1.

The solubility of Ag2Cro4 at 85°c is 8.0 ×10.5 m. Find the solubility product of Ag2Cro4.

Answer»

Ag2Cro4 → 2 Ag+ + cro4-2

using law at mass action

K = \(\frac {[Ag^+]^2[Cro_4^{-2}}{Ag_2Cro_4}\)

= K [Ag2Cro4] = [Ag+]2 [Crp4-2]

= Ksp = [Ag+]2 [Cro4-2] (Ksp = K [Ag2Cro4])

Let say solubility of Ag2Cro4 = S

M = 8.0 x 10-5m

∴ Ksp = [2s]2 [s]

Ksp = 4s3

Ksp = 4 x (8.0 x 10-5)3

Ksp = 2048 x 10-15

Ksp = 2.048 x 10-12

Hence the solubility product at Ag2Cro4 will be 2.048 x 10-12



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