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Question

50.00 cm³ of 0.75 mol dm⁻³ sodium hydroxide was added in 1.00 cm³ portions to 22.50 cm³ of 0.50 mol dm⁻³ chloroethanoic acid.

Calculate the initial pH before any sodium hydroxide was added, using section 21 of the data booklet.

[2]

The concentration of excess sodium hydroxide was 0.362 mol dm⁻³. Calculate the pH of the solution at the end of the experiment.

[1]

Sketch the neutralisation curve obtained and label the equivalence point.

[3]

«Ka = 10^–2.87 = 1.35 × 10^–3 »

«1.35 × 10^–3 = $\frac{[chloroethanoate] \times [H^{+}]}{0.50 mol {dm}^{- 3}} = \frac{x^{2}}{0.50 mol {dm}^{- 3}}$ »

«x = [H⁺] = $\sqrt{1.4 \times 10^{- 3} \times 0.50}$ =» 2.6 × 10^–2 «mol dm^–3» ✔

«pH = –log[H⁺] = –log(2.6 × 10^–2) =» 1.59 ✔

Accept final answer in range 1.58–1.60.

Award [2] for correct final answer.

«pOH = –log(0.362) = 0.441»

«pH = 14.000 – 0.441 =» 13.559 ✔

starts at 1.6 AND finishes at 13.6 ✔

approximately vertical at the correct volume of alkali added ✔

equivalence point labelled AND above pH 7 ✔

Accept any range from 1.1-1.9 AND 13.1-13.9 for M1 or ECF from 11c(i) and 11c(ii).

Award M2 for vertical climb at 28 cm³ OR 15 cm³.

Equivalence point must be labelled for M3.

[N/A]