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Question: The dissociation constant \[{\text{(}}{{\text{K}}_{\text{a}}}{\text{)}}\] and per cent of the degree...

The dissociation constant (Ka){\text{(}}{{\text{K}}_{\text{a}}}{\text{)}} and per cent of the degree of dissociation (α)\left( {{\alpha }} \right){\text{}} of a weak monobasic acid solution of 0.2  M  0.2\;{\text{M}}\; with pH=6  {\text{pH}} = 6\; are respectively :
A.5×10−9,5×10−125 \times {10^{ - 9}},5 \times {10^{ - 12}}
B.5×10−13,5×10−45 \times {10^{ - 13}},5 \times {10^{ - 4}}
C.5×10−12,5×10−45 \times {10^{ - 12}},5 \times {10^{ - 4}}
D.5×10−5,5×10−25 \times {10^{ - 5}},5 \times {10^{ - 2}}

Explanation

Solution

To answer this question you must know the concept of degree of dissociation. It refers to the ratio of the reactant molecules ionized to the number of starting reactant molecules.
Formula used:
pH = − log [H3O + ]{\text{pH }} = {\text{ }} - {\text{ log [}}{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}]
(α)  \left( \alpha \right)\; = (Number of reactant molecules dissociated at the start)(Number of reactant molecules at the start) = {\text{ }}\dfrac{{\left( {{\text{Number of reactant molecules dissociated at the start}}} \right)}}{{\left( {{\text{Number of reactant molecules at the start}}} \right)}}

Complete step by step answer:
For calculation of the degree of dissociation you need to divide the mass of dissociated ions by the total mass of dissociated and undissociated species. This can be done using the balanced reaction too by the use of proper coefficients from the balanced reaction.
From the definition of dissociation, we know that Degree of dissociation or ionization
(α)  \left( \alpha \right)\; = (Number of reactant molecules dissociated at the start)(Number of reactant molecules at the start) = {\text{ }}\dfrac{{\left( {{\text{Number of reactant molecules dissociated at the start}}} \right)}}{{\left( {{\text{Number of reactant molecules at the start}}} \right)}}
The reaction for dissociation can be represented by the equation:
HX→H++X−HX \to {H^ + } + {X^ - }.
Let the amount dissociated be cα{{c\alpha }} where cα{{c\alpha }} is the degree of dissociation.
In the question we have c = 0.2{\text{c = 0}}{\text{.2}}
So the concentration of monobasic acid now left after dissociation will be 0.2−x0.2 - x.
∴\therefore the concentration of H+  {H^ + }\; formed = x and concentration of X−{X^ - } formed = x .
The pH of a given solution is pH=6  {\text{pH}} = 6\; which means that H+  {H^ + }\; concentration in the solution is, x=10−6{\text{x}} = {10^{ - 6}}.
So, Ka=0.2x2{{\text{K}}_{\text{a}}} = 0.2{{\text{x}}^2}
⇒Ka=5×10−12\Rightarrow {{\text{K}}_{\text{a}}} = 5 \times {10^{ - 12}}.
Now ultimately the dissociation percentage = 10 - 60.2×100=5×10 - 4\dfrac{{{\text{1}}{{\text{0}}^{{\text{ - 6}}}}}}{{{\text{0}}{\text{.2}}}}{{ \times 100 = 5 \times 1}}{{\text{0}}^{{\text{ - 4}}}}.

Hence, the correct option is option C.

Note:
You should remember the difference between strong bases and weak bases: a strong base is a base that is ionized in solution but if it is less than ionized in solution, it is a weak base. There are only a few strong bases and certain salts also will affect the acidity or basicity of aqueous solutions because a number of the ions will undergo hydrolysis. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyse, while salts with ions that are part of weak acids or bases will hydrolyse. Various factors affect the degree of dissociation: Nature of the electrolyte: strong, weak, insoluble, Nature of the solvent: High dielectric solvents increase ionization, Dilution: larger the dilution higher the ionization, Temperature: higher the temperature, larger the ionization and presence of common ions between the solvent decreases the ionization of the weak electrolyte.