The solubility product constant for silver chromate, Ag2CrO4, is 1.2 x 1012. If a common ion has been introduced such that [CrO42−]= 0.010 M, what is the molar solubility of the silver chromate?
ANSWER: Molar solubility of Ag2CrO4 =S = 5.5 x 10 -06 M
Consider reaction : Ag2CrO4(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
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