The pH of 0.1MNH4OHKb=1.8×10−5M at 25 °C is

The pH of $0.1\mathrm{M}{\mathrm{NH}}_{4}\mathrm{OH}\left({K}_{\mathrm{b}}=1.8×{10}^{-5}\mathrm{M}\right)$ at 25 °C is

1. A

2.87

2. B

1.8

3. C

11.13

4. D

12.2

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Solution:

$\begin{array}{l}{\mathrm{NH}}_{4}\mathrm{OH}⇌{\mathrm{NH}}_{4}^{+}+{\mathrm{OH}}^{-}\\ {K}_{\mathrm{eq}}=\frac{\left[{\mathrm{NH}}_{4}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_{4}\mathrm{OH}\right]}\simeq \frac{{\left[{\mathrm{OH}}^{-}\right]}^{2}}{{\left[{\mathrm{NH}}_{4}\mathrm{OH}\right]}_{0}}\end{array}$ ( since $\left[{\mathrm{NH}}_{4}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]$)

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