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

  1. 2
  2. 3
  3. 4
  4. 5

1 Answer
Oliver S.
Apr 8, 2018

pH≈4 so option 3.
Disclaimer: Somewhat long answer, but the answer is not as bad as one might think!

Explanation:

To find the pH we must find how far it has dissociated:
Let's set up some equation using the Ka values:

Ka(1)=[H3O+]×[HS−][H2S]

Ka(2)=[H3O+]×[S2−][HS−]

This acid will dissociate in two steps. We are given the concentration of H2S so lets start from the top and work our way down.

10−7=[H3O+]×[HS−][0.1]

10−8=([H3O+]×[HS−])
Then we can assume that both of these species are in a 1:1 ratio in the dissociation, allowing us to take the square root to find the concentration of both species:

√10−8=10−4=([H3O+]=[HS−])

Now in the second dissociation, [HS−] will act as the acid. That means we plug in the concentration found in the first calculation in the denominator of the second dissociation:

10−13=[H3O+]×[S2−][10−4]

Same principle to find the concentration of [H3O+]:

10−17=([H3O+]×[S2−])

Hence:
√10−17=3.16×10−9=[H3O+]=[S2−]

So the combined concentration of H3O+ will be:
10−4+(3.16×10−9)≈10−4

pH=−log[H3O+]
pH=−log[10−4]
pH=4

So the second dissocation was so small it did not really impact the pH. I guess if this was a multiple choice exam then you only needed to look at the first dissociation and find the square root of 10−8 to find the H3O+ concentration, and hence the pH using the log law:

log10(10x)=x

But it is always good to be thorough :)

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