Chemistry 9701 · AS & A Level · Equilibria

Equilibria — practice question

Solution $Y$ is hydrochloric acid, $\text{HCl}(aq)$. Solution $Z$ is aqueous 4-chlorobutanoic acid, $\text{Cl}(\text{CH}_2)_3\text{CO}_2\text{H}(aq)$. The $pK_a$ for $\text{Cl}(\text{CH}_2)_3\text{CO}_2\text{H}(aq)$ is $4.52$. Both solutions have a pH of $4.00$.
(a(i))[1]

Write the expression for the $K_a$ of $\text{Cl}(\text{CH}_2)_3\text{CO}_2\text{H}(aq)$.

(a(ii))[1]

Write a mathematical expression showing the relationship between $K_a$ and $pK_a$.

(a(iii))[1]

Calculate $[\text{H}^+]$ for solutions $Y$ and $Z$.

(a(iv))[2]

Calculate the ratio $\dfrac{[\text{HCl}]\ \text{dissolved in solution } Y}{[\text{Cl}(\text{CH}_2)_3\text{CO}_2\text{H}]\ \text{dissolved in solution } Z}$.

(b)[3]

A buffer with pH $5.00$ is prepared by adding sodium propanoate to $5.00\,\text{g}$ of propanoic acid in $100\,\text{cm}^3$ of distilled water. Calculate the mass of sodium propanoate needed to make this buffer solution. The $K_a$ of propanoic acid is $1.35 \times 10^{-5}\,\text{mol dm}^{-3}$. $[M_r:\ \text{propanoic acid},\ 74.0;\ \text{sodium propanoate},\ 96.0]$.

(c)[2]

A small amount of dilute sulfuric acid is added to a small sample of the buffer described in part (b). The pH of the resulting mixture is close to $1$. Explain this observation.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: So, $K_a = \frac{[\mathrm{H^+}][\mathrm{Cl(CH_2)_3CO_2^-}]}{[\mathrm{Cl(CH_2)_3CO_2H}]}$

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