Question

For the reaction:
$N{H}_4^{+}+2H_{2}O\rightleftharpoons N{H}_{4}OH+H_3{O}^{+}$ $K_1=5.56\times {10}^{-10}$
$H_{2}O+H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$ $K_2=1\times {10}^{-14}$
Hence $K_3$ of the reaction:
$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$ is:

• A. $1.8\times {10}^5$
• B. $1.8\times {10}^{-5}$
• C. $5.56\times {10}^4$
• D. $5.56\times {10}^{-4}$

Answer 1

The correct option is B $1.8\times {10}^{-5}$

$N{H}_4^{+}+2H_{2}O\rightleftharpoons N{H}_{4}OH+H_3{O}^{+}{K}_1=5.56\times {10}^{-10}$

$H_{2}O+H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}{K}_2=1\times {10}^{-14}$

$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}{K}_3=?$

$K_1=\frac{\left[N{H}_{4}OH\right]\left[H_3{O}^{+}\right]}{\left[N{H}_4^{+}\right][H_{2}O{]}^2}$ ...(i)

$K_2=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[H_{2}O\right]}$ ....(ii)

$K_3=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$

divide equation (ii) by (i)

$K_3=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[H_{2}O\right]}.\frac{\left[N{H}_4^{+}\right][H_{2}O{]}^2}{\left[N{H}_{4}OH\right]\left[H_3{O}^{+}\right]}$

$K_3=K_2.\frac{1}{k_1}=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$

$K_3=\frac{1\times {10}^{-14}}{5.56\times {10}^{-10}}=1.8\times {10}^{-5}$

option B is correct option