Question

The solubility product constant for silver chromate, Ag₂​CrO₄​, is 1.2 x 10¹². If a common ion has been introduced such that [CrO_4​²⁻]= 0.010 M, what is the molar solubility of the silver chromate?

Answer 1

ANSWER: Molar solubility of Ag₂CrO₄ =S = 5.5 x 10 ^-06 M

Consider reaction : Ag₂CrO_4(s) 2 Ag ⁺_(aq) + CrO₄^-_(aq)

For above reaction , Ksp = [ Ag ⁺ ] ² [ CrO₄^- ]= 1.2 x 10 ^-12

If S is a Molar solubility of Ag₂CrO₄ then [ Ag ⁺ ] = 2 S and  [ CrO₄^- ] = S+ 0.01

As Ksp is small , S <<<0.01 thus S +0.01 0.01.

Then Ksp = ( 2 S) ² ( 0.01 )

= 4 S ² ( 0.01)

= 0.04 S ² = 1.2 x 10 ^-12

S ² = 1.2 x 10 ^-12 / 0.04

=3 x 10 ^-11

S = 5.5 x 10 ^-06 M