question_answer 1)

The first ionization constant of\[{{H}_{2}}S\]is\[9.1\times {{10}^{-8}}\]. Calculate the concentration of\[H{{S}^{-}}ions\] in its 0.1M solution and how will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of \[{{H}_{2}}S\] is \[1.2\times {{10}^{-13}}\], calculate the concentration of \[{{S}^{2-}}\] under both conditions.

Answer:

\[{{H}_{2}}S\] undergoes ionization as: \[\underset{\begin{smallmatrix} t=0 \\ {{t}_{eq}} \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\,\,\,\underset{\begin{smallmatrix} 0.1 \\ (0.1-x) \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,+\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{H{{S}^{-}}}}\,\] \[{{K}_{a}}=\frac{[{{H}^{+}}][H{{S}^{-}}]}{[{{H}_{2}}S]}\] \[9.1\times {{10}^{-8}}=\frac{x\times x}{0.1}\] \[x=\sqrt{9.1\times {{10}^{-9}}}=9.54\times {{10}^{-5}}\] \[\therefore \] \[[H{{S}^{-}}]=9.54\times {{10}^{-5}}M\] In presence of 0.1 M HCl, the concentration of \[[{{H}^{+}}]\] will be taken equal to 0.1 M because HCl is stronger acid than HCl \[\underset{\begin{smallmatrix} t=0 \\ {{t}_{eq}} \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\underset{\begin{smallmatrix} 0.1 \\ (0.1-y) \\ \approx \,0.1 \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0.1 \\ 0.1+y \\ \approx 0.1 \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,+\underset{\begin{smallmatrix} 0 \\ y \end{smallmatrix}}{\mathop{H{{S}^{-}}}}\,\] \[{{K}_{a}}=\frac{[{{H}^{+}}][H{{S}^{-}}]}{[{{H}_{2}}S]}\] \[9.1\times {{10}^{-8}}=\frac{0.1\times y}{0.1}\] \[i.e.,\] \[y=9.1\times {{10}^{-8}}\] \[\therefore \] \[[H{{S}^{-}}]=9.1\times {{10}^{-8}}M\] Overall dissociation constant of\[{{H}_{2}}S\], \[{{K}_{a}}={{K}_{{{a}_{1}}}}\times {{K}_{{{a}_{2}}}}\] \[=9.1\times {{10}^{-8}}\times 1.2\times {{10}^{-13}}\] \[=1.092\times {{10}^{-20}}\] \[\underset{\begin{smallmatrix} t=0 \\ {{t}_{eq}} \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\underset{\begin{smallmatrix} 0.1 \\ (0.1-x) \\ \approx \,0.1 \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ 2x \end{smallmatrix}}{\mathop{2{{H}^{+}}}}\,+\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{a}}=\frac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{[{{H}_{2}}S]}\] \[1.092\times {{10}^{-20}}=\frac{{{[2x]}^{2}}\times x}{0.1}\] \[4{{x}^{3}}=1.092\times {{10}^{-20}}\] \[x=6.5\times {{10}^{-8}}\] (on solving by logarithm method) \[\therefore \] \[[{{S}^{2-}}]=6.5\times {{10}^{-8}}M\] In presence of 0.1 M HCl, \[\underset{\begin{smallmatrix} t=0 \\ {{t}_{eq}} \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\,\,\underset{\begin{smallmatrix} 0.1 \\ (0.1-x) \\ \approx 0.1 \end{smallmatrix}}{\mathop{{{H}_{2}}S}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0.1 \\ (0.1+2x) \\ \approx 0.1 \end{smallmatrix}}{\mathop{2{{H}^{+}}}}\,+\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{S}^{2-}}}}\,\] \[{{K}_{a}}=\frac{[{{H}^{+}}][{{S}^{2-}}]}{[{{H}_{2}}S]}\] \[1.092\times {{10}^{-20}}=\frac{{{[0.1]}^{2}}[{{S}^{2-}}]}{[0.1]}\] \[[{{S}^{2-}}]=1.092\times {{10}^{-19}}M\]