Question

For which initial concentration of chromate anion would[Ag +] = 6.0 x 10^ -6 M and cause the solution to begin to preciitate Ag2CrO4(s)? Where Ksp = 9.0 x 10^-12

Ag2CrO4 --> 2Ag+ + CrO4(-2)
a. 0.08
b. 0.11
c. 0.21
d. 0.25

Answer 1

d. 0.25.

Step-by-step explanation:

Hello,

In this case, since the equilibrium repression for the considered chemical reaction is:

`Ksp=[Ag^+]^2[CrO_4^(2-)]`

For a concentration of silver of 6.0x10⁻⁶ we need a concentration of chromate anion that makes the reaction quotient greater than the solubility product, thus, we write:

`[CrO_4^(2-)]=(Ksp)/([Ag^+]^2) =(9.0x10^(-12))/((6.0x10^(-6))^2)`

`[CrO_4^(2-)]=0.25M`

It means that at concentrations of chromate anion of 0.25 M or more, the reaction quotient Q becomes greater than the solubility product which means that precipitation will begin occurring, therefore, answer is d. 0.25.

Best regards.