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Question: Selenious acid ( \( {H_2}Se{O_3} \) ), a diprotic acid has \( {K_{a1}} = 3.0 \times {10^{ - 3}} \) a...

Selenious acid ( H2SeO3{H_2}Se{O_3} ), a diprotic acid has Ka1=3.0×10−3{K_{a1}} = 3.0 \times {10^{ - 3}} and Ka2=5.0×10−8{K_{a2}} = 5.0 \times {10^{ - 8}} . What is the [OH−][O{H^ - }] of a 0.30 M solution of a selenious acid?
A) 2.85×10−32.85 \times {10^{ - 3}}
B) 5.0×10−65.0 \times {10^{ - 6}}
C) 3.5×10−123.5 \times {10^{ - 12}}
D) 3.5×10−133.5 \times {10^{ - 13}}

Explanation

Solution

Diprotic acid is the acid that has the ability to donate two hydrogen ions or protons per molecule in an aqueous solution. Diprotic acid is also known as dibasic acid. Diprotic acid is a type of polyprotic acid, i.e. it has the ability to donate more than one proton per molecule.

Complete answer:
First let us see the dissociation of selenious acid in an aqueous solution. Since it is diprotic it will have two steps of dissociation.
H2SeO3⇌H++HSeO3−{H_2}Se{O_3} \rightleftharpoons {H^ + } + HSe{O_3}^ -
HSeO3−⇌H++SeO3−2HSe{O_3}^ - \rightleftharpoons {H^ + } + Se{O_3}^{ - 2}
The acid dissociation constant Ka{K_a} for the steps are given as Ka1=3.0×10−3{K_{a1}} = 3.0 \times {10^{ - 3}} and Ka2=5.0×10−8{K_{a2}} = 5.0 \times {10^{ - 8}} . From the values we can easily say that the value of Ka1>>>>Ka2{K_{a1}} > > > > {K_{a2}} . Hence, the second dissociation will be ignored and the H+{H^ + } ions from only the first dissociation will be considered. The concentration of the acid is 0.30M. The concentrations of the products after dissociation at equilibrium can be considered as ‘x’ . The equilibrium can be shown as:
H2SeO3 ⇌ H+ + HSeO3−{H_2}Se{O_3}{\text{ }} \rightleftharpoons {\text{ }}{H^ + }{\text{ }} + {\text{ }}HSe{O_3}^ -

T=00.30----
T=equilibrium0.30−x0.30 - xxxxx

The dissociation constant Ka{K_a} can be given as: Ka1=(x)(x)(0.30−x)=x20.30−x{K_{a1}} = \dfrac{{(x)(x)}}{{(0.30 - x)}} = \dfrac{{{x^2}}}{{0.30 - x}}
The value of x is very small compared to 0.30, hence can be ignored in the denominator. Therefore, Ka1=x20.30−x=3.0×10−3{K_{a1}} = \dfrac{{{x^2}}}{{0.30 - x}} = 3.0 \times {10^{ - 3}}
Ka1=x20.30=3.0×10−3{K_{a1}} = \dfrac{{{x^2}}}{{0.30}} = 3.0 \times {10^{ - 3}}
x2=(3.0×10−3)(0.30)=9×10−4{x^2} = (3.0 \times {10^{ - 3}})(0.30) = 9 \times {10^{ - 4}}
x=9×10−4=3×10−2x = \sqrt {9 \times {{10}^{ - 4}}} = 3 \times {10^{ - 2}}
The concentration of H+{H^ + } will be equal to [H+]=x=3×10−2M[{H^ + }] = x = 3 \times {10^{ - 2}}M
The relationship between H+{H^ + } and [OH−][O{H^ - }] can be given as: [H+][OH−]=10−14[{H^ + }][O{H^ - }] = {10^{ - 14}}
Therefore, [OH−]=10−14[H+]=10−143×10−2[O{H^ - }] = \dfrac{{{{10}^{ - 14}}}}{{[{H^ + }]}} = \dfrac{{{{10}^{ - 14}}}}{{3 \times {{10}^{ - 2}}}}
[OH−]=3.3×10−13≈3.5×10−13M[O{H^ - }] = 3.3 \times {10^{ - 13}} \approx 3.5 \times {10^{ - 13}}M
Hence the correct answer is Option D.

Note:
The first dissociation constant is highest in Polyprotic acids. The second dissociation is lesser. If we are given a Triprotic acid then the third dissociation will be much less than the second and so on. Since the values are very small, they are neglected keeping in solution only the first dissociation.