Question

Given $K_a$ values of $5.76\times {10}^{-10}$ and $4.8\times {10}^{-10}$ for $N{H}_4^{+}$ and $HCN$ respectively. What is the equilibrium constant for the following reaction?
$N{H}_4^{+}(aq.)+C{N}^{-}(aq.)\rightleftharpoons N{H}_3(aq.)+HCN(aq.)$

• A. $0.83$
• B. $1.2$
• C. $8.0\times {10}^{-11}$
• D. $27.6\times {20}^{-10}$

Answer 1

Answer: (B)

$1.2$

$N{H}_4^{\oplus }+O{H}^{\ominus }\rightleftharpoons N{H}_{4}OHK{a}_1$
$J^{+}+C{N}^{\ominus }\rightleftharpoons HCNK{a}_2$
$H_{2}O+N{H}_4^{\oplus }+C{N}^{\ominus }\rightleftharpoons N{H}_{4}OH+HCNKa$
$K{a}_1=\frac{\left[N{H}_{4}OH\right]}{\left[N{H}_4^{\oplus }\right]\left[O{H}^{\ominus }\right]}$
$K{a}_2=\frac{\left[HCN\right]}{\left[H^{+}\right]\left[C{N}^{\ominus }\right]}$
$Ka=\frac{\left[N{H}_{4}OH\right]\left[HCN\right]}{\left[H^{+}\right]\left[O{H}^{\ominus }\right]\left[N{H}_4^{\oplus }\right]\left[C{N}^{\ominus }\right]}$
$K{a}_1\times K{a}_2=Ka$
$Ka=5.6\times {10}^{-10}\times 4.8\times {10}^{-10}$
$=2.7\times {10}^{-19}$