Question

$ CH_3COOH $ is $ 1\% $ ionised in its aqueous solution of $ 0.1\, M $ strength. Its $ pOH $ will be

• A. $ 11 $
• B. $ 3 $
• C. $ 13 $
• D. $ 2 $

Answer 1

Answer: (A)

$\begin{matrix}&CH_{3}COOH+H_{2}O&\ce{<=>}&CH_{3}COO^{-}&+&H_{3}O^{+}\\ \text{Initial conc.}&C&&0&&0\\ \text{conc. at equilibrium }&C — C\alpha&&C \alpha&&C \alpha\end{matrix}$
Thus, $\left[H^{+}\right]=C\alpha=0.1\times0.01=0.001 \left[\alpha=1\%\right]$
$\Rightarrow 10^{-3}M$
$\Rightarrow \left[OH^{-}\right]$
$=\frac{10^{-14}}{10^{-3}}=10^{-11}$
$\Rightarrow pOH=11$