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# The dissociation constant of $H_2S$ and $HS^-$ are respectively $10^{-7}$ and $10^{-13}$. The $pH$ of 0.1 M aqueous solution of $H_2S$ will be

(a) 4

(b) 3

(c) 5

(d) 2.5

Since, $K_{a2} <<< K_{a1}$ Thus, $[H^+]$ ions are mainly obtained from the first step i.e.,
$H_2S \rightleftharpoons HS^- + H^+$

$\therefore [H^+] = \sqrt{KC} = \sqrt{10^{-7} \times 10^{-1}} = 10^{-4}$

$pH = -log(10^{-4}) = 4$