Answer:
\[N{{H}_{4}}OH\]is a weak mono acidic
base.
\[\therefore \] \[[O{{H}^{-}}]=\sqrt{C{{K}_{b}}}\]
\[pOH=-\log
\sqrt{C{{K}_{b}}}\]
\[=-\log
\sqrt{0.1\times 1.77\times {{10}^{-5}}}=2.876\]
\[pH=14-2.876=11.124\]
When \[HCl\]
is added to ammonia solution, its neutralization takes place.
\[N{{H}_{4}}OH+HCl\to N{{H}_{4}}Cl+{{H}_{2}}O\]
\[{{\,}^{n}}N{{H}_{4}}OH=\frac{MV}{1000}=\frac{0.1\times
50}{1000}=0.005\]
\[{{\,}^{n}}HCl=\frac{MV}{1000}=\frac{0.1\times
25}{1000}=0.0025\]
\[\therefore
\] \[{{\,}^{n}}N{{H}_{4}}Cl=0.0025\]
Remaining moles of
\[N{{H}_{4}}OH=0.005-0.0025=0.0025\]According
to Henderson equation:
\[pOH=p{{K}_{b}}+\log \frac{[Salt]}{[Base]}\]
\[=-\log {{K}_{b}}+\log
\frac{{{\,}^{n}}N{{H}_{4}}Cl}{{{\,}^{n}}N{{H}_{4}}OH}\]
\[=-\log
1.77\times {{10}^{-5}}+\log \frac{0.0025}{0.0025}\]
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