Question

How much water must be added to $300mL$ of $0.2M$ solution of $C{H_3}COOH({K_a}) = 1.8 \times {10^{ - 5}}$ for the degree of dissociation of the acid to double?
(A) $600mL$
(B) $900mL$
(C) $1200mL$
(D) $1500mL$

Answer 1

Degree of dissociation: It is defined as the phenomenon of generating current carrying free ions, which are dissociated in the reactions to achieve the equilibrium, is known as degree of dissociation.

Complete step by step solution:
First of all let us define the degree of dissociation of the acid.
Molar Concentration: It is defined as the amount of substance per unit volume of solution.
Molarity: It is defined as the number of moles in the given volume of the solution. The unit of the molarity is as: $mol/litre$ .
Degree of dissociation: It is defined as the phenomenon of generating current carrying free ions, which are dissociated in the reactions to achieve the equilibrium, is known as degree of dissociation.
Here we are given with $300mL$ of $0.2M$ solution of $C{H_3}COOH({K_a}) = 1.8 \times {10^{ - 5}}$ and we have to find the amount of water which should be added for the degree of dissociation of acid to double the value.
We know that ${K_a} = {\alpha ^2}C$ , where ${K_a}$ is dissociation constant, $\alpha$ is degree of dissociation and $C$ is the concentration of the solute.
Here we know that dissociation is constant for all the acids and it does not change. And in the question it is given that $\dfrac{{{\alpha _2}}}{{{\alpha _1}}} = 2$ (because the degree of dissociation becomes double) so the ratio of concentration will become $\dfrac{{{\alpha _2}^2}}{{{\alpha _1}^2}} = \dfrac{{\dfrac{{{K_a}}}{{{C_2}}}}}{{\dfrac{{{K_a}}}{{{C_1}}}}} = \dfrac{{{C_1}}}{{{C_2}}}$ . So the value of concentration will become ${C_2} = {C_1}\dfrac{{{\alpha _1}^2}}{{{\alpha _2}^2}} = \dfrac{{0.2}}{4}$ .
Now we know that the relation between molarity and volume for two solutions is as:
${M_1}{V_1} = {M_2}{V_2}$ where ${M_1}$ is the molarity of one solution, ${M_2}$ is the molarity of second solution, ${V_1}$ is the volume of one solution and ${V_2}$ is the volume of second solution.
Here we are given with ${M_1} = 0.2$ , ${M_2} = \dfrac{{0.2}}{4}$ and ${V_1}$ is $300mL$ and we have to find the value of ${V_2}$ . Hence ${V_2} = \dfrac{{{M_1}{V_1}}}{{{M_2}}} = \dfrac{{0.2 \times 300 \times 4}}{{0.2}} = 1200mL$ .
So option C is correct.

Note:
Number of moles: It is defined as the ratio of mass of substance or molecule in the given solution to the molar mass of that substance or molecule in the solution.
Mole fraction: It is defined as the ratio of number of moles of the solute or solvent to the total number of moles in the solution.