How much water must be added to 300mL of 0.2M solution of `CH_(3)COOHKa=1.8xx10^(-5)` for the degree of dissociation to double ?

Explanation

Let the degree of dissociation of `CH_(3)COOH` be x. Thus,
`overset(0.2)(CH_(3)COOH) Leftrightarrow overset(0)(CH_(3)COO^(-))+overset(0)(H^+)` Initial concentration
`"0.2 (1-x) 0.2x 0.2x Concentration at eqb."`
`K_(a)=(0.2 x xx 0.2x)/(0.2(1-x))=1.8 xx 10^(-5), x=9.5 xx 10^(-3)`
Now, let the solution be diluted to c mole per litre in order to double the degree of dissociation, x. Since for the degree of dissociation "2x. the dissociation constant will be the same, therefore, we have,
`K_(a)=((c xx 2x) (c xx 2x))/(c(1-2x))=(c(2x)^2)/((1-2x))`
`1.8 xx 10^(-5)=(c xx (2 xx 9.5 xx 10^(-3))^2))/((1-2 xx 9.5 xx 10^(-3)), c=0.05M`
Now suppose V mL of water is to be added to dilute 300 mL of `CH_3COOH` solution to change its concentration from 0.2 M to 0.05 M. As millimoles of `CH_3COOH` before and after dilution will be same,
`0.2 xx 300=0.05 xx (300+V)`
V=900mL