question_answer 24)

Calculate the pH of a 0.1 M ammonia solution. Calculate the pH after 50 mL of this solution is treated with 25 mL of 0.1 M HCl. The dissociation constant of ammonia, \[{{K}_{b}}=1.77\times {{10}^{-5}}\].

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: \[pOH=p{{K}_{b}}+\log \frac{[Salt]}{[Base]}\] \[=-\log {{K}_{b}}+\log \frac{{{\,}^{n}}N{{H}_{4}}Cl}{{{\,}^{n}}N{{H}_{4}}OH}\] \[=-\log 1.77\times {{10}^{-5}}+\log \frac{0.0025}{0.0025}\]