At 25 °C only 0.0610 mol of the generic salt AB2 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?

AB2(s) <- -> A^2+(aq)+2B^-(aq)

Is the answer ksp=9.08*10^-4?

I agree.

To calculate the Ksp (solubility product constant) of the salt AB2 at 25 °C, we need to use the given information that 0.0610 mol of AB2 is soluble in 1.00 L of water.

The Ksp expression for the salt AB2 is:

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

Where [A^2+] represents the concentration of A^2+ ions and [B^-] represents the concentration of B^- ions.

Since the stoichiometry of the reaction is 1:2 (one A^2+ ion and two B^- ions), we can express the concentrations in terms of x:

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

Now, let's substitute these values into the Ksp expression:

Ksp = [A^2+][B^-]^2
Ksp = x * (2x)^2
Ksp = 4x^3

We are given that 0.0610 mol of AB2 is soluble in 1.00 L, so the concentration of AB2 can be calculated:

Concentration of AB2 = 0.0610 mol / 1.00 L = 0.0610 M

Since AB2 dissociates into one A^2+ ion and two B^- ions, the concentration of A^2+ ions will also be equal to 0.0610 M.

Now we can substitute this concentration into the expression for Ksp:

Ksp = 4x^3 = 4 * (0.0610)^3

Evaluating this expression gives:
Ksp ≈ 9.063401 × 10^-4

Therefore, the Ksp of the salt AB2 at 25 °C is approximately 9.06 × 10^-4, which is slightly different from the answer you provided.