Question
At $25^{\circ}C,20.0\text{ }mL$ of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH . The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are :
Given: $K_{a} = 5 \times 10^{- 4},{pK}_{a} = 3.3,\alpha \ll 1$
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Solution
Sol. [Key Concept: Start: $pH = \frac{1}{2}({pK}_{a} - log\ c)$; half-neutralised: Henderson $pH = {pK}_{a} + log\frac{salt}{acid}$.]
For $0.2M,20mlHX$ solution,
$$\begin{matrix} pH\text{~}\text{of weak acid}\text{~} = & \frac{1}{2}\left\lbrack {pk}_{a} - logc \right\rbrack \\ & = \frac{1}{2}\lbrack 4 - log5 - log0.2\rbrack \\ & = \frac{1}{2}\lbrack 4 - log5 + log5\rbrack \\ & = 2 \\ & \\ & \\ & \text{~} \\ \text{~} & \\ & \end{matrix}$$
$\lbrack HX + NaA\rbrack \Rightarrow$ acidic buffer
$${pH = {pK}_{a} + log\frac{\text{~}\text{salt}\text{~}}{\text{~}\text{acid}\text{~}} }{pH = 3.3 + log\frac{2}{2} }{pH = 3.3 }$$
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