The solubility of AB(s) in a 1.000-M solution of C(aq) is found to be 0.121 M. What is the Ksp of AB?

The generic metal A forms an insoluble salt ABsolid and a complex AC5aqueous. The equilibrium concentrations in a solution of AC5 were found to be A is 0.100 M, C is 0.0320 M, and AC5 is 0.100 M. Determine the formation constant Kf of AC5

To find the Ksp of AB, you need to know the solubility product constant expression, which is the equilibrium expression for the dissolution of AB(s) into its constituent ions.

The solubility product constant expression for AB is:

Ksp = [A+]^m [B-]^n

Where [A+] represents the concentration of the cation A+ in solution and [B-] represents the concentration of the anion B- in solution. The exponents m and n represent the stoichiometric coefficients of A+ and B- in the balanced chemical equation for the dissolution of AB.

From the given information, we know that the concentration of AB in solution is 0.121 M. However, we don't have direct information about the concentrations of A+ and B- separately.

Given that AB is a solid, we can assume it fully dissociates into its constituent ions in solution. This means that the concentrations of A+ and B- are equal to each other and can be represented as x in the equation:

AB(s) ⇌ A+(aq) + B-(aq)

Let's assume the stoichiometric coefficients for A+ and B- in the equation are both 1. The Ksp expression for AB becomes:

Ksp = [A+]^1 [B-]^1 = x^2

Since the solubility of AB is given as 0.121 M, we can substitute this value into the Ksp expression:

Ksp = (0.121)^2
Ksp = 0.014641

Therefore, the Ksp of AB is approximately 0.014641.

To find the Ksp (solubility product constant) of AB, we need to use the given solubility information. The solubility of AB(s) in a 1.000-M solution of C(aq) is reported to be 0.121 M.

The general solubility equilibrium expression for an ionic compound is as follows:

AB ⇌ A+ + B-

The solubility product constant (Ksp) expression for the reaction is:

Ksp = [A+]^a [B-]^b,

where a and b are the coefficients of A+ and B- ions, respectively, in the balanced chemical equation.

In this case, since AB is a binary compound, the coefficients for the product ions A+ and B- are both 1. Therefore, we can write the Ksp expression as:

Ksp = [A+] [B-].

We have been provided with the solubility of AB in terms of [AB] concentration, which is 0.121 M.

Since AB dissolves completely and dissociates into A+ and B- ions, we can assume that the concentration of A+ and B- is equal to the solubility of AB:

[A+] = [B-] = 0.121 M.

So, the Ksp of AB is:

Ksp = (0.121 M) (0.121 M) = 0.014641 M^2.

Therefore, the Ksp of AB is 0.014641 M^2.