The acid dissociation constant of $H_{2} S$ $"H"_2"S"$ and ${HS}^{-}$ $"HS"^-$ are $10^{- 7}$ $10^-7$ and $10^{- 13}$ $10^-13$ respectively. The pH of 0.1 M aqueous solution of $H_{2} S$ $"H"_2"S"$ will be?

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The acid dissociation constant of $H_{2} S$ $"H"_2"S"$ and ${HS}^{-}$ $"HS"^-$ are $10^{- 7}$ $10^-7$ and $10^{- 13}$ $10^-13$ respectively. The pH of 0.1 M aqueous solution of $H_{2} S$ $"H"_2"S"$ will be?

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Answer 1 · 2018-04-08T12:48:05

$p H \approx 4$ $pH approx 4$ so option 3.
Disclaimer: Somewhat long answer, but the answer is not as bad as one might think!

Explanation:

To find the $p H$ $pH$ we must find how far it has dissociated:
Let's set up some equation using the $K_{a}$ $K_a$ values:

$K_{a} (1) = \frac{[H_{3} O^{+}] \times [H S^{-}]}{[H_{2} S]}$ $K_a(1)=([H_3O^+] times [HS^-])/([H_2S])$

$K_{a} (2) = \frac{[H_{3} O^{+}] \times [S^{2 -}]}{[H S^{-}]}$ $K_a(2)=([H_3O^+] times [S^(2-)])/([HS^(-)])$

This acid will dissociate in two steps. We are given the concentration of $H_{2} S$ $H_2S$ so lets start from the top and work our way down.

$10^{- 7} = \frac{[H_{3} O^{+}] \times [H S^{-}]}{[0.1]}$ $10^-7=([H_3O^+] times [HS^-])/([0.1])$

$10^{- 8} = ([H_{3} O^{+}] \times [H S^{-}])$ $10^-8=([H_3O^+] times [HS^-])$
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:

$\sqrt{10^{- 8}} = 10^{- 4} = ([H_{3} O^{+}] = [H S^{-}])$ $sqrt(10^-8)=10^-4=([H_3O^+] = [HS^-])$

Now in the second dissociation, $[H S^{-}]$ $[HS^-]$ will act as the acid. That means we plug in the concentration found in the first calculation in the denominator of the second dissociation:

$10^{- 13} = \frac{[H_{3} O^{+}] \times [S^{2 -}]}{[10^{- 4}]}$ $10^-13=([H_3O^+] times [S^(2-)])/([10^-4])$

Same principle to find the concentration of $[H_{3} O^{+}]$ $[H_3O^+]$ :

$10^{- 17} = ([H_{3} O^{+}] \times [S^{2 -}])$ $10^-17=([H_3O^+] times [S^(2-)])$

Hence:
$\sqrt{10^{- 17}} = 3.16 \times 10^{- 9} = [H_{3} O^{+}] = [S^{2 -}]$ $sqrt(10^-17)=3.16 times 10^-9=[H_3O^+] = [S^(2-)]$

So the combined concentration of $H_{3} O^{+}$ $H_3O^+$ will be:
$10^{- 4} + (3.16 \times 10^{- 9}) \approx 10^{- 4}$ $10^-4 + (3.16 times 10^-9) approx 10^-4$

$p H = - log [H_{3} O^{+}]$ $pH=-log[H_3O^+]$
$p H = - log [10^{- 4}]$ $pH=-log[10^-4]$
$p H = 4$ $pH=4$

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 $10^{- 8}$ $10^-8$ to find the $H_{3} O^{+}$ $H_3O^+$ concentration, and hence the $p H$ $pH$ using the log law:

${log}_{10} (10^{x}) = x$ $log_10(10^x)=x$

But it is always good to be thorough :)

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The acid dissociation constant of $H_{2} S$ $"H"_2"S"$ and ${HS}^{-}$ $"HS"^-$ are $10^{- 7}$ $10^-7$ and $10^{- 13}$ $10^-13$ respectively. The pH of 0.1 M aqueous solution of $H_{2} S$ $"H"_2"S"$ will be?

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Explanation:

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The acid dissociation constant of "H"_2"S"H2S"H"_2"S" and "HS"^-HS−"HS"^- are 10^-710−710^-7 and 10^-1310−1310^-13 respectively. The pH of 0.1 M aqueous solution of "H"_2"S"H2S"H"_2"S" will be?

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The acid dissociation constant of $H_{2} S$ $"H"_2"S"$ and ${HS}^{-}$ $"HS"^-$ are $10^{- 7}$ $10^-7$ and $10^{- 13}$ $10^-13$ respectively. The pH of 0.1 M aqueous solution of $H_{2} S$ $"H"_2"S"$ will be?

2

3

4

5