Chromate ion is added to a saturated solution of Ag₂CrO₄ to reach 0.78 M CrO₄²^-.Calculate the final concentration of silver ion at equilibrium (K_sp for Ag₂CrO₄ is 9.0 $\times$10^-¹²). A) M B) M C) M D) M E) M

Question

Quizplus · Accepted Answer

The addition of chromate ion will react with silver ions in the solution to form silver chromate precipitate:
Ag+ + CrO42-  -> Ag2CrO4(s)
This reaction will shift the equilibrium to the left, decreasing the concentration of silver ion. Therefore, the final concentration of silver ion will be lower than its initial concentration. We can use the solubility product constant (Ksp) of Ag2CrO4 to calculate the concentration of silver ion at equilibrium:
Ksp = [Ag+]2[CrO42-] = 9.0 x 10-12
Substituting [CrO42-] = 0.78 M, we get:
[Ag+]2 = 9.0 x 10-12 / 0.78 = 1.15 x 10-12
[Ag+] = 3.39 x 10-7 M
Therefore, the final concentration of silver ion at equilibrium is 3.39 x 10-7 M, which corresponds to choice B.

Chromate Ion Is Added to a Saturated Solution of Ag2CrO4 ×\times×

Chromate Ion Is Added to a Saturated Solution of Ag₂CrO₄ $\times$