Question

calculate the equilibrium constant for

$N H_{3} \, + \, H_{2}O \, \rightleftharpoons NH_{4}^{+}+OH^{-}$

if, for the equilibrium,

$NH_{4}^{+}+H_{2}O\rightleftharpoons \, N H_{4}OH \, + \, H^{+}$

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)

$NH_{4}^{+}+H_{2}O\rightleftharpoons \, N H_{4}OH \, + \, H^{+}$

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

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

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

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