Chemistry 9701 · AS & A Level · Equilibria

Equilibria — practice question

(a)[2]

HF, hydrogen fluoride, acts as a weak acid in water and has $K_a = 5.6 \times 10^{-4}\,\text{mol dm}^{-3}$. Find the pH of a $0.050\,\text{mol dm}^{-3}$ HF solution.

(b(i))[1]

Gaseous ammonia and hydrogen fluoride react to produce solid ionic ammonium fluoride: $\text{NH}_3(g) + \text{HF}(g) \rightleftharpoons \text{NH}_4\text{F}(s)$, $\Delta H = -147\,\text{kJ mol}^{-1}$. What kind of reaction is this?

(b(ii))[4]

Draw dot-and-cross diagrams for the outer shells only to show the bonding in the three compounds taking part in this reaction: $\text{NH}_3$, HF and $\text{NH}_4\text{F}$.

(b(iii))[3]

In $\text{NH}_4\text{F}$ there are three bonding types. Name each one and state the part of the compound where it is found.

(b(iv))[9]

The reaction between $\text{NH}_3$ and HF is reversible. Which temperature and pressure conditions would favour the reverse reaction, that is, the dissociation of $\text{NH}_4\text{F}$? Explain your reasoning.

(c(i))[1]

Many commercial copper and brass polishes contain ammonia. The tarnish that appears on copper is often copper sulfide, CuS. In the presence of $\text{O}_2$ from the air, $\text{NH}_3$ can react with this copper sulfide to form the soluble cuprammonium sulfate, $[\text{Cu}(\text{NH}_3)_4]\text{SO}_4$. Write an equation for this reaction.

(c(ii))[1]

State the colour of cuprammonium sulfate solution.

(c(iii))[3]

Describe what you would observe if a solution of cuprammonium sulfate were diluted with water. Explain your answer.

(d)[3]

On addition of sulfuric acid to $\text{Cu}^{2+}(aq)$, no colour change is seen, but adding concentrated hydrochloric acid to $\text{Cu}^{2+}(aq)$ turns the solution yellow-green. When water is added to dilute it, the solution returns to its original colour. Suggest the type of reaction taking place with $\text{HCl}(aq)$, say what is formed in the process, and write an equation for the change.

Worked solution & mark scheme

This 27-mark question has a full step-by-step worked solution and mark scheme. One marking point: Therefore, [$H^+$] = $\sqrt{0.05 \times 5.6 \times 10^{-4}} = 5.29 \times 10^{-3}\ \text{mol dm}^{-3}$

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