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}}} $
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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}}} $
