Question

What molar [] of silver ion could exist in a solution in which the [] of CrO4^2- is 1.0*10^-4 (Ksp of Ag2CrO4= 1.1*10^-12)

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

Answer 1

My work:
Ksp= [Ag]^2 [CrO4^2-]
= (2x^2)(x) = 4x^3


1.1*10^-12= 2x^2(1.0*10^-4)
1.1*10^-8= 2x^2
x= 7.4*10^-5

but the correct answer is 1.1*10^-4

Answer 2

You are trying to find [Ag]

[Ag]=sqrt 1.1E-12/1.0E-4

Answer 3

My work:
Ksp= [Ag]^2 [CrO4^2-]
<b>You are ok to here. This is a common ion type problem. The problem tells you CrO4^-2 = 1 x 10^-4. So substitute 1 x 10^-4 for CrO4^-2 and solve for Ag^+ and that will be the maximum amount of Ag^+ that can exist in solution with that amount of chromate ion. </b>
= (2x^2)(x) = 4x^3


1.1*10^-12= 2x^2(1.0*10^-4)
1.1*10^-8= 2x^2
x= 7.4*10^-5

but the correct answer is 1.1*10^-4

Answer 4

so [Ag]= x^2 not 2x^2?

Answer 5

Actually, Ag^+ = x (not x^2) if that's what you let it stand for. The squared part of it comes from the Ksp expression which is
(Ag^+)^2(CrO4^-2) = 1.1 x 10^-12
(Ag^+)^2 = (1.1 x 10^-12/1.0 x 10^-4) =
(Ag^+) = sqrt of ...........
(Ag^) = 1.05 x 10^-4 but you are allowed only two s.f.; we round that to 1.0 x 10^-4 M.
Remember, when the CrO4^-2 is given, that sets the limit on what Ag^+ can be. Ag^+ squared x CrO4^- can't be larger than Ksp. And the problem asks exactly that; i.e. what is the maximum (Ag^+) that can exist when the chromate is 1 x 10^-4M.

Answer 6

now it is clear, thank you =)