A Level Chemistry Acid and Bases

35.0 cm³ of 0.150 mol dm^–3 aqueous sodium hydroxide are mixed with 20.0 cm³ of a 0.100 mol dm^–3 solution of hydrochloric acid.

The temperature of the solution formed is 40 °C

Calculate the pH of the solution formed. Give your answer to 2 decimal places. (5)

n_OH^- = 5.25 × 10^-3 and n_H⁺ = 2.00 × 10^-3 mol (1)

excess OH^- = 3.25 × 10^-3 mol (1)

[OH^-]=3.25×10^-3 /55.0×10^-3 = 0.0591 mol dm^-3 (1)

[H⁺] = 2.92×10^-14 /0.0591=4.94×10^-13 (1)

12.31 (1)

Suggest why a student doing an experiment to produce the curve in the figure above would add the sodium hydroxide solution dropwise around the equivalence point. (1)

As there is a large pH change (for a small addition of alkali) (1)

When sodium hydroxide solution is added to aqueous propanoic acid, the solution formed acts as a buffer when between 5 cm³ and 15 cm³ have been added.

Explain why the pH stays approximately constant during this part of the experiment. (2)

OH^- reacts with propanoic acid OR reacts with H⁺(1)

ratio of [CH₃CH₂COOH] to [CH₃CH₂COO^- ] remains almost constant OR Equilibrium for dissociation of CH₃CH₂COOH moves right to maintain [H⁺] /replace H⁺ (1)

Methyl orange and universal indicator are not suitable indicators for the titration of solutions of propanoic acid with sodium hydroxide.

State the reason why each indicator is not suitable (2)

Methyl orange - would not change colour at the equivalence point/pH range does not match rapid pH change (1)

Universal indicator - idea of range of colours during titration/no distinct colour change (at equivalence/end-point) (1)

Give an expression for K_a for propanoic acid (CH₃CH₂COOH). Use this expression to show that pH = pK_a when half of the propanoic acid has reacted with sodium hydroxide.(3)

See image (1)

[CH₃CH₂COOH] = [CH₃CH₂COO^- ] (1)

K_a=[H⁺] (1)

The final mixture contains a saturated solution of Ca(OH)₂ at 293 K

At 293 K

• the solubility of Ca(OH)₂ in this solution is 0.400 g dm–3

• Kw = 6.80 × 10^–15 mol²dm^–6

Calculate the pH of this solution. Give your answer to two decimal places.(5)

0.400/74.1(=0.00540mol dm^-3) (1)

[OH^-]=0.00540×2=0.0108mol dm^-3(1)

[H⁺]=6.80×10^-15 /0.0108 (=6.30×10^-13mol dm^-3) (1)

pH=-log₁₀[H⁺] (1)

-log₁₀(6.30×10^-13)(= 12.20)(1)

The pH of a barium hydroxide solution is lower at 50 °C than at 10 °C

At 50 °C a 25 cm³ sample of this barium hydroxide solution was neutralised by 22.45 cm³ of hydrochloric acid added from a burette.

Deduce the volume of this hydrochloric acid that should be added from a burette to neutralise another 25 cm ⁽ sample of this barium hydroxide solution at 10 °C Explain your answer (2)

22.45 or same (1)

same [OH^–] or amount/number of OH^– ions (1)

State how a buffer solution can be made from solutions of potassium hydroxide and ethanoic acid.

Give an equation for the reaction between potassium hydroxide and ethanoic acid. State how this buffer solution resists changes in pH when a small amount of acid is added. (3)

add excess ethanoic acid to KOH (1)

KOH + CH₃COOH ⟶ CH₃COOK + H₂O (1)

CH₃COO^– (from salt) reacts with (added) acid/H⁺ (1)

A buffer solution is made by adding 2.00 g of sodium hydroxide to 500 cm³ of 1.00 mol dm^–3 ethanoic acid solution.

Calculate the pH of this buffer solution at 25 °C Give your answer to 2 decimal places. For ethanoic acid, Ka = 1.74 x 10^–5 mol dm^–3 at 25 °C (5)

n(NaOH) =2/40=0.05 mol and n(CH₃COOH) =500×1/1000=0.5 mol (1)

(after adding NaOH) n(CH₃COOH) = (0.50 − 0.05) = 0.45 (mol) (1)

n(CH3COO^–) = (n NaOH) = 0.05 (mol) (1)

[H⁺]= (K_ax[CH₃COOH])/[CH₃COO^-] (1)

pH = 3.80 (1)

A student plans to titrate butanoic acid solution with a solution of ethylamine. Explain why this titration could not be done using an indicator.(2)

This is a weak acid and weak base/alkali titration (1)

pH change is too gradual/not sharp (at the equivalence point so colour change of indicator is difficult to judge) (1)