You have a solution consisting of 0.10 M Cl^- and 0.10 M CrO₄²^-.You add 0.10 M silver nitrate dropwise to this solution.Given that the K_sp for Ag₂CrO₄ is 9.0 $\times$ 10^-¹²,and that for AgCl is 1.6 $\times$10^-¹⁰,which of the following will precipitate first? A)silver chloride B)silver chromate C)silver nitrate D)cannot be determined by the information given E)none of these

Question

Quizplus · Accepted Answer

To determine which compound will precipitate first, we need to compare the solubility product (Ksp) values of each compound. The compound with the lowest Ksp value will be the least soluble and therefore will precipitate first.

Using the given Ksp values, we can calculate the ion product (Qsp) of each compound in the solution:
- Qsp of AgCl = [Ag+][Cl-] = (0.10 M) x (0.10 M) = 1.0 x 10^-2
- Qsp of Ag2CrO4 = [Ag+][CrO42-] = (0.10 M) x (0.10 M) x (9.0 x 10^-10) = 9.0 x 10^-12

Since the ion product of AgCl (1.0 x 10^-2) is greater than its Ksp (1.6 x 10^-10), it exceeds the solubility limit and will precipitate first. Therefore, the answer is A) silver chloride.

You Have a Solution Consisting of 0 ×\times× 10-12,and That for AgCl Is 1

You Have a Solution Consisting of 0 $\times$ 10^-¹²,and That for AgCl Is 1