Solutions Question 360

Question: 0.010M solution an acid HA freezes at $ -0.0205{}^\circ C $ . If $ ~K_{f} $ for water is $ 1.860Kkgmo{l^{-1}} $ , the ionization constant of the conjugate base of the acid will be (assume $ 0.010M=0.010m $ )

Options:

A) $ 1.1\times {10^{-4}} $

B) $ 1.1\times {10^{-3}} $

C) $ 9.0\times {10^{-11}} $

D) $ 9.0\times {10^{-12}} $

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Answer:

Correct Answer: C

Solution:

$ \Delta T_{f}( normal )=K_{f}m=1.86\times 0.01=0.0186; $

$ i=\frac{\Delta {T_{f(obs)}}}{\Delta {T_{f(nor)}}}=\frac{0.0205}{0.0186}=1.10=1+\alpha ; $

$ \alpha =0.1 $

$ K_{a}=\frac{Ca^{2}}{1-\alpha }=\frac{0.01\times {{0.1}^{2}}}{1-0.1}=\frac{1}{9}\times {10^{-3}}; $

$ K_{a}=\frac{K_{w}}{K_{a}}=10\times {10^{-14}}\times 9\times 10^{3}=9\times {10^{-11}} $



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