer Preparation and Capacity

Question

(a) Describe how you would prepare 250 cm$^{3}$ of an ethanoic acid / sodium ethanoate buffer with pH = 5.00, using 0.50 mol/dm$^{3}$ ethanoic acid and solid sodium ethanoate ($M = 82.0$ g/mol). ($K_{a} = 1.8 \times 10^{-5}$) [4 marks]

(b) Calculate the mass of sodium ethanoate required. [2 marks]

(c) Explain what is meant by buffer capacity and state how it can be increased. [2 marks]

Worked Solution

(a) Using the Henderson-Hasselbalch equation:

$pH = pK_{a} + \log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$5.00 = 4.74 + \log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$\log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]} = 0.26$

$\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]} = 10^{0.26} = 1.82$

The buffer needs $[CH_{3}COO^{-}]/[CH_{3}COOH] = 1.82$.

To prepare 250 cm$^{3}$ (0.250 dm$^{3}$): choose $[CH_{3}COOH] = 0.20$ mol/dm$^{3}$ (from the 0.50 mol/dm$^{3}$ stock by dilution).

Volume of stock needed: $V = \frac{0.20 \times 250}{0.50} = 100$ cm$^{3}$ of 0.50 mol/dm$^{3}$ $CH_{3}COOH$Diluted to 250 cm$^{3}$.

$[CH_{3}COO^{-}] = 1.82 \times 0.20 = 0.364$ mol/dm$^{3}$

$n(CH_{3}COO^{-}) = 0.364 \times 0.250 = 0.0910$ mol

(b) $m(CH_{3}COONa) = 0.0910 \times 82.0 = 7.46 \text{ g}$

Procedure: Dissolve 7.46 g of sodium ethanoate in approximately 150 cm$^{3}$ of distilled water. Add 100 cm$^{3}$ of 0.50 mol/dm$^{3}$ ethanoic acid. Transfer to a 250 cm$^{3}$ volumetric flask and make up to the mark with distilled water. Mix thoroughly.

(c) Buffer capacity is the amount of strong acid or strong base that can be added to a buffer solution before the pH changes significantly ( defined as the amount needed to change the pH by 1 unit).

Buffer capacity can be increased by:

1. Increasing the total concentration of the weak acid and its conjugate base (higher concentrations provide more $H^{+}$ or $OH^{-}$ to absorb).
2. Keeping the ratio $[A^{-}]/[HA]$ close to 1 (maximum buffer capacity occurs when $pH = pK_{a}$).