Equivalent condictance of `BaCI_(2), H_(2)SO_(4)` and `HCI` are `x_(1) , x_(2)` and `x_(3) S cm^(2) "equiv"^(-1)` at infinite dilution , if specific condictance of structured `BaSO_(4)` solution is of `y S cm^(-1)` then `K_(p)` of `BaSO_(4)` is

Options

A. `(10^(6)y^(2))/(2(x_(1)+x_(2)-2x_(3)))`
B. `(10^(9)y^(3))/(8(x_(1)+x_(2)-2x_(3))^(3))`
C. `(10^(3)y)/(2(x_(1)+x_(2)-2x_(3)))`
D. `(10^(6)y^(2))/(4(x_(1)+x_(2)-2x_(3))^(2))`

Answer

D

Explanation

`wedge_(eq)^(@)(BaSO_(4))=wedge_(eq)^(@)(BaCl_(2))+wedge_(eq)^(@)(H_(2)SO_(4))-2wedge_(eq)^(@)(HCl)`
`=x_(1)+x_(2)-2x_(3)`
`wedge_(eq)^(@)(BaSO_(4))=(1000xx x)/("Normality "("solubility in g eq "L^(-1)))`
`therefore` Solubility in g eq `L^(-1)=(1000 xx y)/(x_(1)+x_(2)+2x_(3))`
Solubility in mol `L^(-1)=(1000y)/(2(x_(1)+x_(2)+2x_(3)))`
`BaSO_(4)hArrBa^(2+)+SO_(4)^(2-)`
`K_(sp)` for `BaSO_(4)=[Ba^(2+)][SO_(4)^(2-)]=sxxs=s^(2)`
`=[(1000y)/(2(x_(1)+x_(2)+2x_(3)))]^(2)=(10^(6)y^(2))/(4(x_(1)+x_(2)-2x_(3))^(2))`