One mole of ammonia was completely absorbed in one litre solution each of
(a) \( 1 \mathrm{M} \mathrm{HCl} \),
(b) \( 1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} \) and
(c) \( 1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} \) at \( 298 \mathrm{~K} \).
The decreasing order for the \( \mathrm{pH} \) of the resulting solutions is
(Given \( \mathrm{K}_{b}\left(\mathrm{NH}_{3}\right)=4.74 \) )

The reactions involved are as follows:
(I) \(\mathrm{HCl}+\mathrm{NH}_{3}(\mathrm{aq}) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}\)
(II) \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3}(\mathrm{aq}) \rightarrow \mathrm{CH}_{3} \mathrm{COONH}_{4}\)
(III) \(\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NH}_{3}(\mathrm{aq}) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)
Now, according to \(\mathrm{Henderson}\) Haselbach equation
\(p \mathrm{H}=p \mathrm{~K}_{a}+\log \left(\frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\right)\)
\(=14-\mathrm{pK}_{b}+\log \left(\frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\right)\)
\(=9.26+\log \left(\frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\right)\)
Thus, \(\mathrm{pH}\) depends on the strength of conjugate acid. The order acid strength of the reactants is as follows:
\(\mathrm{CH}_{3} \mathrm{COOH} < \mathrm{HCl} < \mathrm{H}_{2} \mathrm{SO}_{4}\). Also \(\mathrm{NH}_{3}\) (aq) is a strong base. As conjugate base of weak acid is strongly acidic, we
have the acidic strength of products as
\(\mathrm{CH}_{3} \mathrm{COONH}_{4}>\mathrm{NH}_{4} \mathrm{Cl} < \left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)
That is, (II) \(>(\mathrm{I})>(\mathrm{III})\).