Question

do not understand this problem

The Ksp of CaSO4 is 4.93× 10–5. Calculate the solubility (in g/L) of CaSO4(s) in 0.500 M Na2SO4(aq) at 25 °C.

Answer 1

This is a solubility product problem with a common ion. The common ion is sulfate (SO4 in CaSO4 and SO4 in Na2SO4).l

let x = solubility in mols/L.
.........CaSO4 ==> Ca^2+ + SO4^2-
.........solid......x........x

Ksp = 4.93E-5 = (Ca^2+) (SO4^2-)

For Na2SO4 ==> 2Na^+ + SO4^2-
...............2*0.5M...0.5M

Ksp you have in the problem.
(Ca^2+) substitute x
(SO4^2-) substitute x+0.5 (NoteZ: x is from CaSO4 and 0.5 from Na2SO4.)
Solve for x = (Ca^2+) = (CaSO4) in mols/L.
You want grams/L; therefore, x in mols/L * molar mass CaSO4 = solubility CaSO4 in g/L.

Answer 2

thank you so much! helped me alot