Question

The pH of a 0.1 M aqueous solution of $NH_4OH$ is : $(K_b = 1.0 \times 10^{-5})$

• A. 3
• B. 10.5
• C. 11
• D. 7.5

Answer 1

Answer: (C)

11

Answer 2

$\begin{matrix}&NH_{4}OH& {<=>}&NH^{+}_{4}&+&OH^{-}\\\ \text{Initial }&0.1&&0&&0\\\ \text{At eqm. }&0.1(1-\alpha)&&0.1\alpha&&0.1 \alpha\end{matrix}$ $\therefore K_{b}=\frac{0.1\alpha\times0.1\alpha}{0.1}=0.1\,\alpha^{2}$ $0.1 \alpha^{2}=1\times10^{-5}$ $\alpha^{2}=10^{-4} \Rightarrow \alpha=10^{-2}$ $\left[OH^{-}\right]=c\alpha=0.1\times10^{-2}=10^{-3}M$ $\left[H^{+}\right]=10^{-14}10^{-3}=10^{-11}$ $\therefore pH=11$