Chemistry 9701 · AS & A Level · Equilibria

Equilibria — practice question

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

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

(a(ii))[1]

Write a mathematical expression showing how $K_a$ and $pK_a$ are related.

(a(iii))[1]

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

(a(iv))[2]

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

(b)[3]

A buffer solution of pH $5.00$ is made 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]

Some dilute sulfuric acid is added to a small sample of the buffer solution described in (b). The final pH of the 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: $K_a = \frac{[\mathrm{H}^+][\mathrm{Cl(CH_2)_3CO_2^-}]}{[\mathrm{Cl(CH_2)_3CO_2H}]}$

  • Full mark scheme, point by point
  • Step-by-step worked solution
  • Write your answer & get it marked instantly by AI