question_answer 1)

The ionization constant of benzoic acid is\[6.46\times {{10}^{-5}}\]and\[{{K}_{sp}}\] for silver benzoate is\[2.5\times {{10}^{-13}}\]. How many times the silver benzoate is more soluble in buffer of pH 3.19 compared to its solubility in pure water?

Answer:

Solubility of silver benzoate in pure water \[=\sqrt{{{K}_{sp}}}=\sqrt{2.5\times {{10}^{-13}}}\] \[=5\times {{10}^{-7}}mol{{L}^{-1}}\] \[[{{H}^{+}}]\]Buffer =Antilog \[(-3.19)=6.457\times {{10}^{-4}}M\] \[{{K}_{a}}=\frac{[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}][{{H}^{+}}]}{[{{C}_{6}}{{H}_{5}}COOH]}\] \[\frac{[{{C}_{6}}{{H}_{5}}COOH]}{[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]}=\frac{[{{H}^{+}}]}{{{K}_{a}}}=\frac{6.457\times {{10}^{-4}}}{6.46\times {{10}^{-5}}}=10\]Let solubility in buffer is \['x'\] \[[A{{g}^{+}}]=x=[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]+[{{C}_{6}}{{H}_{5}}COOH]\] \[=[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]+10[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]\] \[=11[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]\] \[[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}]=\frac{x}{11}\] Most of benzoate ion is converted to benzoic acid \[{{K}_{sp}}=[{{C}_{6}}{{H}_{5}}CO{{O}^{-}}][A{{g}^{+}}]\] \[=x\times \frac{x}{11}=\frac{{{x}^{2}}}{11}\] \[x=1.66\times {{10}^{-6}}\] Ratio of solubility will be = \[\frac{1.66\times {{10}^{-6}}}{5\times {{10}^{-7}}}=3.32\]