Question

calculate the equilibrium constant for

$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

if, for the equilibrium,

$N{H}_4^{+}+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{+}$

the equilibrium constant is $5.5\times 10^{-10}$

• A. $1.8\times 10^{-4}$
• B. $1.8\times 10^{-5}$
• C. $1.8\times 10^{-6}$
• D. $1.8\times 10^{-7}$

Answer 1

Answer: (B)

$N{H}_4^{+}+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{+}$

${\mathrm{K}}_{\mathrm{eq}}=\frac{\left({\mathrm{NH}}_4\mathrm{OH}\right)\left[{\mathrm{OH}}^{-}\right]\left({\mathrm{H}}^{+}\right)}{\left({\mathrm{H}}_2\mathrm{O}\right)\left({\mathrm{NH}}_4^{+}\right)\left[{\mathrm{OH}}^{-}\right]}$

${\text{K}}_{\text{e}\text{q}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}$

$\Rightarrow 5.5\times 10^{-10}=\frac{10^{-14}}{{\text{K}}_{\text{b}}}$

${\text{K}}_{\text{b}}=\frac{10^{-14}}{5.5\times 10^{-10}}=1.8\times 10^{-5}$