`0.01(M)` solution an acid HA freezes at `-0.0205^(@)C`. If `K_(f)` for water is `1.86K kg mol^(-1)`, the ionization constant of the conjugate base of the acid will be (consider molarity)`~=` molarity)

Options

A. `1.1xx10^(-4)`
B. `1.1xx10^(-3)`
C. `9xx10^(-11)`
D. `9xx10^(-12)`

Answer

C

Explanation

`DeltaT_(f)=k_(f)xxm=1.86xx0.01=0.0186`
`i=(0.0205)/(0.0186)=1.10=1+alpha`or`alpha=0.1`
`K_(a)=((Calpha)^(2))/(1-alpha)=(0.01xx(0.1)^(2))/(1-0.1)=(1)/(9)xx10^(-3)`
or `K_(b)=9xx10^(-11)`