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The dissociation constants for acetic acid and HCN\ at\ 25^oC are 1.5 \times 10^{-5}  and 4.5 \times 10^{-10} respectively. The equilibrium constant for the equilibrium \text{CN}^{-}+\text{CH}_{3}\text{COOH}\rightleftharpoons \text{HCN}+\text{CH}_{3}\text{COO}^{-} would be:

Option: 1

3.0 \times 10^{- 5}


Option: 2

3.0 \times 10^{- 4}


Option: 3

3.0 \times 10^4


Option: 4

3.0 \times 10^5


Answers (1)

best_answer

The Dissociation of CH_3COOH\ and\ HCN are

CH_3COOH \rightleftharpoons CH_3COO^-+H^+\\ K_a_1=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}

HCN \rightleftharpoons CN^-+H^+\\ K_a_2=\frac{[CN^-][H^+]}{[HCN]}

For The reaction 

CN^-+CH_3COOH\rightleftharpoons CH_3COO^-+HCN \\ K=\frac{K_a_1}{K_a_2}

K=\frac{1.5\times10^{-5}}{4.5\times 10^{-10}}=\frac{10^5}{3}=3.33\times 10^4

The Correct answer is option 3

 

Posted by

Irshad Anwar

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