Consider the following data. Electrolyte ^ _m (S cm^2 mol^ -1 ) BaCl _2 x_1 H _2 SO _4 x_2 HCl x_3 BaSO _4 is sparingly soluble in water. If the conductivity of the saturated BaSO _4 solution is x S cm^ -1 then the solubility product of BaSO _4 can be given as (Here _m = ^ _m)
A. 10^6 x^2 ^2(x_1 + x_2 - 2x_3)^2
Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of BaSO _4 is given by: ^ _m( BaSO _4) = ^ _m( BaCl _2) + ^ _m( H _2 SO _4) - 2 ^ _m( HCl ) Substituting the given values: ^ _m( BaSO _4) = x_1 + x_2 - 2x_3 For a sparingly soluble salt, the solubility S (in mol L^ -1 ) is related to its conductivity (in S cm^ -1 ) and molar conductivity _m (in S cm^2 mol^ -1 ) by the relation: _m = 1000 S Given that = x and _m = ^ _m (which implies complete dissociation, meaning the degree of dissociation = 1), we get: S = 1000 x x_1 + x_2 - 2x_3 The solubility product K_ sp for BaSO _4 (which dissociates into Ba ^ 2+ and SO _4^ 2- ) is: K_ sp = S^2 = ( 1000 x x_1 + x_2 - 2x_3 )^2 = 10^6 x^2 (x_1 + x_2 - 2x_3)^2 Since = 1, this expression is equivalent to 10^6 x^2 ^2(x_1 + x_2 - 2x_3)^2 . Answer: 10^6 x^2 ^2(x_1 + x_2 - 2x_3)^2
Related: Chemistry — Electrochemistry · All PYQ Banks