Question

The acid dissociation constant of `"H"_2"S"` and `"HS"^-` are `10^-7` and `10^-13` respectively. The pH of 0.1 M aqueous solution of `"H"_2"S"` will be?

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

Answer 1

`pH approx 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 `K_a` values:

`K_a(1)=([H_3O^+] times [HS^-])/([H_2S])`

`K_a(2)=([H_3O^+] times [S^(2-)])/([HS^(-)])`

This acid will dissociate in two steps. We are given the concentration of `H_2S` so lets start from the top and work our way down.

`10^-7=([H_3O^+] times [HS^-])/([0.1])`

`10^-8=([H_3O^+] times [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:

`sqrt(10^-8)=10^-4=([H_3O^+] = [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=([H_3O^+] times [S^(2-)])/([10^-4])`

Same principle to find the concentration of `[H_3O^+]`:

`10^-17=([H_3O^+] times [S^(2-)])`

Hence:
`sqrt(10^-17)=3.16 times 10^-9=[H_3O^+] = [S^(2-)]`

So the combined concentration of `H_3O^+` will be:
`10^-4 + (3.16 times 10^-9) approx 10^-4`

`pH=-log[H_3O^+]`
`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 `H_3O^+` concentration, and hence the `pH` using the log law:

`log_10(10^x)=x`

But it is always good to be thorough :)