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 1.00 M potassium chromate aqueous solution?
This is an example of the common ion effect. The common ion in this case is CrO4^2- from K2CrO4.
............Ag2CrO4 ==> 2Ag^+ + CrO4^2-
solubility.....x........2x.........x
.......K2CrO4 ==> 2K^+ + CrO4^2-
initial..1M.........0......0
change...-1M........1M.....1M
equil.....0.........1M.......1M
Ksp Ag2CrO4 = (Ag^+)^2(CrO4^2-)
(Ag^+) = 2x from Ag2CrO4--see chart above.
(CrO4^2-) = x from Ag2CrO4 + 1M from K2CrO4
Substitute into Ksp expression and solve for x.
Post your work if you get stuck.
I got to 4x^3 + 6x^2=1.12x10^-12 (my moles of potassium chromate is 1.5 instead of 1) but now I'm stuck
To determine the solubility of silver chromate (Ag2CrO4) in a 1.00 M potassium chromate (K2CrO4) solution, we need to consider the common ion effect.
First, let's write the balanced chemical equation for the dissolution of silver chromate:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
The solubility product expression (Ksp) can be written as:
Ksp = [Ag+]^2 [CrO4^2-]
Since we are given the Ksp value of Ag2CrO4 as 1.12× 10–12, we can set up the following equation:
1.12× 10–12 = (2x)^2 (x)
Where x represents the molar solubility of Ag2CrO4.
Since the molar solubility of Ag2CrO4 is small compared to the 1.00 M concentration of K2CrO4, we can assume that the change in concentration of K+ and CrO4^2- ions from the dissociation of K2CrO4 is negligible.
Therefore, we can use the initial concentration of CrO4^2- ions from the 1.00 M K2CrO4 solution as the concentration of CrO4^2- ions in the equilibrium expression:
Ksp = (2x)^2 (1.00 - x)
Simplifying the equation:
1.12× 10–12 = 4x^3 - 4x^2
4x^3 - 4x^2 - 1.12× 10–12 = 0
Now, you can solve this cubic equation using numerical methods or calculators specifically designed for solving equations. Additionally, you can use software like MATLAB or Wolfram Alpha to solve the equation.
To determine the solubility of silver chromate (Ag2CrO4) in 1.00 M potassium chromate (K2CrO4) solution, we need to use the solubility product constant (Ksp) and the information provided.
The solubility product constant (Ksp) is an equilibrium constant that represents the extent of dissolution of a sparingly soluble salt. It is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients in the balanced equation. For silver chromate, the Ksp expression is:
Ksp = [Ag+]^2 [CrO42-]
Given that the Ksp of Ag2CrO4 is 1.12 × 10^–12, we can assume that when silver chromate dissolves, it dissociates fully into Ag+ and CrO42- ions. Therefore, the concentrations of Ag+ and CrO42- are equal and can be represented as "x."
Substituting these values into the Ksp expression, we get:
1.12 × 10^–12 = x^2 * x
Simplifying the equation, we have:
1.12 × 10^–12 = x^3
To solve for "x," take the cubic root of both sides:
x = (1.12 × 10^–12)^(1/3)
Using a calculator, the value of x is approximately 1.616 × 10^–4 mol/L.
Therefore, the solubility of silver chromate in 1.00 M potassium chromate solution is approximately 1.616 × 10^–4 mol/L.