Please help me in the equilibrium question.........plsssssss?

Question

How much water is to be added to 400ml of 0.25M solution of acetic acid for its degree of dissociation to become double ?
(Ka=dissociation or ionisation constant=1.8*10^(-5)

Answer is = 1200ml

Answer 1

a=sqrt(Ka/C)
Let concentration be c1 when degree of dissociation a1
Let concentration be c2 when degree of dissociation a2
a1/a2 = sqrt[(Ka*C2)/(Ka*C1)] = sqrt(C2/C1)

but a2=2*a1
Thus
0.5 = sqrt(C2/0.25)
0.25 = C2/0.25
C2=0.0625
M1*V1=M2*V2
0.25*400=0.0625*V2
V2=1600
Thus Volume of water to be added = 1600-400 = 1200ml

Answer 2

Mass action law:

HAc + H2O -> H3O+ + Ac-

Ka ignores the H2O concentration, so the equation is

Ka = c[H3O+]*c[Ac-]/c[HAc]

with the side condition that

c[H3O+] = c[Ac-]

Added water will primarily affect the concentration of HAc.

You calculate the degree of dissociation (D) as
c[Ac-]/c[HAc]

so

Ka = c[Ac-]^2/c[HAc]

Ka/c[Ac-] = D

Oh well. Any more arithmetics without numbers and I'll stop understanding what I do.

Calculate D in the example above.
Then calculate the new c[Ac-] for doubling D, and then use the equation with the squared concentration to calculate c[HAc].

You will find that for doubling D you need to quarter c[HAc] (which is what adding 3 times the amount of water will do). That's because c[Ac-] comes in squared.