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} $

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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} $



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