50 ml of a solution which is 0.05 M in the acid $HA$ $(p K_{a}$ = 3.80) and 0.1 Min $HB$ $(p K_{a}$ = 8.20) is titrated with $0.2 M - NaOH$ solution. Calculate the pH of solution at the first equivalence point. Give your answer after multiplying with 100

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Detailed Solution

$\begin{array}{l} HA + {OH}^{-} ⟶ A^{-} + H_{2} O \\ 50 \times 0.05 = 0.2 \times V \\ V = 12.5 mL \end{array}$

$\begin{array}{lccc} A^{-} & + HB & ⇌ & H A + B^{-} \\ 2.5 & 5 & - - \\ 5 - x & 5 - x & x x \end{array}$

$\frac{x^{2}}{(2.5 - x) (5 - x)} = \frac{10^{- 8.2}}{10^{- 3.8}} = 4 \times 10^{- 5}$

$\begin{array}{l} x = 0.0224 \\ 10^{- 8.2} = \frac{[H^{+}] \times \sqrt{5 \times 10^{- 4}}}{5} \\ [H^{+}] = \sqrt{5} \times 10^{- 6.2} \\ pH = 6.2 - \frac{1}{2} \times 0.7 = 5.85 \end{array}$

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