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}} $
Show Answer
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}} $
