Question

The dissociation constants for acetic acid and $HCN$ at $25°C$ are $1.5\times {10}^{-5}$ and $4.5\times {10}^{-10}$ respectively. The equilibrium constant for the equilibrium,
$C{N}^{-}+C{H}_{3}COOH\rightleftharpoons HCN+C{H}_{3}CO{O}^{-}$ would be:

• A. $3.33\times {10}^{-5}$
• B. $3.33\times {10}^{-4}$
• C. $3.33\times {10}^4$
• D. $3.33\times {10}^{-5}$

Answer 1

The correct option is C $3.33\times {10}^4$

Given,
$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$
$K_1=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}=1.5\times {10}^{-5}$
$HCN\rightleftharpoons H^{+}+C{N}^{-}$
$K_2=\frac{\left[C{N}^{-}\right]\left[H^{+}\right]}{\left[HCN\right]}=4.5\times {10}^{-10}$
$C{N}^{-}+C{H}_{3}COOH\rightleftharpoons HCN+C{H}_{3}CO{O}^{-}$
$K=\frac{\left[HCN\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{N}^{-}\right]\left[C{H}_{3}COOH\right]}$
$K=\frac{K_1}{K_2}=\frac{1.5\times {10}^{-5}}{4.5\times {10}^{-10}}=3.33\times {10}^4$