Question

The $p H$ at the equivalent point for the titration of $0.10\, M\,KH _{2} BO _{3}$ with $0.1\, M\, Hl$ is $\left(\mathrm{K}_{\mathrm{a}}\right.$ of $\left.\mathrm{H}_3 \mathrm{BO}_3=12.8 \times 10^{-10}\right)$ Report your answer by rounding it up to nearest whole number.

Answer 1

First there will be an acid-base reaction between $KH_{2}BO_{3}$ and $HCl$ .

conc. will become half after reaction of an equal volume of $K H_{2}B O_{3}$ and $HCl$
So $\frac{0.1}{5}=0.05$
Now,

$\left[H^{+}\right]=\sqrt{K_{a} C}=\sqrt{12.8 \times 10^{- 10} \times 0.05}=8\times 10^{- 6}$
$\mathrm{pH}=-\log \left(8 \times 10^{-6}\right) \approx 5.0$