A) \[3.0\times {{10}^{4}}\]
B) \[3.0\times {{10}^{-5}}\]
C) \[3.0\times {{10}^{5}}\]
D) \[3.0\times {{10}^{-4}}\]
Correct Answer: A
Solution :
\[C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}COO{{H}^{-}}+{{H}^{+}}\] \[K{{a}_{1}}=1.5\times {{10}^{-5}}=\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\] \[HCN\rightleftharpoons {{H}^{+}}+C{{N}^{-}}\] \[{{K}_{{{a}_{2}}}}=4.5\times {{10}^{-10}}=\frac{[{{H}^{+}}][C{{N}^{-}}]}{[HCN]}\] (ii) By (i)(ii) \[K=\frac{{{K}_{{{a}_{1}}}}}{{{K}_{{{a}_{2}}}}}=\frac{[HCN][C{{H}_{2}}CO{{O}^{-}}]}{[C{{N}^{-}}][C{{H}_{3}}COOH]}\] \[=\frac{1.5\times {{10}^{-5}}}{4.5\times {{10}^{-10}}}=3\times {{10}^{4}}\]You need to login to perform this action.
You will be redirected in
3 sec