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