What is the solubility of AgCrO4 (Ksp= 9.0 x 10^-12) in distilled water?

Ag2CrO4 ==> 2Ag^+ + CrO4^2-

...x........2x.......x

Ksp = (Ag^+)^2(CrO4^2-)
Substitute and solve for x = (CrO4^2-) = (Ag2CrO4)

To determine the solubility of AgCrO4 (silver chromate) in distilled water, you need to use the solubility product constant (Ksp) provided.

The Ksp value for AgCrO4 indicates the equilibrium constant for the dissolution of AgCrO4 in water. The expression for the solubility product constant is as follows:

AgCrO4 (s) <=> Ag+ (aq) + CrO4^2- (aq)

The Ksp expression for this reaction is:

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

In this reaction, one molecule of AgCrO4 dissolves to form one Ag+ ion and one CrO4^2- ion. Therefore, the stoichiometric coefficient for Ag+ and CrO4^2- is both 1.

Now, substitute the solubility of AgCrO4 with 'x', which indicates the concentration of Ag+ and CrO4^2- ions formed in the solution.

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

The Ksp value given in the question is 9.0 x 10^-12. Therefore, we can write the equation as:

9.0 x 10^-12 = x^2

To solve for 'x', take the square root of both sides of the equation:

sqrt(9.0 x 10^-12) = x

Hence, the solubility of AgCrO4 in distilled water is approximately equal to the square root of the Ksp value, which is 3.0 x 10^-6 M (moles per liter).