Question

$0.01M$ HA (aq) is $2%$ ionized: $\left[ O{{H}^{-}} \right]$ of the solution is?
A.$0.5\times {{10}^{-10}}$
B.${{10}^{-8}}$
C.$2\times {{10}^{-11}}$
D.$2\times {{10}^{-12}}$

Answer 1

Hint:Degree of dissociation is defined as the fraction of dissociated moles to the initial moles. It produces free ions carrying current. Dissociation in chemistry is defined as breaking down of larger particles into smaller particles such as ions, atoms and radicals.

Complete step-by-step answer:In ionic equilibrium, the ionic substances dissociate into ions and always remain in equilibrium with its undissociated solute.
Firstly let us see the dissociation reaction of HA:
                               $H{{A}_{\left( aq \right)}}\rightleftharpoons H_{\left( aq \right)}^{+}+A_{\left( aq \right)}^{-}$
At equilibrium $C\left( 1-\alpha \right)$ $C\alpha $ $C\alpha $
The concentration of $\left[ {{H}^{+}} \right]=C\alpha $
Here, $\alpha $ is degree of dissociation
$C$is the initial moles $\left( C=O.O1M \right)$
In this question, it is given that the concentration of acid $\left[ HA \right]=0.01M$
The percentage of degree of dissociation $\left( \alpha \right)=2%$
The degree of dissociation $\left( \alpha \right)=0.02$
As we know that,
$\left[ {{H}^{+}} \right]=C\alpha $
On substituting the values in the above formula we get,
$\left[ {{H}^{+}} \right]=0.01\times 0.02$
$\left[ {{H}^{+}} \right]=2\times {{10}^{-4}}$
At equilibrium, ${{K}_{W}}={{10}^{-14}}$
${{K}_{W}}=\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]$
Where, ${{K}_{W}}$ is the ionic product of water
On substituting the value we get,
${{10}^{-14}}=2\times {{10}^{-4}}\left[ O{{H}^{-}} \right]$
On further solving we get,
$\left[ O{{H}^{-}} \right]=\dfrac{{{10}^{-14}}}{2\times {{10}^{-4}}}$
$\left[ O{{H}^{-}} \right]=0.5\times {{10}^{-10}}$
The $\left[ O{{H}^{-}} \right]$ in the solution is $0.5\times {{10}^{-10}}$

Additional information:
Electrolytes are the substances that dissociate into ions and conduct electricity in aqueous solution.
Electrolytes are of two types: strong electrolytes and weak electrolytes
Strong electrolytes are the substances that dissociate completely in aqueous solution. For example: $NaCl$
Weak electrolytes are the substances that do not dissociate completely in aqueous solution. For example: $C{{H}_{3}}COOH$

Therefore, the correct option is A.

Note:Factors on which degree of dissociation depends upon:
*Degree of dissociation depends upon the nature of electrolyte. If it is a strong electrolyte, it dissociates completely otherwise in weak electrolytes, it partially dissociates.
*It also depends upon the nature of solvent. High dielectric solvent increases dissociation.
*It also depends upon dilution. If dilution is more, dissociation will also increase.
*It depends upon the temperature. If temperature increases, dissociation will also increases.