Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12× 10–12. What is the solubility (in mol/L) of silver chromate in 1.4 M potassium chromate aqueous solution? In 1.4 M silver nitrate aqueous solution? In pure water?

6.54x10^-5 for pure water

To find the solubility of silver chromate (Ag2CrO4) in various solutions, we need to consider the solubility product constant (Ksp) and the concentrations of the ions present in the solution.

The solubility product constant (Ksp) expression for silver chromate is:

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

Let's find the solubility of silver chromate in each of the given solutions:

1) In 1.4 M potassium chromate (K2CrO4) solution:
Silver chromate will react with the potassium chromate to form insoluble silver chromate. However, since the concentration of chromate ions is high (1.4 M), it will shift the equilibrium towards the dissolution of the silver chromate. Therefore, we need to calculate the ion concentrations using the given molarity.

When silver chromate dissolves, it forms two silver ions (Ag+) and one chromate ion (CrO4^2-).

Ksp = (2x)^2 * (x)
1.12 x 10^-12 = 4x^2 * x
4x^3 = 1.12 x 10^-12
x^3 = 2.8 x 10^-13
x = ∛(2.8 x 10^-13)
x = 9.33 x 10^-5 M

Therefore, the solubility of silver chromate in 1.4 M potassium chromate solution is 9.33 x 10^-5 M.

2) In 1.4 M silver nitrate (AgNO3) solution:
The silver nitrate solution provides a high concentration of silver ions (Ag+). The concentration of chromate ions will be negligible compared to the concentration of silver ions. In this case, we have an excess of silver ions, so the dissolution will be limited by the solubility of silver chromate.

Ksp = (2x)^2 * (x)
1.12 x 10^-12 = 4x^2 * x
4x^3 = 1.12 x 10^-12
x^3 = 2.8 x 10^-13
x = ∛(2.8 x 10^-13)
x = 9.33 x 10^-5 M

Therefore, the solubility of silver chromate in 1.4 M silver nitrate solution is 9.33 x 10^-5 M.

3) In pure water (H2O):
In pure water, there are no additional ions present to react with silver chromate. We only need to consider the dissociation of silver chromate.

Ksp = (2x)^2 * (x)
1.12 x 10^-12 = 4x^2 * x
4x^3 = 1.12 x 10^-12
x^3 = 2.8 x 10^-13
x = ∛(2.8 x 10^-13)
x = 9.33 x 10^-5 M

Therefore, the solubility of silver chromate in pure water is 9.33 x 10^-5 M.

To determine the solubility of silver chromate (Ag2CrO4) in various solutions, we need to compare the solubility product constant (Ksp) with the concentrations of the relevant ions in the solution.

The solubility product constant (Ksp) expression for Ag2CrO4 is:

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

Let's consider each solution separately:

1. Solubility in 1.4 M potassium chromate (K2CrO4) solution:

Since we have a 1.4 M potassium chromate solution, we can assume that [CrO4^2-] is also 1.4 M.

To find the solubility of Ag2CrO4, we need to calculate the concentration of Ag+ ions using the Ksp expression. Since Ag2CrO4 dissociates into two Ag+ ions and one CrO4^2- ion, we can assume that [Ag+] is twice the solubility (S):

Ksp = (2S)^2 * (1.4)

Simplify the equation:

1.12 x 10^-12 = 4S^2 * 1.4

Divide the Ksp by 4 and then by 1.4:

S^2 = (1.12 x 10^-12) / (4 * 1.4)

S^2 ≈ 2.52 x 10^-14

Take the square root to solve for S:

S ≈ 1.59 x 10^-7 M

Therefore, the solubility of silver chromate in a 1.4 M potassium chromate solution is approximately 1.59 x 10^-7 M.

2. Solubility in 1.4 M silver nitrate (AgNO3) solution:

Similar to the previous calculation, we assume that [Ag+] is 1.4 M.

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

1.12 x 10^-12 = (1.4)^2 * [CrO4^2-]

Divide the Ksp by (1.4)^2:

[CrO4^2-] ≈ (1.12 x 10^-12) / (1.4)^2

[CrO4^2-] ≈ 5.00 x 10^-13 M

Therefore, the solubility of silver chromate in a 1.4 M silver nitrate solution is approximately 5.00 x 10^-13 M.

3. Solubility in pure water:

In pure water, there is no additional source of Ag+ or CrO4^2- ions other than those from the dissociation of Ag2CrO4. Therefore, the solubility of Ag2CrO4 equals the concentration of Ag+ or CrO4^2- ions when the compound is in equilibrium.

Since Ag2CrO4 dissociates into two Ag+ ions and one CrO4^2- ion, and the initial concentration of Ag2CrO4 is S:

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

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

1.12 x 10^-12 = (2S)^2 * S

Simplify the equation:

1.12 x 10^-12 = 4S^3

Divide the Ksp by 4:

S^3 = (1.12 x 10^-12) / 4

S^3 ≈ 2.80 x 10^-13

Take the cube root to solve for S:

S ≈ 1.42 x 10^-4 M

Therefore, the solubility of silver chromate in pure water is approximately 1.42 x 10^-4 M.

.......Ag2CrO4 ==> 2Ag^+ + CrO4^2-

I.......solid.......0........0
C.......solid.......2x.......x
E.......solid.......2x.......x

For pure water. Substitute E line into Ksp expression and solve for x. That is the solubility of Ag2CrO4 in mol/L.

In 1.4M AgNO3.
Substitute E line into Ksp expression EXCEPT (Ag^+) is not x but 1.4+2x. Solve for x and that is solubility Ag2CrO4 in mols/L.

For K2CrO4 solution.
Substitute E line into Ksp expression EXCEPT (CrO4^2-) is not x + 1.4+x. Solve for x which is solubility Ag2CrO4 in mols/L.