Redox Reactions And Electrochemistry Ques 716
Question: Equivalent conductance of saturated $ BaSO_4 $ is $ 400,{{\Omega }^{-1}},cm^{2}eq{u^{-1}} $ and specific conductance is $ 8\times {10^{-5}},{{\Omega }^{-1}},c{m^{-1}} $ . Hence, $ K_{sp} $ of $ BaSO_4 $ is
Options:
A) $ 4\times {10^{-8}}M^{2} $
B) $ 1\times {10^{-8}}M^{2} $
C) $ 2\times {10^{-4}}M^{2} $
D) $ 1\times {10^{-4}}M^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ {\lambda_0}(BaSO_4)=\frac{1000\times sp,\text{.}conductance}{conc\text{.},\text{(Normality)}} $
$ \therefore $ Normality $ =\frac{1000\times 8\times {10^{-5}}}{400} $ $ Molarity=\frac{Normality}{2}={10^{-4}}M $
$ \therefore $ $ K_{sp}=S^{2}={{({10^{-4}})}^{2}}={10^{-8}}M^{2} $
