How much water must be added to 300 ml of 0.2 M solution of acetic acid for the degree of dissociation of the acid to double Ka=1.8*10 raised -5

Acetic acid is a weak acid. So, according to Ostwald dilution law, when the dilution is increased, degree of dissociation increases.
Initial concentration =C= 0.2 MDegree of dissociation = α                                                                CH3COOH              H+           +      CH3COO-Initial concentration at time, t=0             0.2M                          0                         0At time t=teq., concentration                  0.2(1-α)                 0.2×α               0.2× αKeq=[H+][CH3COO][CH3COOH] =  0.2×α× 0.2×α0.2(1-α)=1.8 × 10-5for weak electrolytes, 1-α  1α2 =KeqC =1.8 × 10-50.2 =9 × 10-5Now, the degree of dissociation is doubled and we have to find out the concentration.So, α'=2α(α')2=(2α)=4α2 = 4 × 9 × 10-5(α')2=KeqC'=36 ×10-5C'=Keq(α')2 =1.8 × 10-536 ×10-5 = 0.05 MNow apply the normality equation, N1V1=N2V20.2×300=0.05× V     (V is the total volume after dilution)V=1200 mlSo, initial volume=300Added water =1200-300 = 900 ml
 

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