Dissociation constants of $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{NH}_4 \mathrm{OH}$ in aqueous s

Question

Dissociation constants of $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{NH}_4 \mathrm{OH}$ in aqueous solution are $10^{-5}$. If pH of a $\mathrm{CH}_3 \mathrm{COOH}$ solution is 3 , What will be the pH of $\mathrm{NH}_4 \mathrm{OH}$ ?

Professor PC Thomas Classes · Answer

(d): pH of $\mathrm{CH}_3 \mathrm{COOH}=3$ [ $\mathrm{H}^{+}$] 
$=10^{-3} \mathrm{M}$ from $\mathrm{CH}_3 \mathrm{COOH}$ 
Since, dissociation constants of $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{NH}_4 \mathrm{OH}$ are equal \ $ \begin{aligned} & {\left[\mathrm{NH}_4 \mathrm{OH}ight]=\left[\mathrm{CH}_3 \mathrm{COOH}ight]=10^{-3} \mathrm{M}} \ & {\left[\mathrm{OH}^{-}ight]=10^{-3} \mathrm{M}} \ & {\left[\mathrm{H}^{+}ight]\left[\mathrm{OH}^{-}ight]=10^{-14} ;\left[\mathrm{H}^{+}ight]=\frac{10^{-14}}{10^{-3}}=10^{-11}} \end{aligned} \ $ pH of $\mathrm{NH}_4 \mathrm{OH}=-\log \left(10^{-11}ight)=11$