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Knowing that the concentration of Ag+ in a silver chromate solution is 1.5 x 10-3 M, calculate its solubility product at 25°C.

To calculate the solubility product (Ksp) of a compound, we need to know the concentrations of the ions in a saturated solution. In the case of silver chromate (Ag2CrO4), it dissociates as follows:

Ag2CrO4 ⇌ 2 Ag+ + CrO4^2-

Given that the concentration of Ag+ in the solution is 1.5 x 10^(-3) M, we can assume that it's the same as the concentration of CrO4^2- ions because they have a 1:1 stoichiometric ratio.

Now, let's denote the solubility of Ag2CrO4 as "s". This means that the concentration of Ag+ (and CrO4^2-) ions is also "s". Therefore, we can express the dissociation of Ag2CrO4 as follows:

Ag2CrO4 ⇌ 2s Ag+ + s CrO4^2-

According to the law of chemical equilibrium, the solubility product expression for Ag2CrO4 is:

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

Substituting the known concentrations, we have:

Ksp = (1.5 x 10^(-3))^2 (1.5 x 10^(-3))

Simplifying this expression, we can calculate the solubility product Ksp:

Ksp = 3.375 x 10^(-9)

Therefore, the solubility product of silver chromate at 25°C is 3.375 x 10^(-9).

A straight solubility product problem and one almost the same as the TlCl problem.