Question

The $pH$ at the equivalence point for the titration of $0.10M$ $K{H}_{2}B{O}_3$ with $0.1MHCl$ 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

Answer: 5

First there will be an acid-base reaction between $K{H}_{2}B{O}_3$ 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,

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