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 equilibriumH2S(Co-xaq)⇌HS-x-y(aq)+H+(x+yaq) HSx-y-(aq)⇌S-2y(aq)+H+y+x(aq)  Ka1=10-7=(x)(x+y)C0-x→1 Ka2=10-13=(x+y)(y)x-y→2Ka1 is very small and Ka2 is even smaller. Also the difference in the vlaues of Ka1 and Ka2 is huge.C0 - x ~ C0x - y ~ x;             x >>> yx + y ~ x(1) ⇒ 10-7 =x×xC0→(1)  ⇒ x2 = 10-8 ⇒ x = 10-4(2) ⇒ 10-7 =x×yx=y [H+] = x + y ~ x = 10-4 PH = -log[H+] = 4