Question

A) 1 L of solution contains 0.10 mol Ag+ and 0.10 mol of Ba2+. If a highly concentrated solution of K2CrO4 is added drop by drop, which will precipitate first, Ag2CrO4 or BaCrO4?

Ksp for AgCrO4 = 2.0 x 10^-12; Ksp for BaCrO4= 1.2 x 10^-10

B) If the solution in part "A" contains 0.0010 Ag+ and 0.0010 mol of Ba2+ instead, which will precipiate first, Ag2CrO4 or BaCrO4?

Answers: A) Ag2CrO4 and B) BaCrO4

Answer 1

we know that

precipitation occurs is

ionic product > solubility product

now

for Ag2Cr04

Ag2Cr04 ----> 2Ag+ + Cr042-

ionic product = [Ag+]2 [Cr042-]

so

precipitate forms when

[Ag+]^2 [Cr042-] > 2 x 10-12

given

[Ag+] = 0.1

so

[0.1]^2 [Cr042-] > 2 x 10-12

[Cr042-] > 2 x 10-10

For BaCr04 :

BaCr04 ---> Ba+2 + Cr042-

ionic product = [Ba+2] [Cr042-]

so

precipitate forms when

[Ba2] [Cr042-] > 1.2 x 10-10

given

[Ba+2] = 0.1

so

[0.1] [Cr042-] > 1.2 x 10-10

[Cr042-] > 1.2 x 10-9

now

Ag2Cr04 will precipitate if [Cr042-] > 2 x 10-10

BaCr04 will precipitate if [Cr042-] > 1.2 x 10-9

so

Ag2Cr04 will precipitate first

B)

now

replace the procedure with changed values

for Ag2Cr04

Ag2Cr04 ----> 2Ag+ + Cr042-

ionic product = [Ag+]2 [Cr042-]

so

precipitate forms when

[Ag+]^2 [Cr042-] > 2 x 10-12

given

[Ag+] = 0.001

so

[0.001]^2 [Cr042-] > 2 x 10-12

[Cr042-] > 2 x 10-6

For BaCr04 :

BaCr04 ---> Ba+2 + Cr042-

ionic product = [Ba+2] [Cr042-]

so

precipitate forms when

[Ba2] [Cr042-] > 1.2 x 10-10

given

[Ba+2] = 0.001

so

[0.001] [Cr042-] > 1.2 x 10-10

[Cr042-] > 1.2 x 10-7

now

Ag2Cr04 will precipitate if [Cr042-] > 2 x 10-6

BaCr04 will precipitate if [Cr042-] > 1.2 x 10-7

so

BaCr04 will precipitate first