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

NH4OH⇌NH4++OH−Keq=NH4+OH−NH4OH≃OH−2NH4OH0 ( since NH4+=OH−)or OH−=KeqNH4OH0OH−=1.8×10−5×0.1M2OH−=1.34×10−3MpOH=−log⁡OH−/M=2.87; pH=14−pOH=11.13

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

NH4OH⇌NH4++OH−Keq=NH4+OH−NH4OH≃OH−2NH4OH0 ( since NH4+=OH−)or OH−=KeqNH4OH0OH−=1.8×10−5×0.1M2OH−=1.34×10−3MpOH=−log⁡OH−/M=2.87; pH=14−pOH=11.13

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The pH of $0.1 M {NH}_{4} OH (K_{b} = 1.8 \times 10^{- 5} M)$ at 25 °C is

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1.8

11.13

12.2

2.87

answer is C.

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Detailed Solution

$\begin{array}{l} {NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-} \\ K_{eq} = \frac{[{NH}_{4}^{+}] [{OH}^{-}]}{[{NH}_{4} OH]} ≃ \frac{{[{OH}^{-}]}^{2}}{{[{NH}_{4} OH]}_{0}} \end{array}$ ( since $[{NH}_{4}^{+}] = [{OH}^{-}]$ )

or $\begin{matrix} [{OH}^{-}] = \sqrt{K_{eq} [{NH}_{4} {OH}_{0}} \\ [{OH}^{-}] = \sqrt{1 .8 \times 10^{- 5} \times 0 .1 M^{2}} \\ [{OH}^{-}] = 1 .34 \times 10^{- 3} M \\ pOH = - \log \{[{OH}^{-}] / M\} = 2 .87; \\ pH = 14 - pOH = 11.13 \end{matrix}$