Question

Calculate the molar solubility, S, of Ag2(Cros) in a 0.414 MK (CrO4) solution. Note that K2 (Cro) is highly soluble and the Ksp of Ag2 (CrO ) is 9.00 x 10-12 S 3.73e-12

Answer 1

Dissociation of Ag2CrO4 can be represented as

Ag2CrO4 $\rightleftharpoons$ 2Ag++CrO4-2

2S S where S=solubility

solubility product of  Ag2CrO4 ,Ksp=9.00 x 10-12

K2CrO4 Dissociates to give K+ and CrO4-2 ions

K2CrO4    $\rightarrow$ 2 K++ CrO4-2

initial concentration 0.414M 0 0

change 0.414 - 0.414 0+2x0.414 0+0.414

final concentration 0 0.828 0.414

solubility of Ag2CrO4 will be decreased when  K2CrO4 is added .it is due to common ion effect .common ion is

CrO4-2

when K2CrO4 is added   

Ag2CrO4 $\rightleftharpoons$ 2Ag++CrO4-2

2S S   

   S+0.414   $\therefore$  [CrO4-2 ] = S+0.414 M

  Ksp= [Ag+]2[CrO4-2 ] =  9.00 x 10-12​​​​​​​

  9.00 x 10-12​​​​​​​ = S X [S+0.414] Solubility,S=   9.00 x 10-12​​​​​​​ / [S+0.414]

=9.00 x 10-12​​​​​​​ / 0.414 $\because$ S<<<< 0.414

Solubility,S= 2.1739 X 10-11 M