The ionisation constant of ammonium hydroxide is $1.77 \times 10^{-5}$ at $298 \mathrm{~K}$. Hydrolysis…
The ionisation constant of ammonium hydroxide is $1.77 \times 10^{-5}$ at $298 \mathrm{~K}$. Hydrolysis constant of ammonium chloride is
- $5.65 \times 10^{-10}$
- $6.50 \times 10^{-12}$
- $5.65 \times 10^{-13}$
- $5.65 \times 10^{-12}$
Solution
Given, $\mathrm{K}_{\mathrm{a}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5}$
$\begin{aligned}
& \mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\
& \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4 \mathrm{OH}\right]}=1.77 \times 10^{-5}
\end{aligned}$
Hydrolysis of $\mathrm{NH}_4 \mathrm{Cl}$ takes place as,
$\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4 \mathrm{OH}+\mathrm{HCl}$
or $\mathrm{NH}_4^{+}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{NH}_4 \mathrm{OH}+\mathrm{H}^{+}$
Hydrolysis constant,
$\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{NH}_4 \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_4^{+}\right]} \ldots$
or
$\mathrm{K}_{\mathrm{h}}=\frac{\left[\mathrm{NH}_4 \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_4^{+}\right]\left[\mathrm{OH}^{-}\right]}$
$\begin{aligned}
\mathrm{K}_{\mathrm{h}} & =\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}} \quad\left[\because\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}\right] \\
& =\frac{10^{-14}}{1.77 \times 10^{-5}} \\
& =5.65 \times 10^{-10}
\end{aligned}$
Asked in: NEET 2009 (Screening)
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