A) \[3.0\times {{10}^{5}}\]
B) \[3.0\times {{10}^{-5}}\]
C) \[3.0\times {{10}^{-4}}\]
D) \[3.0\times {{10}^{4}}\]
Correct Answer: D
Solution :
Given, \[C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}O{{O}^{-}}+{{H}^{+}};\] \[{{K}_{a1}}=1.5\times {{10}^{-5}}\] (i) \[{{K}_{a2}}=4.5\times {{10}^{-10}}\] (ii) \[HCN\rightleftharpoons {{H}^{+}}+C{{N}^{-}};\] \[C{{N}^{-}}+C{{H}_{3}}COOH\rightleftharpoons \] \[HCN+C{{H}_{3}}CO{{O}^{-}}\] \[K=?\] On subtracting Eq. (ii) from Eq. (i), we get \[C{{H}_{3}}COOH+C{{N}^{-}}\rightleftharpoons HCN+C{{H}_{3}}CO{{O}^{-}}\] \[K=\frac{{{K}_{a1}}}{{{K}_{a2}}}=\frac{1.5\times {{10}^{-5}}}{4.5\times {{10}^{-10}}}=\frac{{{10}^{5}}}{3}=3.33\times {{10}^{4}}\]You need to login to perform this action.
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