Question

A dilute HCl solution saturated with ${{H}_{2}}S$(0.1M) has pH value of 3. The $[{{S}^{2-}}]$ is
(given the dissociation constant of ${{H}_{2}}S$are ${{K}_{1}}=1X{{10}^{-7}},{{K}_{2}}=1.3X{{10}^{-13}}$)
(A).$2 \times {{10}^{-13}}$
(B).$2.4 \times {{10}^{-13}}$
(C).$3 \times {{10}^{-15}}$
(D).$1.3 \times {{10}^{-15}}$

Answer 1

Hint: pH is defined as the power of hydrogen. To solve this question first we have to write the expression for ${{K}_{1}}$and ${{K}_{2}}$and then use its value to calculate the concentration of $[{{S}^{2-}}]$.

Complete answer:
Given in the question:
The pH of the solution = 3
The value of first dissociation constant for ${{H}_{2}}S$= ${{K}_{1}}=1X{{10}^{-7}}$
The value of second dissociation constant for ${{H}_{2}}S$=${{K}_{2}}=1.3X{{10}^{-13}}$
Let's write the steps for dissociation of ${{H}_{2}}S$
In the first step ${{H}_{2}}S$ will dissociates to hydrogen ion and hydrogen sulphide ion as shown in the below mentioned equation:
${{H}_{2}}S\rightleftharpoons {{H}^{+}}+H{{S}^{-}}$---1
In the second step the $H{{S}^{-}}$ ions dissociates into hydrogen ion and sulphide ion as shown in the below mentioned equation:
$H{{S}^{-}}\rightleftharpoons {{H}^{+}}+{{S}^{-}}$---2
The expression for first dissociation constant for equation 1,${{H}_{2}}S$= ${{K}_{1}}=\dfrac{[{{H}^{+}}][H{{S}^{-}}]}{[{{H}_{2}}S]}$
The expression for first dissociation constant for equation 2,${{H}_{2}}S$= ${{K}_{2}}=\dfrac{[{{H}^{+}}][{{S}^{2-}}]}{[HS]}$
Now we can write that:
\[\begin{align}
  & {{K}_{1}}.{{K}_{2}}=\dfrac{{{[{{H}^{+}}]}^{2}}[{{S}^{2-}}]}{[{{H}_{2}}S]} \\
 & 1.3({{10}^{-20}})=\dfrac{{{[{{10}^{-3}}]}^{2}}[{{S}^{2-}}]}{[0.1]} \\
 & [{{S}^{2-}}]=(1.3)({{10}^{-15}})M \\
\end{align}\]

Hence the correct answer for the question mentioned above is option (D).

Additional information:
Dissociation is basically the separation of ions that occurs when a solid compound dissolves. Strong acids and the strong bases completely dissociates and the value of degree of dissociation for strong acids and strong bases is approximately equal to 1. The ionization reaction of the acids and bases are reversible which means that the products can combine again and form the reactants.

Note:
While solving this question it is important to correctly write the equation of \[{{K}_{1}}\] and \[{{K}_{2}}\]it is defined as the product of concentration of product to the product of concentration of reactants each raise to the power of their respective stoichiometric coefficient.