What concentration of silver chromate (Ksp = 9.0 x 10-12) will dissolve to make a saturated solution in water?

*For this one the formula is Ag2CrO4, which is [Ag]^2[CrO4]. I plugged x in for CrO4, so x^3 is equal to Ksp. I then got that x = 2.0e-4. However, that x is for CrO4, not for silver chromate. Is the answer for silver chromate just the Ksp value?*

Answered an earlier post.

To find the concentration of silver chromate that will dissolve to make a saturated solution, you need to use the solubility product constant (Ksp) and the stoichiometric relationship in the balanced equation for the solubility equilibrium.

The balanced equation for the solubility equilibrium of silver chromate (Ag2CrO4) is:

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)

From this equation, we can see that each formula unit of silver chromate dissociates into two silver ions (Ag+) and one chromate ion (CrO4^2-).

Based on the equation, we can write the expression for the solubility product constant (Ksp) as:

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

Given that the Ksp value for silver chromate is 9.0 x 10^(-12), this means that in a saturated solution, the product of the concentration of the silver ions squared and the concentration of the chromate ions must equal this value.

Plugging in the values into the equation, we get:

9.0 x 10^(-12) = [Ag+]^2[CrO4^2-]

Because the stoichiometric ratio between Ag+ and CrO4^2- is 2:1, we can substitute [CrO4^2-] by 2[Ag+]:

9.0 x 10^(-12) = [Ag+]^2(2[Ag+])

Rearranging the equation, we have:

[Ag+]^3 = (9.0 x 10^(-12))/(2)

[Ag+]^3 = 4.5 x 10^(-12)

To find the concentration ([Ag+]), we take the cube root of both sides:

[Ag+] = (4.5 x 10^(-12))^(1/3)

[Ag+] ≈ 1.57 x 10^(-4) M

Therefore, the concentration of silver ions in a saturated solution of silver chromate is approximately 1.57 x 10^(-4) M. This concentration applies specifically to the silver ions, not the entire silver chromate compound.