Question

A Calculate the molar solubility of PbCl2 in a 0.2340 M lead(II) perchlorate, Pb(ClO )2 solution Solubility = B. Let's say we have a beaker where a saturated solution of lead(II) chloride is in equilibrium with solid lead(II) chloride in which of these cases will the molar solubility be lowest after equilibrium is reestablished? After the addition of solid NaNO3. After the addition of 0.120 moles of Clion. Not enough information given, After letting some of the solvent evaporate None of these things will do it. Upon the addition of more water, After the addition of 0.120 moles of Pb2+ ion. The Ksp of lead(II) chloride is 1.70 x 10-5

Answer 1

Ans A:

Lead(II) perchlorate (Pb(ClO₄)₂) is a highly soluble salt in water. Whereas PbCl₂ is a sparingly soluble salt. Solubility of PbCl₂ in 0.2340M Pb(ClO₄)₂ will be affected by the common ion ‘Pb²⁺’ which is furnished by the complete dissociation of Pb(ClO₄)₂.

+ Pb(C104)2(aq) 0.2340 M Pb2+ (aq) 0.2340 M common ion 2 (C104) (aq) (2 x 0.2340) M If 'S' M is the solubility of PbCl2 in water, then Ksp = 1.70 x 10-5 PbCl2(S) - Pb2+ (aq) + 2Cl(aq) (S+ 0.2340) M 2SM According to law of chemical equilibrium, Ksp = [Pb"] [C1-12 = (S+0.2340) (28)2 = 1.70 x 10-5 Since Ksp is very less of the order 10), S also must be a very small number in comparison to 0.2340. So (S+ 0.2340) can be approximated to 0.2340. Now the equation takes the form, (0.2340)(25)2 = 1.70 x 10-5 S = 0.00426 M

Ans B: Ksp = 1.70 x 10-5 PbCl2() = Pb2+ (aq) + 2Cl(aq) According to law of chemical equilibrium, Ksp = [Pb2][Cl] Addition of both Pb2+ & Cl- will lower the solubility of PbCl2 in water. But we can see from Ksp expression that Cl contribution is to the power '2'. So addition of CI lowers solubility more than addition of Pb2+