Question

The $pH$ at the equivalence point for the titration of $0.10 \, M$ $KH_{2}BO_{3}$ with $0.1 \, M \, HCl$ is (Report the answer in the nearest integer value)
$\left(K_{a} \, of \, H_{3} \left(BO\right)_{3} = 12 . 8 \left(\times 10\right)^{- 10}\right)$

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 $\frac{0 .1}{5}=0.05M$
Now,

$\left[H^{+}\right]=\sqrt{K_{a} C}=\sqrt{12 .8 \times 10^{- 10} \times 0 .05}=8\times 10^{- 6}M$
$pH=-\log \left(8 \times \left(10\right)^{- 6}\right)\approx5.0$