calculate the solunility of silver chromate, Ag2CrO4, in 0.005 M NA2CrO4- Ksp=2.6x10-12
Please help.....
f(x)= 6/x
Well, at least this is not a buffer problem. This is a problem that illustrates the common ion effect on solubility. The effect is to make the solubility of a slightly soluble salt even less soluble by using a common ion. In this case the common ion is CrO4^2- from the Na2CrO4.
Let x = solubility of Ag2CrO4.
Ag2CrO4(s) ==> 2Ag^+ + CrO4^2-
...x............2x.......x
...........Na2CrO4 ==> 2Na^+ + CrO4^2-
initial....0.005M.......0........0
change....-0.005........0.005...0.005
equil.......0..........0.005......0.005
Ksp Ag2CrO4 = (Ag^+)^2(CrO4^2-)
(Ag^+) = 2x from the Ag2CrO4
(CrO4^2-) = x from Ag2CrO4 and 0.005 from Na2CrO4.
Substitute and solve for x which is the solubility of Ag2CrO4 in moles/L = M.
Post your work if you get stuck.
To calculate the solubility of silver chromate (Ag2CrO4) in 0.005 M Na2CrO4, you can use the concept of the solubility product constant (Ksp).
The balanced equation representing the dissolution of silver chromate is:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
The Ksp expression for this reaction is:
Ksp = [Ag+]^2 [CrO4^2-]
We are given that the Ksp value for silver chromate is 2.6 x 10^-12. Since the concentration of the sodium chromate (Na2CrO4) is 0.005 M, we can assume that the concentration of the chromate ion (CrO4^2-) is also 0.005 M.
Let's assume that the initial solubility of silver chromate is "x mol/L". This means that the initial concentrations of silver ion (Ag+) and chromate ion (CrO4^2-) are both zero.
At equilibrium, the concentrations of Ag+ and CrO4^2- will be equal to the initial solubility "x mol/L". Thus, the concentration of Ag+ will be [Ag+] = 2x, and the concentration of CrO4^2- will be [CrO4^2-] = x.
Substituting these values into the Ksp expression:
Ksp = (2x)^2 * x
2.6 x 10^-12 = 4x^3
We can solve this equation for x:
x^3 = (2.6 x 10^-12) / 4
x = (2.6 x 10^-12 / 4)^(1/3)
x ≈ 5.15 x 10^-5 mol/L
Therefore, the solubility of silver chromate in 0.005 M Na2CrO4 is approximately 5.15 x 10^-5 mol/L.
To calculate the solubility of silver chromate (Ag2CrO4) in a solution of sodium chromate (Na2CrO4), you need to use the concept of the solubility product constant (Ksp). The Ksp represents the equilibrium constant for the dissolution of a slightly soluble salt in water.
The solubility product constant (Ksp) for Ag2CrO4 is given as 2.6x10^-12. It expresses the relationship between the concentrations of the silver ion (Ag+) and chromate ion (CrO4^2-) in a saturated solution of silver chromate.
Let's denote the solubility of silver chromate as 'x'. Since silver chromate is a 1:1 ratio compound, the concentration of Ag+ and CrO4^2- ions in the solution is also 'x'.
The balanced chemical equation for the dissolution of silver chromate is:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
Now, using the stoichiometry of the reaction, we can write the expression for the Ksp:
Ksp = [Ag+]^2 * [CrO4^2-]
Substituting the equilibrium concentrations of Ag+ and CrO4^2- with 'x', the Ksp expression becomes:
2.6x10^-12 = (x)^2 * (x)
Simplifying the equation further:
2.6x10^-12 = x^3
Now, solve for 'x' by taking the cubic root of both sides:
x = (2.6x10^-12)^(1/3)
Using a calculator, evaluate the cube root to find the solubility of silver chromate in the given sodium chromate solution.
Simply substitute the value of Ksp into the equation and calculate the root value.