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-]
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
You are trying to find [Ag]
[Ag]=sqrt 1.1E-12/1.0E-4
My work:
Ksp= [Ag]^2 [CrO4^2-]
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.
= (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
so [Ag]= x^2 not 2x^2?
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.
now it is clear, thank you =)
To find the molar concentration of silver ion ([Ag]) that could exist in the solution, we need to use the solubility product constant (Ksp) expression and solve for [Ag].
Given:
Ksp = [Ag]^2 [CrO4^2-] = 1.1*10^-12
[CrO4^2-] = 1.0*10^-4
First, we rearrange the Ksp expression to solve for [Ag]:
[Ag]^2 = Ksp / [CrO4^2-]
Now, substitute the given values:
[Ag]^2 = (1.1*10^-12) / (1.0*10^-4)
To solve for [Ag], we need to take the square root of both sides:
[Ag] = √[(1.1*10^-12) / (1.0*10^-4)]
Plug in the values and calculate:
[Ag] = √(1.1*10^-12 / 1.0*10^-4) ≈ 1.05*10^-4 M
Therefore, the molar concentration of silver ion ([Ag]) that could exist in the solution is approximately 1.05*10^-4 M.