You made up a saturated solution of calcium sulfate (CaSO4). The temp is 25 degrees Celsius. You then add 5.00*10^-3 of sodium sulfate (NaSO4). What are the concentrations of calcium and sulfate after equilibrium is reached? The pKs of CaSO4 is 4.58.

I know the Ks of CaSO4 to be 10^(-4.58) which equals 2.63E-5 but then what?

To determine the concentrations of calcium and sulfate after equilibrium is reached, we need to take into account the solubility of calcium sulfate and the stoichiometry of the reaction.

The solubility product constant expression (Ksp) for calcium sulfate (CaSO4) is given by:
Ksp = [Ca2+][SO42-]

Based on the pKs value provided (pKs = -log(Ksp)), the Ksp for CaSO4 is found to be:
Ksp = 10^(-pKs) = 10^(-4.58) = 2.63E-5

Since calcium sulfate is considered sparingly soluble, we can assume that the concentration of Ca2+ at equilibrium will be significantly smaller than the initial concentration of sodium sulfate (Na2SO4) you added. Therefore, we can also assume that adding 5.00*10^-3 of sodium sulfate will have a negligible effect on the initial concentration of sulfate ions.

Let's assume the initial concentration of calcium and sulfate ions before adding sodium sulfate is x M.

At equilibrium, the concentration of calcium ions ([Ca2+]) will be less than x M. Let's say it is y M.

The concentration of sulfate ions ([SO42-]) after equilibrium can be assumed to be x M because the addition of 5.00*10^-3 of sodium sulfate will have a negligible effect on the initial concentration of sulfate ions.

Using the solubility product constant expression and the stoichiometry of the reaction, we can write:

Ksp = [Ca2+][SO42-]
2.63E-5 = y * x

Since the concentration of calcium sulfate is assumed to be low, we can approximate y M as x M. Therefore, we can write:

2.63E-5 = x * x
x^2 = 2.63E-5
x = √(2.63E-5)
x ≈ 0.005125 M

So, after equilibrium is reached, the approximate concentration of calcium ions ([Ca2+]) is 0.005125 M, and the concentration of sulfate ions ([SO42-]) is 0.005 M.

To find the concentrations of calcium and sulfate after equilibrium is reached, we need to consider the solubility product constant (Ksp) for calcium sulfate (CaSO4).

The Ksp expression for CaSO4 is:
Ksp = [Ca2+][SO4 2-]

We know that the pKs of CaSO4 is 4.58, which means that the negative logarithm of the Ksp is 4.58. So, we can calculate the Ksp as follows:
Ksp = 10^(-pKs) = 10^(-4.58) = 2.63E-5

Now, let's assume that x moles of CaSO4 dissolve and dissociate to give x moles of Ca2+ and x moles of SO4 2-. Since the stoichiometry of CaSO4 is 1:1, we can consider the concentrations of Ca2+ and SO4 2- to be also x.

At equilibrium, the solubility product constant expression is satisfied. Therefore, we can write:
Ksp = [Ca2+][SO4 2-]

Substituting the concentrations of Ca2+ and SO4 2- as x, we have:
2.63E-5 = x * x

Simplifying the equation further:
x^2 = 2.63E-5

Taking the square root of both sides:
x = sqrt(2.63E-5)

Calculating the square root:
x ≈ 0.0051194

So, the concentration of both calcium (Ca2+) and sulfate (SO4 2-) ions after equilibrium is approximately 0.0051194 M.

CaSO4 ==> Ca^2+ + SO4^2-

Ksp = (Ca^2+)(SO4^2-) = 2.63E-5
From CaSO4 alone;
(Ca^2+) = x
(SO4^2-) = x

Na2SO4 ==> 2Na^+ + SO4^2-
You added 5.00E-3 WHAT? Is that molar? If so, then Na^2+ = 2*5.00E-3M
and (SO4^2-) = 0.005M
Plug all of this into Ksp for CaSO4.
Ksp = (x)(x+0.005) = 2.63E-5
Solve for x, then x + 0.005.