Sodium sulfate is slowly added to a solution containing 0.0500 M Ca2 (aq) and 0.0290 M Ag (aq). What will be the concentration of Ca2 (aq) when Ag2SO4(s) begins to precipitate?

I worked this below.

To determine the concentration of Ca2+(aq) when Ag2SO4(s) begins to precipitate, we can set up an equilibrium expression using the solubility product constant, Ksp.

The balanced equation for the precipitation reaction is:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO4^(2-)(aq)

The solubility product constant is given by:
Ksp = [Ag+]^2 * [SO4^(2-)]

In this case, the silver ion concentration, [Ag+], is directly related to the concentration of Ag(aq) in the solution, given as 0.0290 M.

Now, we need to determine the concentration of SO4^(2-) ions in the solution. Sodium sulfate, Na2SO4, dissociates completely in water, producing 2 moles of SO4^(2-) ions for every mole of Na2SO4.

Since we are slowly adding sodium sulfate, we can assume that the concentration of Ca2+(aq) remains constant throughout the process. Thus, the concentration of Ca2+ in the solution is 0.0500 M.

Let's assume that x M of Ag2SO4(s) precipitates. Then, the initial concentration of SO4^(2-) ions is also x M.

Using the stoichiometry of the balanced equation, we can write the expression for [SO4^(2-)]:
[SO4^(2-)] = 2x

Substituting these values into the solubility product expression, we have:
Ksp = (0.0290)^2 * (2x)

Now we can solve for x:
Ksp = 0.0290^2 * (2x)
x = Ksp / (0.0290^2 * 2)

Given that the solubility product constant, Ksp, of Ag2SO4 is 1.20 x 10^(-5), we can substitute this value into the equation:
x = (1.20 x 10^(-5)) / (0.0290^2 * 2)
x ≈ 0.0214 M

Therefore, when Ag2SO4(s) begins to precipitate, the concentration of Ca2+(aq) in the solution will still be approximately 0.0500 M.