At \(T(\mathrm{~K})\), if the ionisation constant of ammonia in solution is \(2.5 \times 10^{-5}\), the…
At \(T(\mathrm{~K})\), if the ionisation constant of ammonia in solution is \(2.5 \times 10^{-5}\), the \(\mathrm{pH}\) of \(0.01 \mathrm{M}\) ammonia solution and the ionisation constant of its conjugate acid respectively at that temperature are \((\log 2=0.30)\)
\(10.7,4.0 \times 10^{-8}\)
\(10.7,4.0 \times 10^{-10}\)
\(3.3,4.0 \times 10^{-8}\)
\(3.3,4.0 \times 10^{-10}\)
Solution
\(K_a \times K_b=K_w=10^{-14}\) (at \(25^{\circ} \mathrm{C}\))
where, \(K_a=\) ionisation constant for acid \(\left(\mathrm{NH}_4^{+}ight)\)
and \(K_b=\) ionisation constant for base \(\left(\mathrm{NH}_3ight)\)
\(=2.5 \times 10^{-5}\)
i.e. \(\quad K_a\left(\mathrm{NH}_4^{+}ight) \cdot K_b\left(\mathrm{NH}_3ight)=10^{-14}\)
\(\begin{gathered}
\therefore \quad K_a\left(\mathrm{NH}_4^{+}ight)=\frac{10^{-14}}{K_a\left(\mathrm{NH}_4^{+}ight)} \\
K_a=\frac{10^{-14}}{2.5 \times 10^{-5}}=4 \times 10^{-10}
\end{gathered}\)
i.e. ionisation constant of conjugate acid \(=4 \times 10^{-8}\)
For \(\mathrm{pH}\) of \(0.01 \mathrm{M}\) ammonia \(\left(\mathrm{NH}_3ight)\)
\(\therefore \mathrm{NH}_4 \mathrm{OH}\) dissociate as :
\(\begin{array}{llll}
& \mathrm{NH}_4 \mathrm{OH} ightleftharpoons & \mathrm{NH}_4^{+} & +\mathrm{OH}^{-} \\
\text {At equilibrium } & 0.01 (1-\alpha) & 0. \alpha & 0.01 \alpha\end{array}\)
(\(\therefore\) Concentration of ammonia solution \(=0.01 \mathrm{M}\))
Also, \(\alpha=\sqrt{\frac{K_b}{C}}=\sqrt{\frac{2.5 \times 10^{-5}}{0.01}}=0.05\)
and \(\left[\mathrm{OH}^{-}ight]=\mathrm{C} \alpha=(0.01) \alpha\) or \(\left[\mathrm{OH}^{-}ight]=0.0005\)
\(\therefore \quad \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}ight]\)
or, \(\mathrm{pH}=14-\mathrm{pOH}=14+\log [(0.01) \alpha]=14-3.3\)
Thus, \(\mathrm{pH}=10.7\) and also ionisation constant \(\left(K_aight)=4 \times 10^{-10}\)
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