For H2S, K_a1 = 1.0x10^-7 and K_a2 = 1.0x10^-14. Neglecting the second ionisation, the pH of 0.1 M H2S is

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For $H X_{2} S$ , $K_{a 1} = 1.0 \times 1 0^{- 7}$ and $K_{a 2} = 1.0 \times 1 0^{- 14}$ . Neglecting the second ionisation, the pH of $0.1 M$ $H X_{2} S$ is

Only the first step matters:

[H X^{+}] = K_{a 1} C = 1 0^{- 7} \times 0.1 = 1 0^{- 4} M

, so the pH is

4