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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).