Question

Selenious acid $ \left[ {{H_2}Se{O_3}} \right] $ , a diprotic acid has $ {K_{{a_1}}}{\text{ }} = {\text{ }}3.0{\text{ }} \times {\text{ }}{10^{ - 3}} $ and $ {K_{{a_2}}}{\text{ }} = {\text{ 5}}.0{\text{ }} \times {\text{ }}{10^{ - 8}} $ What is the $ \left[ {O{H^ - }} \right] $ of a $ 0.30{\text{ M}} $ solution of a selenious acid?
 $ (i){\text{ 2}}{\text{.85 }} \times {\text{ 1}}{{\text{0}}^{ - 3}} $
 $ (ii){\text{ 5}}{\text{.0 }} \times {\text{ 1}}{{\text{0}}^{ - 6}} $
 $ (iii){\text{ 3}}{\text{.5 }} \times {\text{ 1}}{{\text{0}}^{ - 12}} $
 $ (iv){\text{ 3}}{\text{.5 }} \times {\text{ 1}}{{\text{0}}^{ - 13}} $

Answer 1

Hint: Here we are provided with $ {K_{{a_1}}} $ and $ {K_{{a_2}}} $ for the selenious acid. We can observe that the value of $ {K_{{a_2}}} $ is much smaller than that of $ {K_{{a_1}}} $ . Thus we will write the equilibrium concentration of the selenious acid and with the help of $ {K_{{a_1}}} $ we can find the concentration of $ \left[ {O{H^ - }} \right] $ .
Ionic product of water:
 $ \left[ {{H^ + }} \right]{\text{ }}\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}{10^{ - 14}} $ .

Complete answer:
Selenious acid $ \left[ {{H_2}Se{O_3}} \right] $ produces two hydrogen atoms and thus we are given with $ {K_{{a_1}}} $ and $ {K_{{a_2}}} $ for the selenious acid. But we can observe that the value of $ {K_{{a_1}}} $ is far greater than that of $ {K_{{a_2}}} $ . Therefore we can ignore the dissociation constant $ {K_{{a_2}}} $ . It can be written as:
Since, $ {K_{{a_2}}} \ll {K_{{a_1}}} $ , hence $ {K_{{a_2}}} $ can be ignored or neglected.
Now the dissociation of $ \left[ {{H_2}Se{O_3}} \right] $ can be represented as:
 $ {H_2}Se{O_3} \rightleftharpoons HSe{O_3}^{ - 1} + {H^ + } $
Now it is given that the initial concentration of selenious acid is $ 0.30{\text{ M}} $ . Let's assume $ x $ concentration of selenious acid is converted into products at equilibrium. It can be represented as:

Time	$ \left[ {{H_2}Se{O_3}} \right] $	$ \left[ {HSe{O_3}^{ - 1}} \right] $	$ \left[ {{H^ + }} \right] $
$ t = 0 $	$ 0.30 $	$ 0 $	$ 0 $
$ t = {t_{eq.}} $	$ 0.30 - x $	$ x $	$ x $

Let us assume $ x $ is very less than $ 0.30 $ , then $ 0.30 - x{\text{ }} \approx {\text{ }}0.30 $ .
Now the dissociation constant can be written as:
 $ {K_{{a_1}}}{\text{ }} = {\text{ }}\dfrac{{\left[ {{H^ + }} \right]{\text{ }}\left[ {HSe{O_3}^{ - 1}} \right]}}{{\left[ {{H_2}Se{O_3}} \right]}} $
On substituting the values we get,
 $ {\text{3}}{\text{.0 }} \times {\text{ 1}}{{\text{0}}^{ - 3}}{\text{ }} = {\text{ }}\dfrac{{x{\text{ }} \times x}}{{0.30}} $
 $ \Rightarrow {x^2}{\text{ }} = {\text{ 9 }} \times {\text{ 1}}{{\text{0}}^{ - 4}} $
By taking square root both side it can be solved as:
 $ \Rightarrow x{\text{ }} = {\text{ 3 }} \times {\text{ 1}}{{\text{0}}^{ - 2}} $
We know that the ionic product of water can be written as:
 $ \Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ }}\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}{10^{ - 14}} $
 $ \Rightarrow {\text{ }}\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}\dfrac{{{{10}^{ - 14}}}}{{\left[ {{H^ + }} \right]}} $
On substituting values we get the result as:
 $ \Rightarrow {\text{ }}\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}\dfrac{{{{10}^{ - 14}}}}{{3{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}}} $
 $ \Rightarrow {\text{ }}\left[ {O{H^ - }} \right]{\text{ }} = {\text{ 3}}{\text{.3 }} \times {\text{ 1}}{{\text{0}}^{ - 13}} $
Thus the concentration of $ \left[ {O{H^ - }} \right] $ will be $ {\text{3}}{\text{.3 }} \times {\text{ 1}}{{\text{0}}^{ - 13}}{\text{ M}} $ . It can be approximated to $ {\text{3}}{\text{.5 }} \times {\text{ 1}}{{\text{0}}^{ - 13}} $ . Therefore the correct option is $ (iv){\text{ 3}}{\text{.5 }} \times {\text{ 1}}{{\text{0}}^{ - 13}} $ .

Note:
It must be noted that diprotic acids are those acids which can produce two hydrogen ions one after another. Since the value of the second dissociation constant of selenious acid is too small compared to the first dissociation constant, we can neglect it. It would not affect our calculations. Since selenious acid produces hydrogen acid, that’s why we have used an ionic product of water to find the concentration of hydroxide ions.