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(D) The reactions are: $(a)$ $NH_{3} + HCl \rightarrow NH_{4}Cl$ (Salt of strong acid and weak base,$pH < 7$). $(b)$ $NH_{3} + CH_{3}COOH \rightarrow

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$b > a > c$

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(D) The reactions are: $(a)$ $NH_{3} + HCl \rightarrow NH_{4}Cl$ (Salt of strong acid and weak base,$pH For $(a)$,$pH = 7 - \frac{1}{2}(pK_{b} + \log C) = 7 - \frac{1}{2}(4.74 + 0) = 4.63$. For $(b)$,$pH = 7 + \frac{1}{2}(pK_{a} - pK_{b}) = 7 + \frac{1}{2}(4.74 - 4.74) = 7.0$. For $(c)$,$H_{2}SO_{4}$ is a strong diprotic acid. $1 \ M \ H_{2}SO_{4}$ provides $2 \ M \ H^{+}$,which reacts with $1 \ M \ NH_{3}$ to leave $1 \ M \ H^{+}$ and $0.5 \ M \ (NH_{4})_{2}SO_{4}$. This results in a highly acidic solution $(pH Thus,the order of $pH$ is $b (7.0) > a (4.63) > c ( a > c$.

One mole of ammonia was completely absorbed in one litre solution each of $(a)$ $1 \ M \ HCl$,$(b)$ $1 \ M \ CH_{3}COOH$,and $(c)$ $1 \ M \ H_{2}SO_{4}$ at $298 \ K$. The decreasing order for the $pH$ of the resulting solutions is (Given $pK_{b}(NH_{3}) = 4.74$)

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