Question

What is pH of $0.1{\text{ M }}{{\text{H}}_{\text{2}}}{\text{S}}$ solution? Given that:;
${{\text{K}}_{{\text{a1}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right) = 1.0 \times {10^{ - 7}}$
${{\text{K}}_{{\text{a2}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right) = 1.3 \times {10^{ - 14}}$

Answer 1

Hint:TWe know that pH measures the acidity of an acid or it measures how easily an acid gives away a proton. We are given ${{\text{H}}_{\text{2}}}{\text{S}}$ which is a diprotic acid having two ionisable hydrogen atoms. The pH of ${{\text{H}}_{\text{2}}}{\text{S}}$ includes the equilibrium constants of both the deprotonation reactions.

 Complete step by step answer:We are given a diprotic acid. Diprotic acid means there are two ionisable hydrogen atoms and the acid donates two protons when dissolved in water. It is also known as diprotic acid.
The dissociation reactions of ${{\text{H}}_{\text{2}}}{\text{S}}$ are as follows:
${{\text{H}}_{\text{2}}}{\text{S}} \to {{\text{H}}^ + } + {\text{H}}{{\text{S}}^ - }{\text{ }}{{\text{K}}_{{\text{a1}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right) = 1.0 \times {10^{ - 7}}$
${\text{H}}{{\text{S}}^ - } \to {{\text{H}}^ + } + {{\text{S}}^ - }{\text{ }}{{\text{K}}_{{\text{a2}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right) = 1.3 \times {10^{ - 14}}$
The pH of ${{\text{H}}_{\text{2}}}{\text{S}}$ includes the equilibrium constants of both the deprotonation reactions.
We are given that ${{\text{K}}_{{\text{a1}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right) = 1.0 \times {10^{ - 7}}$ and ${{\text{K}}_{{\text{a2}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right) = 1.3 \times {10^{ - 14}}$ where ${{\text{K}}_{{\text{a1}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right)$ is the first dissociation constant of ${{\text{H}}_{\text{2}}}{\text{S}}$ and ${{\text{K}}_{{\text{a2}}}}$ is the second dissociation constant of ${{\text{H}}_{\text{2}}}{\text{S}}$.
From the given values, we can see that the value of second dissociation constant of ${{\text{H}}_{\text{2}}}{\text{S}}$ is very low compared to the value of first dissociation constant of ${{\text{H}}_{\text{2}}}{\text{S}}$. Thus, the value of the second dissociation constant is considered as negligible. Thus, the pH of ${{\text{H}}_{\text{2}}}{\text{S}}$ is calculated from the first dissociation constant.
From the reaction, the expression for the dissociation constant is,
${{\text{K}}_{{\text{a1}}}} = \dfrac{{[{{\text{H}}^ + }][{{\text{S}}^ - }]}}{{[{{\text{H}}_{\text{2}}}{\text{S}}]}}$
And, $[{{\text{H}}^ + }] = [{{\text{S}}^ - }]$. Thus,
${{\text{K}}_{{\text{a1}}}} = \dfrac{{{{[{{\text{H}}^ + }]}^2}}}{{[{{\text{H}}_{\text{2}}}{\text{S}}]}}$
${[{{\text{H}}^ + }]^2} = {{\text{K}}_{{\text{a1}}}} \times [{{\text{H}}_{\text{2}}}{\text{S}}]$
We are given that ${{\text{K}}_{{\text{a1}}}}\left( {{{\text{H}}_{\text{2}}}{\text{S}}} \right) = 1.0 \times {10^{ - 7}}$ and the concentration of ${{\text{H}}_{\text{2}}}{\text{S}}$ solution is $0.1{\text{ M}}$. Thus,
${[{{\text{H}}^ + }]^2} = 1.0 \times {10^{ - 7}} \times 0.1{\text{ M}}$
${[{{\text{H}}^ + }]^2} = 1.0 \times {10^{ - 8}}{\text{ M}}$
$[{{\text{H}}^ + }] = 1.0 \times {10^{ - 4}}{\text{ M}}$
Thus, the concentration of hydrogen ion is $1.0 \times {10^{ - 4}}{\text{ M}}$.
We know that pH is the negative logarithm of the hydrogen ion concentration. Thus,
${\text{pH}} = - \log [{{\text{H}}^ + }]$
Substitute $1.0 \times {10^{ - 4}}{\text{ M}}$ for the concentration of hydrogen ion. Thus,
${\text{pH}} = - \log \left( {1.0 \times {{10}^{ - 4}}{\text{ M}}} \right)$
${\text{pH}} = - \left( { - 4} \right)$
${\text{pH}} = 4$
Thus, the pH of $0.1{\text{ M }}{{\text{H}}_{\text{2}}}{\text{S}}$ solution is 4.

 Note: The diprotic acid loses its first proton more easily than it loses its second proton. With each ionization step, the removal of protons becomes difficult. This is because the negative charge increases and the effective nuclear charge increases. The pH of $0.1{\text{ M }}{{\text{H}}_{\text{2}}}{\text{S}}$ solution is 4 which indicates that $0.1{\text{ M }}{{\text{H}}_{\text{2}}}{\text{S}}$ solution is acidic in nature.