Question

Approximate pH of 0.1 M aqueous $H_{2}S$ solution when $K_1$ and $K_2$ for $H_{2}S$ at 25${}^{0}C$ are $1\text{x 1}{\text{0}}^{-7}$ and $1.3\text{x 1}{\text{0}}^{-13}$ respectively?

Answer 1

Hint: $H_{2}S$ is a dibasic acid i.e. can release two protons per molecule of $H_{2}S$. The pH of $H_{2}S$ will include the equilibrium constant of both the deprotonation reaction. However, it is important to decide the significance of second equilibrium constant as it has a very low and insignificant value. Based on this you can find the final $K_a$ for $H_{2}S$ and then the pH as well.

Complete step-by-step answer:
Hydrogen sulphide is the chemical compound having the molecular formula $H_{2}S$. It is a colourless chalcogen hydride gas having the characteristic smell of rotten eggs. It is a highly corrosive and poisonous substance which is easily flammable. Hydrogen sulphide is slightly denser than air. A mixture of hydrogen sulphide and air can be highly explosive. It acts as a reducing agent especially in presence of its conjugate base. Hydrogen sulphide exists as a gas at 298 K temperature. When we observe the values of first dissociation constant and second dissociation constant, it is seen that the main dissociation happens at the first deprotonation only. So, we will consider the second dissociation as insignificant as it is lower than the concentration of hydrogen ions in neutral water.
$H_{2}S\to [H^{+}]+[H{S}^{-}]$
The dissociation in terms of concentration of products and reactants can be written as,
$K_1={\displaystyle \frac{[H^{+}][H{S}^{-}]}{[H_{2}S]}}={\displaystyle \frac{{[H^{+}]}^2}{[H_{2}S]}}$
$[H^{+}]=\sqrt{K_1[H_{2}S]}={10}^{-4}$
$pH=-\log [H^{+}]=4$
From the above set of reactions, we can conclude that the pH for 0.1 M hydrogen sulfide is 4.

Therefore, the correct answer is option (A).

Note: It is important to know that we do not consider the dissociation reactions, whose reaction constant is less than that of water. This is because the reaction will never take place.