Consider the following data. arraycc Electrolyte & m ( S cm2 mol-1) \ BaCl2 & x1 \ H2SO4 & x2 \ HCl & x3 array BaSO4 is sparingly soluble in water. If the conductivity of the saturated BaSO4 solution is x S cm-1 then the solubility product of BaSO4 can be given as (Here m = m)
- 106x2 2(x1 + x2 - 2x3)2
- x2(x1 + x2 - 2x3)2
- 2(x1 + x2 - 2x3)2106x2
- x2(x1 + x2 + 2x3)2
Answer
(1) 106x2 2(x1 + x2 - 2x3)2
Solution
By Kohlrausch's law,
m(BaSO4) = m(BaCl2) + m(H2SO4) - 2 m(HCl)
m(BaSO4) = x1 + x2 - 2x3
For the saturated solution (κ = x S cm-1, s in mol L-1), m = 1000 s, so s = 1000 x m; and m = m (α = degree of dissociation; here m = m, i.e. α = 1),
s = 1000 x (x1 + x2 - 2x3)
For BaSO4(s) Ba2+(aq) + SO42-(aq), Ksp = s2, so
∴ Ksp = 106x2 2(x1 + x2 - 2x3)2 (with α = 1)
Answer: (1)