Question

You have a 0.850 M solution of Na2CrO4 at a given temperature. At what concentration will silver ions need to be added in order for a participate to form? The Ksp for Ag2CrO4 at this temperature is 2.00 × 10^-12.

A) 2.35 × 10^-12 M
B) 1.37 × 10^-5 M
C) 5.02 × 10^-10 M
D) 1.53 × 10^-6 M

Answer 1

We can use the solubility product constant (Ksp) expression for Ag2CrO4 to determine the concentration of silver ions (Ag+) required for a precipitate to form:

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

Since we know the Ksp and the concentration of Na2CrO4 (which dissociates to form CrO4^2- ions), we can solve for the concentration of Ag+ ions:

Ksp = 2.00 × 10^-12 (mol/L)^2
[CrO4^2-] = 0.850 M

2.00 × 10^-12 = [Ag+]^2 (0.850)
[Ag+]^2 = (2.00 × 10^-12) / (0.850)
[Ag+]^2 = 2.35 × 10^-12
[Ag+] = 1.53 × 10^-6 M

Therefore, the concentration of silver ions required for a precipitate to form is 1.53 × 10^-6 M. The answer is D.