question_answer 1)

The ionisation constant of phenol is\[1\times {{10}^{-10}}\]. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionisation if the solution is also 0.01 M in sodium phenolate?

Answer:

Phenol undergoes ionization as: \[\underset{{{t}_{\begin{smallmatrix} 0 \\ {{t}_{eq}} \end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{smallmatrix} C \\ C-C\alpha \end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}OH}}\,\rightleftharpoons \underset{\begin{smallmatrix} Phenolate\,ion \\ 0 \\ C\alpha \end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}{{O}^{-}}}}\,+\underset{\begin{smallmatrix} 0 \\ C\alpha \end{smallmatrix}}{\mathop{{{H}^{+}}}}\,\] \[{{K}_{a}}=\frac{[{{C}_{6}}{{H}_{5}}{{O}^{-}}][{{H}^{+}}]}{[{{C}_{6}}{{H}_{5}}OH]}=\frac{C\alpha \times C\alpha }{(C-C\alpha )}\] \[=\frac{C{{\alpha }^{2}}}{1-\alpha }\approx C{{\alpha }^{2}}\] \[\alpha =\sqrt{\frac{{{K}_{a}}}{C}}=\sqrt{\frac{{{10}^{-10}}}{0.05}}=4.47\times {{10}^{-5}}\]   Concentration of phenolate ion = \[C\alpha \] \[=0.05\times 4.47\times {{10}^{-5}}\] \[=2.235\times {{10}^{-8}}M\] In presence of 0.01 M sodium phenolate, the degree of ionization may be calculated as: \[{{C}_{6}}{{H}_{5}}{{O}^{-}}N{{a}^{+}}\to {{C}_{6}}{{H}_{5}}{{O}^{-}}+N{{a}^{+}}\] \[[{{C}_{6}}{{H}_{5}}{{O}^{-}}]=0.01=C\alpha \] Sodium phenolate is strong electrolyte then its phenolate ion concentration will dominate the concentration of phenolate ion from phenol. \[{{K}_{a}}=C{{\alpha }^{2}}\] \[=C\alpha \times \alpha \] \[{{10}^{-10}}=0.01\times \alpha \] \[\alpha ={{10}^{-8}}\]