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Q.

0.01 (M) solution of an acid HA freezes at −0.0205oC. If Kf for water is 1.86 K kg mol−1, the ionisation constant of the conjugate base of the acid will be (consider molarity ≃ molality)

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a

1.1×10−4

b

1.1×10−3

c

9×10−11

d

9×10−12

answer is A.

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Detailed Solution

ΔTf=kf×m=1.86×0.01=0.0186i=0.02050.0186=1.10 i = 1+αn-1 α =i-1n-1 α =1.10-12-1  α=0.1Ka=Cα21−α=0.01×(0.1)21−0.1=1×10-40.9 Ka =1.1×10-4
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