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# Ammonium acetate which is 0.01 M, is hydrolysed to 0.001 M concentration. Calculate the change in $pH$ in 0.001 M solution, if initially $pH = pK_a$

(a) 5

(b) 10

(c) 100

(d) 1

$CH_3COO^- + NH^+_4 + H_2O \rightleftharpoons CH_3COOH + NH_4OH$
Initially [$CH_3COO^-$] = 0.01 M, [$NH^+_4$] = 0.01 M
At equilibrium, [$CH_3COO^-$] = 0.001 M, [$NH^+_4$] = 0.001 M, [$CH_3COOH$] = (0.01-0.001) M = 0.009 M, [$NH_4OH$] = (0.01-0.001) M = 0.009 M

$K_h = \frac{[CH_3COOH][NH_4OH]}{[CH_3COO^-][NH^+_4]} = \frac{(0.009)^2}{(0.001)^2} = 10^2$
$K_h = \frac{K_w}{K_a \times K_b}$
$\therefore K_b = \frac{K_w}{K_a \times K_h} = \frac{10^{-14}}{K_a \times 10^2} = \frac{10^{-16}}{K_a}$

$[H^+] = \sqrt{\frac{K_a \times K_w}{K_b}} = \sqrt{\frac{K_a \times 10^{-14}}{10^{-16} / K_a}} = 10 K_A$

$(pH)_{initial} = pK_a$
$(pH)_{final} = -log [H^+] = - log (10 K_a) = pK_a - 1$

Change in $pH = (pH)_{final} - (pH)_{initial} = pK_a - 1 - pK_a = -1$
Thus the $pH$ of the solution will decrease by 1.