A) \[7.00\]
B) \[7.85\]
C) \[8.00\]
D) \[8.95\]
Correct Answer: D
Solution :
\[N{{H}_{3}}+{{H}_{2}}ONH_{4}^{+}+O{{H}^{-}}\] The ionisation constant of \[N{{H}_{3}}\]: \[{{K}_{b}}\]=antilog \[(-p{{K}_{b}})\]i.e., \[{{K}_{b}}=10-4.75\] \[=1.77\times {{10}^{-5}}M\] \[\begin{align} & Intial\,conc.\,(M) \\ & At\,equili.\underset{0.10-x}{\mathop{\underset{0.10}{\mathop{N{{H}_{3}}}}\,}}\,+{{H}_{2}}O\underset{(0.20+x)}{\mathop{\underset{0.20}{\mathop{NH_{4}^{+}}}\,}}\,+\underset{x}{\mathop{\underset{0}{\mathop{O{{H}^{-}}}}\,}}\, \\ \end{align}\] \[{{K}_{b}}=\frac{[NH_{4}^{+}][O{{H}^{-}}]}{N{{H}_{3}}}=\frac{(0.20+x).x}{(0.10-x)}\] \[1.77\times {{10}^{-5}}=\frac{(0.20+x).x}{0.10-x}\] (As \[{{K}_{b}}\] is small we can neglect x in comparison to \[0.1M\] and \[0.2M\]) \[x=[O{{H}^{-}}]=0.88\times {{10}^{-5}}\] Therefore, \[[{{H}^{+}}]=\frac{1\times {{10}^{-14}}}{0.805\times {{10}^{-5}}}=1.12\times {{10}^{-9}}\] \[pH=-\log [{{H}^{+}}]=8.95\]
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