Ag2SO4 has a solubility of 4,496 g/L. What's its Ksp (product of solubility)?

Silver sulfate has the formula `Ag_2SO_4` . The number you gave for solubility is not within the range of solubility of silver sulfate, so I'm assuming that the comma should be a decimal point and you're trying to calculate Ksp based on a solubility of 4.496 grams per liter.

First, we need to change the grams to moles to get the molar solubility of `Ag_2SO_4` :

(4.496 grams Ag2SO4)/1 Liter x (1 mole/311.8 grams) = 0.01442 moles/L

Here's the equation for the dissociation of `Ag_2SO_4` :

`Ag_2SO_4 -> 2 Ag^+ + SO_4^(2-)`

The expression for the solubility product constant is:

`K_(sp) = [Ag^+]^2[SO_4^(2-)]`

For each mole that dissolves, two moles of `Ag^+` and 1 mole of `SO_4^(2-)` are produced. We can substitute the molarity previously calculated, giving us:

`K_(sp) = [2(0.01442]^2[0.01442] = 1.199 X 10^(-5)`

This is the accepted value of Ksp for silver sulfate at 25ºC as per the attached link, so this verifies that the solubility in the problem is 4.496 g/L and not 4,496 g/L.