Question

The solubility product of silver chromate is 1.12*10^-12 at 25 degrees Celsius. What are the concentrations of Ag+ and CrO4^2- at equilibrium? What is the solubility of silver chromate in g/L? What is one way that you could increase the solubility of this slightly soluble salt?

Answer 1

In equilbirium:

Ag2CrO4(s) <-> 2Ag+ + CrO4-2

Ksp = [Ag+]^2[CrO4-2]

1.12*10^-12 = (2S)^2(S)

4*S^3 = 1.12*10^-12

S = ((1.12*10^-12)/(4)) ^(1/3) = 6.5421*10^-5 ;

change to g/L

mass = mol*MW =  6.5421*10^-5 * 331.73 = 0.0217021 g of Ag2CrO4 per liter

a)

one way t increas solublity of salt is:

addicion of HCl --> H+ and Cl- is added

Ag+ + Cl- --> AgCl(s) so more Ag2CrO4 will go to soluble state

H+ + CrO4-2 <--> HCrO4- will form,   so more Ag2CrO4 will go to soluble state

answered by: anonymous