Question

Given that $K_a$ for $HClO$ and $HCN$ are $3.0\times {10}^{-8}$ and $4.8\times {10}^{-10}$, respectively, the equilibrium constant for the reaction
$HClO\left(aq\right)+C{N}^{-}\left(aq\right)\rightleftharpoons Cl{O}^{-}\left(aq\right)+HCN\left(aq\right)$, is:

• A. $0.625\times {10}^2$
• B. $1.6\times {10}^{12}$
• C. $1.6\times {10}^{-2}$
• D. $1.4\times {10}^{-17}$

Answer 1

The correct option is A $0.625\times {10}^2$

$HClO\left(aq\right)+C{N}^{-}\rightleftharpoons HCN+Cl{O}^{-}$
$K=\frac{\left[HCN\right]\left[Cl{O}^{-}\right]}{\left[HClO\right]\left[C{N}^{-}\right]}\times \frac{\left[H^{+}\right]}{\left[H^{+}\right]}=\frac{\left[H^{+}\right]\left[Cl{O}^{-}\right]}{\left[HClO\right]}\times \frac{1}{\left\{\frac{\left[H^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}\right\}}$
$=\frac{K_a\left(HClO\right)}{K_a\left(HCN\right)}=\frac{3\times {10}^{-8}}{4.8\times {10}^{-10}}=0.625\times {10}^2$