Redox Reactions And Electrochemistry Ques 712

Question: Equivalent conductivity of $ BaCl_2,H_2SO_4 $ and HCl are $ y_1,y_2 $ and $ y_3S,c{m^{-1}}e{q^{-1}} $ at infinite dilution. If conductivity of saturated $ BaSO_4 $ solution is $ y,S,c{m^{-1}}, $ then find $ K_{sp} $ of $ BaSO_4 $ .

Options:

A) $ \frac{y^{2}}{{{(y_1+y_2-y_3)}^{2}}} $

B) $ \frac{2.5{y^{-2}}}{{{(y_1+y_2-y_3)}^{2}}} $

C) $ \frac{500}{y_1+y_2-y_3} $

D) $ \frac{2.5\times 10^{5}y^{2}}{{{(y_1+y_2-y_3)}^{2}}} $

Show Answer

Answer:

Correct Answer: D

Solution:

[d] $ \Delta _m^{\infty }BaSO_4=2\Delta _eq^{\infty }(BaSO_4) $

$ \Delta _eq^{\infty }(BaSO_4)=\Delta _eq^{\infty }(B{a^{2+}})+\Delta _eq^{\infty }(SO_4^{-2}) $

$ =\Delta _eq^{\infty }(BaCl_2)+\Delta _eq^{\infty }(H_2SO_4)-\Delta _eq^{\infty }(HCl) $

$ \Delta _eq^{\infty }(BaSO_4)y_1+y_2-y_3 $

$ \Delta _m^{\infty }=2(y_1+y_2-y_3) $

For sparingly soluble salt, $ \Delta _m^{\infty }=\frac{k}{m}\times 100 $

or $ M=\frac{y}{2(y_1+y_2-y_3)}\times 1000 $

$ =\frac{500}{y_1+y_2-y_3} $

$ K_{sp}=M^{2}=\frac{2.5\times 10^{5}y^{2}}{{{(y_1+y_2-y_3)}^{2}}} $



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