What hydroxide concentration is required to
(a) Initiate precipitation of Ag2CrO4 from a solution that is 2.12 x 10^-3 M in Ag+?
(b) lower the Al^3+ concentration in the foregoing solution to 1.00 x 10^-9
How did Al^+3 get in that solution and what does it have to do with the problem?
Sorry it was
(a) initiate precipitation of Al^3+ from a 5.00 x 10^-3 M solution of Al2(SO4)3?
(b) lower the Al^3+ concentration in the foregoing solution to 1.00 x 10^-9 M?
To determine the hydroxide concentration required to initiate precipitation, you need to consider the solubility product constant (Ksp) of Ag2CrO4 and the reaction equation.
(a) The balanced equation for the precipitation reaction is:
2Ag+ + CrO4 2- -> Ag2CrO4
The corresponding expression for Ksp is:
Ksp = [Ag+]^2 [CrO4 2-]
Given that the solution is 2.12 x 10^-3 M in Ag+, the concentration of [Ag+] = 2.12 x 10^-3 M.
To determine the [CrO4 2-] concentration needed for precipitation, we can rearrange the Ksp expression as follows:
[CrO4 2-] = Ksp / [Ag+]^2
Now, substitute the given values to find the required [CrO4 2-] concentration:
[CrO4 2-] = Ksp / [Ag+]^2
[CrO4 2-] = Ksp / (2.12 x 10^-3)^2
You would need to know the value of the Ksp constant for Ag2CrO4 to calculate the [CrO4 2-] concentration required.
(b) To lower the Al^3+ concentration in the solution to 1.00 x 10^-9, you need to consider the reaction between Al^3+ and hydroxide (OH-) ions to form Al(OH)3 precipitate.
The balanced equation for this reaction is:
Al^3+ + 3OH- -> Al(OH)3
To calculate the hydroxide concentration needed to achieve the desired Al^3+ concentration, you would need to know the initial concentration of Al^3+ and the Ksp constant for Al(OH)3.
Without these values, it is not possible to determine the exact concentration of hydroxide required to lower the Al^3+ concentration to 1.00 x 10^-9.
To determine the hydroxide concentration required to initiate precipitation of Ag2CrO4 and to lower the Al^3+ concentration in the solution, we need to consider the solubility product constant, Ksp, and the hydroxide ion concentration required to reach that equilibrium.
(a) To initiate precipitation of Ag2CrO4:
1. Write the balanced chemical equation for the dissolution of Ag2CrO4:
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^2-(aq)
The solubility product expression is: Ksp = [Ag+]^2[CrO4^2-]
2. Ag2CrO4 is considered insoluble because it has a low Ksp value.
The Ksp for Ag2CrO4 is given by: 1.1 x 10^-12
The balanced equation indicates that for every mole of Ag2CrO4 that dissolves, 2 moles of Ag+ ions are released. Therefore, the concentration of Ag+ introduced in the problem (2.12 x 10^-3 M) is equal to twice the concentration of Ag2CrO4 that dissolves.
Let x be the concentration of Ag2CrO4 that will dissolve, then the concentration of Ag+ ions will be 2x.
Substituting the concentrations into the Ksp expression:
Ksp = (2x)^2 * [CrO4^2-]
Simplifying, we have: 1.1 x 10^-12 = 4x^2 * [CrO4^2-]
Since Ksp is very small, we can assume that the concentration of Ag2CrO4 that dissolves (x) is negligible compared to the starting concentration (2.12 x 10^-3 M).
Therefore, we can simplify the equation to: 1.1 x 10^-12 ≈ 4x^2 * [CrO4^2-]
Solving for x, we get: x ≈ √(1.1 x 10^-12 / 4) ≈ 2.96 x 10^-7 M
This concentration represents the hydroxide ion concentration required to initiate precipitation of Ag2CrO4.
(b) To lower the Al^3+ concentration to 1.00 x 10^-9 M:
1. Write the balanced chemical equation for the dissolution of Al(OH)3:
Al(OH)3(s) + 3OH-(aq) ⇌ Al(OH)4^-(aq)
The equilibrium expression is: Ksp = [Al(OH)4^-] / [OH-]^3
2. The solubility product constant Ksp for Al(OH)3 is given as: 1.9 x 10^-33
Since the Ksp value for Al(OH)3 is extremely small, we can assume that the concentration of Al(OH)3 that dissolves (x) is entirely consumed by the hydroxide ions to form Al(OH)4^-. Therefore, the concentration of OH- ions introduced in the problem is equal to the concentration of Al(OH)3 that dissolves.
Let x be the concentration of OH- ions, then the concentration of Al^3+ ions will be 3x.
Substituting the concentrations into the Ksp expression:
Ksp = (3x) / (x^3)
Simplifying, we have: 1.9 x 10^-33 = 3 / x^2
Rearranging the equation, we get: x^2 = 3 / (1.9 x 10^-33)
Solving for x, we find: x ≈ √(3 / (1.9 x 10^-33)) ≈ 5.49 x 10^-12 M
This concentration represents the hydroxide ion concentration required to lower the Al^3+ concentration in the solution to 1.00 x 10^-9 M.