Question

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​

Answer 1

The pH of 0.1M  aqueous solution of H2S will be​ 4.

Explanation:

Here we see that, K2 is much smaller than K1 and so, it is expected that [H3O+ ]-hydronium ion is formed in first dissociation as:

H2S + H2O ⇄ H3O+ + HS-  

So, K1 =  [H3O+ ]  [HS- ] / [H2S]

Or, K1 =  [H3O+ ]2 / [H2S]

Or, [H3O+ ] = √(K1.[H2S]) = √(1.0 x 10-7 x 0.1) = 1.0 x 10-4  

Now, pH = - log [H3O+ ] = -log [ 1.0 x 10-4 ] = 4

Thus the pH of 0.1 M  aqueous solution of H2S will be​ 4.

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