Answer to Question #186832 in General Chemistry for Jose

Question

2. 

PbF2(s)  ⇌ Pb2+(aq) + 2 F-(aq)

At a constant temperature, in a saturated solution of PbF2 the concentration of Pb2+ is 2.0x10-3 M. The equilibrium is represented by the equation above.

a.   Write the expression for the solubility-product constant, Ksp (2pt)

b.   Calculate the value of Ksp at this temperature. (3pt)

c.   You have a 1.00 L saturated solution of PbF2 and 0.0428 moles of solid NaF is added and dissolves completely. Calculate the equilibrium concentration of Pb2+ under these conditions and assume no change in volume occurred. (2pt)

d.   Is the concentration of Pb2+ in part (c) greater, less than, or equal to the original concentration. Explain why this is the case. (2pt)

Answer 1

(a) Ksp = [Pb²⁺][I^-]²

7x10^-9 = (s)(2s)² = 4s³

(b) There is a common ion effect because of the added I^- from NaI. According to Le Chatelier (and the Ksp expression) this will reduce the solubility of PbI₂ so [Pb²⁺] will be LESS than in part (a)

(c) PbCl₂ will have the highest concentration of Pb²⁺ because it has the largest Ksp meaning that the molar solubility of Pb²⁺ will be 2x10^-5 M compared to the others where the molar solubility of Pb²+ will be only 7x10^-9 M or 3x10^-13 M