Calculate the solubility of silver chromate, Ag2CrO4- in 0.005M Na22Cr)4- Ksp=2.6X10-12
HELP PLEASE......
To calculate the solubility of silver chromate (Ag2CrO4) in 0.005M Na2CrO4, we will use the solubility product constant (Ksp) value provided.
The balanced equation for the dissolution of silver chromate is:
Ag2CrO4 ⇌ 2 Ag+ + CrO4^2-
According to the equation, one mol of Ag2CrO4 will produce two mols of Ag+ ions and one mol of CrO4^2- ions.
Let's define "x" as the solubility of Ag2CrO4 (in mol/L). Then, the concentration of Ag+ ions would be 2x (since two moles of Ag+ ions are formed per one mole of Ag2CrO4), and the concentration of CrO4^2- ions would be x.
The expression for the solubility product constant (Ksp) can be written as:
Ksp = [Ag+]^2 [CrO4^2-]
Substituting the concentrations we obtained:
Ksp = (2x)^2 * x = 4x^3
Ksp value given is 2.6 x 10^-12, so we can set up the equation:
2.6 x 10^-12 = 4x^3
Now, let's solve for x:
Divide both sides of the equation by 4:
6.5 x 10^-13 = x^3
Take the cube root of both sides to solve for x:
x = ∛(6.5 x 10^-13)
Using a scientific calculator, we find:
x ≈ 1.21 x 10^-4
Therefore, the solubility of silver chromate (Ag2CrO4) in 0.005M Na2CrO4 solution is approximately 1.21 x 10^-4 M.
To calculate the solubility of silver chromate (Ag2CrO4) in the given solution, you need to use the solubility product constant (Ksp) and the balanced chemical equation for the dissociation of the compound.
The balanced chemical equation for the dissociation of silver chromate is:
Ag2CrO4 ⇌ 2Ag+ + CrO4^2-
According to the equation, 1 mole of silver chromate (Ag2CrO4) dissociates to produce 2 moles of silver ions (2Ag+) and 1 mole of chromate ions (CrO4^2-).
The solubility product constant (Ksp) expression for silver chromate can be written as:
Ksp = [Ag+]^2 * [CrO4^2-]
Given:
Ksp = 2.6 × 10^-12 M^3
Let's assume x mol/L is the solubility of silver chromate in the solution. Since 1 mole of silver chromate dissociates to produce 2 moles of silver ions:
[Ag+] = 2x mol/L
Similarly, for chromate ions:
[CrO4^2-] = x mol/L
Substituting these values into the Ksp expression, we get:
Ksp = (2x)^2 * x = 4x^3
Now, we can insert the given Ksp value and solve for x:
2.6 × 10^-12 = 4x^3
To solve for x, rearrange the equation:
x^3 = (2.6 × 10^-12) / 4
Now, take the cube root of both sides of the equation:
x = ∛((2.6 × 10^-12) / 4)
Calculating this expression will give you the solubility of silver chromate (Ag2CrO4) in the given solution.
Let x = solubility Ag2CrO4 at equilibrium.
.........Ag2CrO4(s) --> 2Ag^+ + CrO4^2-
equil.......x............2x.......x
..........Na2CrO4==> 2Na^+ + CrO4^2-
initial.....0.005M.....0.......0
change.....-0.005...2*0.005...0.005
equil.......0........2*0.005..0.005
Ksp = (Ag^+)^2(CrO4^2-)
Substitute from the ICE charts above.
(Ag^+) = 2x from Ag2CrO4.
(CrO4^2-) = x from Ag2CrO4 + 0.005 from Na2CrO4.
Solve for x = solubility Ag2CrO4.