Question

$100$ ml of $0.1$ M aqueous solution of $HN{O}_2(K_a={10}^{-5})$ is titrated using $0.1$ M barium hydroxide solution, then :

(Given that, $log3=0.48,log2=0.30$)

• A. pH of solution at equivalent point of titration is $8.85$
• B. pH of solution at equivalent point of titration is $8.91$
• C. pH of solution after adding $20$ ml of $Ba(OH{)}_2$ solution is $4.82$
• D. pH of solution after adding $100$ ml of $Ba(OH{)}_2$ solution is $12.7$

Answer 1

The correct option is A pH of solution at equivalent point of titration is $8.85$

Millimoles of nitrous acid $=100\times 0.1=10mmol=\text{millimoles of barium hydroxide}$
Hence, volume of barium hydroxide added $=\frac{10}{0.1\times 2}=50ml$
Total volume $=150ml$
$\left[N{O}_2^{-}\right]=\frac{10}{150}=0.067M$
Let $\left[O{H}^{-}\right]=x$
$\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-5}}=\frac{x^2}{0.067}$
$\left[O{H}^{-}\right]=x=8.165\times {10}^{-6}$
$pOH=5.15,pH=14-5.15=8.85$