calculate the minimum concentration of ag+ ion that must be added to (or built up in) a 0.140 m Na2CrO4 solution in order to initiate a precipitation of silver chromate. The ksp of ag2cro4 is 1.2 × 10^-12.
I will assume you have writer's cramp and you can't use the caps key. Sentences begin with a capital letter, you know, and it's important that you remember that m stands for molality and M stands for molarity. There's a difference. So that must be 0.140 M. Also, ag2cro4 means nothing to me especially when you realize CO stands for carbon monoxide, Co stands for cobalt, and co stands for company.
Ag2CrO4 ==> 2Ag^2+ + CrO4^2-
Ksp = (Ag^2+)^2(CrO4^2-)
You know CrO4^2-, plug in and solve for Ag^+.
To calculate the minimum concentration of Ag+ ion required to precipitate silver chromate (Ag2CrO4) from a 0.140 M Na2CrO4 solution, we need to use the solubility product constant (Ksp) of Ag2CrO4.
The equation for the dissolution of Ag2CrO4 is:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
From this equation, we see that for every 1 mole of Ag2CrO4 that dissolves, we get 2 moles of Ag+ ions.
Given:
Ksp = 1.2 × 10^-12
[Molarity of Na2CrO4] = 0.140 M
Let x be the concentration of Ag+ ions formed in the solution.
So, the concentration of CrO4^2- ions would also be x since we started with a 1:1 molar ratio in the Ag2CrO4 equation.
Using the Ksp expression for Ag2CrO4:
Ksp = [Ag+]^2 * [CrO4^2-]
1.2 × 10^-12 = (2x)^2 * x
1.2 × 10^-12 = 4x^3
Solving for x:
x^3 = (1.2 × 10^-12) / 4
x^3 = 3 × 10^-13
x = (3 × 10^-13)^(1/3)
x ≈ 6.08 × 10^-5
Therefore, the minimum concentration of Ag+ ions that must be added to the 0.140 M Na2CrO4 solution in order to initiate the precipitation of silver chromate is approximately 6.08 × 10^-5 M.
To calculate the minimum concentration of Ag+ ion needed to initiate the precipitation of silver chromate in a 0.140 M Na2CrO4 solution, we can use the concept of the solubility product constant (Ksp). Ksp is an equilibrium constant that represents the solubility of a slightly soluble salt.
The balanced chemical equation for the precipitation reaction is:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
The Ksp expression for this reaction is:
Ksp = [Ag+]^2 [CrO4^2-]
Given that the Ksp value for Ag2CrO4 is 1.2 × 10^-12, we can set up the equation as follows:
1.2 × 10^-12 = [Ag+]^2 [CrO4^2-]
Since the precipitation of silver chromate occurs when Ag+ concentration exceeds the concentration required to saturate the solution, we can assume that the concentration of Ag+ ions is 2x (where x is the minimum concentration of Ag+ ion needed).
Re-arranging the equation, we get:
1.2 × 10^-12 = (2x)^2 [CrO4^2-]
1.2 × 10^-12 = 4x^2 [CrO4^2-]
Next, we can substitute the concentration of Na2CrO4 solution:
0.140 M = [CrO4^2-]
Now, the equation becomes:
1.2 × 10^-12 = 4x^2 (0.140)
Simplifying the equation:
8.57 × 10^-12 = x^2
x = √(8.57 × 10^-12)
x ≈ 9.25 × 10^-6 M
Therefore, the minimum concentration of Ag+ ion that must be added to the 0.140 M Na2CrO4 solution in order to initiate precipitation of silver chromate is approximately 9.25 × 10^-6 M.