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Question: The acid dissociation constant of \( {{\text{H}}_2}\;{\text{S}} \) and \( {\text{H}}{{\text{S}}^ - }...

The acid dissociation constant of H2  S{{\text{H}}_2}\;{\text{S}} and HS−{\text{H}}{{\text{S}}^ - } are 10−7{10^{ - 7}} and 10−13{10^{ - 13}} respectively. The pH of 0.1M0.1{\text{M}} aqueous solution of H2  S{{\text{H}}_2}\;{\text{S}} will be
(A) 22
(B) 33
(C) 44
(D) 55

Explanation

Solution

The negative logarithm of H+{{\rm H}^ + } ion concentration is defined as pH. The meaning of the pH name is therefore justified by the power of hydrogen. We shall calculate the molarity of the hydrogen ions from the acid dissociation constant given using the equilibrium constant and thus, find the pH.

Formula Used
We can find out the pH value of any substance by the following formula
pH=−log⁡[H+]pH = - \log \left[ {{H^ + }} \right]
Where
pHpH is the required pH value
[H+]\left[ {{H^ + }} \right] is the concentration of H+{H^ + } ions.

Complete Step-by-Step Solution
According to the question, the following information is provided to us:
The acid dissociation constant of H2  S{{\text{H}}_2}\;{\text{S}} is 10−7{10^{ - 7}}
The acid dissociation constant of HS−{\text{H}}{{\text{S}}^ - } is 10−13{10^{ - 13}}
The molarity of the aqueous H2  S{{\text{H}}_2}\;{\text{S}} is 0.1M0.1 M
The following compound H2  S{{\text{H}}_2}\;{\text{S}} can be broken down as
H2S→HS−+H+{H_2}S \to H{S^ - } + {H^ + }
Ka=x2(10−1−x){K_a} = \dfrac{{{x^2}}}{{\left( {{{10}^{ - 1}} - x} \right)}}
Where
Ka{K_a} is the acid dissociation constant
We can see that x<<10−1x < < {10^{ - 1}}
Therefore, we can neglect xx since it is very small quantity
Such that 10−1−x≈10−1{10^{ - 1}} - x \approx {10^{ - 1}}
Now,
10−1×10−7=x2{10^{ - 1}} \times {10^{ - 7}} = {x^2}
Upon further solving, we get
∴x=10−4\therefore x = {10^{ - 4}}
Now, we will determine the value of pH of H2  S{{\text{H}}_2}\;{\text{S}}
That is,
pH=−log⁡[H+]pH = - \log \left[ {{H^ + }} \right]
Upon substituting the values, we get
pH=−log⁡[10−4]pH = - \log \left[ {{{10}^{ - 4}}} \right]
⇒pH=4log10\Rightarrow pH = 4log10
On solving, we get
∴pH=4\therefore pH = 4
Hence, the correct option is (C).

Additional Information
pH is a measure of the acidic/basic nature of substances. With 77 being neutral, the range goes from 00 to 1414 . Acidity is indicated by a pH of less than 77 , whereas a base is indicated by a pH greater than 77 . Actually, pH is a measure of the relative amount in the water of free hydrogen and hydroxyl ions. Acidic water with more free hydrogen ions is acidic, while water with more free hydroxyl ions is basic.

Note
The pH is a measure of the hydrogen ion concentration, the acidity or alkalinity of a solution. Normally, the pH-scale is between 00 and 1414 . Aqueous solutions are acidic at 25∘C{25^ \circ }C with a pH of less than 77 and basic or alkaline solutions are those with a pH of more than 77 .