Question

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

Answer 1

Hint:We know from Ostwald dilution law that dissociation constant of a weak acid is given as-
$ \Rightarrow {K_\alpha } = C{\alpha ^2}$
Where C is the concentration of acid, ${K_\alpha }$ is the dissociation constant and $\alpha $is the degree of dissociation.
Use this to find the degree of dissociation of acetic acid and the solution of acetic acid and water. Then use the formula-${M_1}{V_1} = {M_2}{V_2}$ where ${M_1}$ is the molar concentration of acetic acid, ${M_2}$ is the molar concentration of solution of acid and water, ${V_1}$ is the volume of acetic acid and ${V_2}$ is the volume of solution. Put the values in the formula to get${V_2}$. Then find the amount of water added by subtracting the volume of acetic acid from the volume of solution.

Complete step by step answer:
Given, molarity of acetic acid is=$0.2$M
The volume of acetic acid=$300$ mL
And dissociation constant of acetic acid $C{H_3}COOH{\left( K \right)_\alpha } = 1.8 \times {10^{ - 5}}$
We have to find the amount of water added to make the degree of dissociation of the solution double the given degree of dissociation.
Let the degree of dissociation of the solution is represented as ${\alpha _2}$
Then according to the question, ${\alpha _2} = 2{\alpha _1}$ where ${\alpha _1}$ is the degree of dissociation of acetic acid.
Then we can write it as-
$ \Rightarrow \dfrac{{{\alpha _2}}}{{{\alpha _1}}} = 2$ --- (i)
Now, according to Oswald dilution law,
$ \Rightarrow {K_\alpha } = C{\alpha ^2}$
Where C is the concentration of acid, ${K_\alpha }$ is the dissociation constant and $\alpha $is the degree of dissociation.
Then for ${\alpha _1}$, we can write-
$ \Rightarrow {\alpha _1} = \sqrt {\dfrac{{{C_1}}}{{{K_\alpha }}}} $-- (ii)
Where ${C_1}$ is the concentration of acetic acid.
And for ${\alpha _2}$, we can write-
$ \Rightarrow {\alpha _2} = \sqrt {\dfrac{{{C_2}}}{{{K_\alpha }}}} $--- (iii)
Where ${C_2}$ is the concentration of solution.
Now substituting the value of eq. (ii) and eq. (iii) in eq. (i), we get-
$ \Rightarrow \dfrac{{{\alpha _2}}}{{{\alpha _1}}} = \dfrac{{\sqrt {{C_1}/{K_\alpha }} }}{{\sqrt {{C_2}/{K_\alpha }} }} = 2$
On solving and squaring both sides, we get-
$ \Rightarrow \dfrac{{0.2}}{{{C_2}}} = 4$
On rearranging, we get-
$ \Rightarrow \dfrac{{0.2}}{4} = {C_2}$
Now, we have ${M_1} = 0.2$M, ${V_1} = 300$ mL and ${M_2} = \dfrac{{0.2}}{4}$ Now, we have to find ${V_2}$
So we will use the formula-
$ \Rightarrow {M_1}{V_1} = {M_2}{V_2}$
The, we can write-
 $ \Rightarrow {V_2} = \dfrac{{{M_1}{V_1}}}{{{M_2}}}$
On putting the given values in the formula, we get-
$ \Rightarrow {V_2} = \dfrac{{300 \times 4 \times 0.2}}{{0.2}}$
On solving, we get-
$ \Rightarrow {V_2} = 1200$
Now, we know the amount of acetic acid in this solution so we can easily find the amount of water added in the solution by subtracting the volume of acetic acid from the volume of solution.
$ \Rightarrow $ Volume of water added=$1200 - 300 = 900$ mL.

The correct answer is option B.

Note:
The uses of acetic acid are-
-The acetic acid is used as an antiseptic as it has antibacterial properties.
-It is used as a major constituent of vinegar.
-It is used in preparation of many chemical compounds like VAM( vinyl acetate monomer).
-It is used as a solvent for polar molecules.