The dissociation constant of H₂S and HS^- are respectively 10^-7 and 10^-13 . The pH of 0.1M aqueous solution of H₂S will be

Let us consider a solution of H₂S(g)

Following equilibria gets established in the given solution.

At equilibrium

$\underset{{\mathrm{C}}_{\mathrm{o}}-\mathrm{x}}{{\mathrm{H}}_2\mathrm{S}(}\mathrm{aq})\rightleftharpoons \underset{\mathrm{x}-\mathrm{y}}{{\mathrm{HS}}^{-}}(\mathrm{aq})+\underset{\mathrm{x}+\mathrm{y}}{{\mathrm{H}}^{+}(}\mathrm{aq})$

 ${\underset{\mathrm{x}-\mathrm{y}}{\mathrm{HS}}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons \underset{\mathrm{y}}{{\mathrm{S}}^{-2}}\left(\mathrm{aq}\right)+\underset{\mathrm{y}+\mathrm{x}}{{\mathrm{H}}^{+}}\left(\mathrm{aq}\right)$

  ${\mathrm{K}}_{{\mathrm{a}}_1}={10}^{-7}=$$\frac{\left(\mathrm{x}\right)(\mathrm{x}+\mathrm{y})}{{\mathrm{C}}_0-\mathrm{x}}\to 1$

 ${\mathrm{K}}_{{\mathrm{a}}_2}={10}^{-13}=$$\frac{(\mathrm{x}+\mathrm{y})\left(\mathrm{y}\right)}{\mathrm{x}-\mathrm{y}}\to 2$

K_a1 is very small and K_a2 is even smaller. Also the difference in the vlaues of K_a1 and K_a2 is huge.

C₀ - x ~ C₀

x - y ~ x;             x >>> y

x + y ~ x

(1) ⇒ 10^-7 =$\frac{\mathrm{x}\times \mathrm{x}}{{\mathrm{C}}_0}\to \left(1\right)$

  ⇒ x² = 10^-8 ⇒ x = 10^-4

(2) ⇒ 10^-7 =$\frac{\mathrm{x}\times \mathrm{y}}{\mathrm{x}}=\mathrm{y}$

 [H⁺] = x + y ~ x = 10^-4

 P^H = -log[H⁺] = 4