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?

2 3 4 5

Apr 8, 2018

$p H \approx 4$ so option 3.

Explanation:

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

${K}_{a} \left(1\right) = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[H {S}^{-}\right]}{\left[{H}_{2} S\right]}$

${K}_{a} \left(2\right) = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[{S}^{2 -}\right]}{\left[H {S}^{-}\right]}$

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

${10}^{-} 7 = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[H {S}^{-}\right]}{\left[0.1\right]}$

${10}^{-} 8 = \left(\left[{H}_{3} {O}^{+}\right] \times \left[H {S}^{-}\right]\right)$
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:

$\sqrt{{10}^{-} 8} = {10}^{-} 4 = \left(\left[{H}_{3} {O}^{+}\right] = \left[H {S}^{-}\right]\right)$

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

${10}^{-} 13 = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[{S}^{2 -}\right]}{\left[{10}^{-} 4\right]}$

Same principle to find the concentration of $\left[{H}_{3} {O}^{+}\right]$:

${10}^{-} 17 = \left(\left[{H}_{3} {O}^{+}\right] \times \left[{S}^{2 -}\right]\right)$

Hence:
$\sqrt{{10}^{-} 17} = 3.16 \times {10}^{-} 9 = \left[{H}_{3} {O}^{+}\right] = \left[{S}^{2 -}\right]$

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

$p H = - \log \left[{H}_{3} {O}^{+}\right]$
$p H = - \log \left[{10}^{-} 4\right]$
$p H = 4$

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 ${H}_{3} {O}^{+}$ concentration, and hence the $p H$ using the log law:

${\log}_{10} \left({10}^{x}\right) = x$

But it is always good to be thorough :)