Question

As HA is stronger acid, it will react first. For first equivalent point,

$V_{NaOH} \times 0.2 = 50 \times 0.05 \Rightarrow V_{NaOH} = 12.5\text{ ml}$

At first equivalent point,

$\left\lbrack A^{-} \right\rbrack = \frac{50 \times 0.05}{62.5} = 0.04\text{ M} $$$\lbrack HB\rbrack = \frac{50 \times 0.08}{62.5} = 0.064\text{ M}$$

Now, $\underset{0.04 - x \simeq 0.04}{A^{-}} + \underset{0.064 - x \simeq 0.064}{HB} \rightleftharpoons \underset{x}{B^{-}} + \underset{x}{HA};$

$K_{eq} = \frac{K_{a}(HB)}{K_{a}(HA)} = \frac{10^{- 8.2}}{10^{- 3.8}} = 10^{- 4.4} = 4.4 \times 10^{- 5}$

$4 \times 10^{- 5} = \frac{x.x}{0.04 \times 0.064} \Rightarrow x = 3.2 \times 10^{- 4}$

Now, $K_{a}(HA) = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack A^{-} \right\rbrack}{\lbrack HA\rbrack} \Rightarrow 1.6 \times 10^{- 4} = \frac{\left\lbrack H^{+} \right\rbrack \times 0.04}{3.2 \times 10^{- 4}}$

$\therefore\left\lbrack H^{+} \right\rbrack = 1.28 \times 10^{- 6} \Rightarrow P^{H} = 5.9$