Answer to Question #186832 in General Chemistry for Jose

Question #186832

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)




1
Expert's answer
2021-04-29T07:17:54-0400

(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


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