1.

Equivalentconductance of BaCI_(2), H_(2)SO_(4) and HCIare x_(1)x_(2) and x_(3) S cm^(2) "equiv"^(-1)atinfinitedilution if specific conductance of saturated BaSO_(4)solutionis of y S cm d^(-1) then k_("SP") of BaSO_(4) is

Answer»

`(10^(3)y)/(2(x_(1)+x_(2)-2x_(3))`
`(10^(6)y^(2))/(x_(1)+x_(2)-2x_(3))^(2)`
`(10^(6)y^(2))/(4(x_(1)+x_(2)-2x_(3))^(2)`
`(x_(1)+x_(2)-2x_(3))/(10^(6)y^(2))`

Solution :`wedge_(BaSO_(4))^(@)=wedge_(BaCI_(2)^(@)+wedge_(H_(2)SO_(4)^(@)-2 wedge_(H_(2)SO_(4))^(@)-2wedge_(HCI)^(@)=(X_(1)+X_(2)-2x_(3))`
`wedge_(BaSO_(4))^(@)=(1000xx"specific conductance")/("solubility" ("in SATURATED solution" ))`
`X_(1)+x_(2)-2x_(3)=(1000Y)/("solubility")`
`therefore`solubility `(BaSO_(4))=(1000y)/(x_(1)+x_(2)-2x_(3))N=(1000y)/(2(x_(1)+x_(2)-2x_(3))M`
`K_(SP)(BaSO_(4))=[Ba^(2+)][SO_(4)^(2-)]M^(2)=(10^(6)y^(2))/(4(x_(1)+x_(2)-2_(3))^(2)`


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