Question

At ${25}^{o}C$, the dissociation constants of ${CH}_{3}COOH$ and ${NH}_{4}OH$ in aqueous solution are almost the same. The pH of a solution of $0.01N$ ${CH}_{3}COOH$ is $4.0$ at ${25}^{o}C$. The pH of $0.01N$ ${NH}_{4}OH$ solution at the same temperature would be:

• A. $3.0$
• B. $4.0$
• C. $10.0$
• D. $10.5$

Answer 1

The correct option is C $10.0$

$pH=4$ means $\left[H^{+}\right]={10}^{-4}M=C\alpha$

As dissociation constants are same for both so $\left[O{H}^{-}\right]={10}^{-4}M$

So,

$pOH=4$ or $pH=14-pOH=10$