An aqueous solution of Na2CrO4 at 25°C is slowly added to an aqueous solution containing 0.001...

Question

Question

Answer 1

Answer #1

Correct option is : PbCrO4(s)

Explanation

Case 1 : PbCrO4

Given : concentration of Pb2+ = [Pb2+] = 0.001 M

PbCrO4 Ksp = [Pb2+][CrO42-]

2.80 x 10-13 = (0.001 M) * [CrO42-]

[CrO42-] = (2.80 x 10-13) / (0.001 M)

[CrO42-] = 2.80 x 10-10 M

Case 2 : BaCrO4

Given : concentration of Ba2+ = [Ba2+] = 0.100 M

BaCrO4 Ksp = [Ba2+][CrO42-]

1.17 x 10-10 = (0.100 M) * [CrO42-]

[CrO42-] = (1.17 x 10-10) / (0.100 M)

[CrO42-] = 1.17 x 10-9 M

Since less concentration of CrO42- is required in Case 1 to initial precipitation, therefore, PbCrO4 will precipitate first

Answer 2

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