🤓 Based on our data, we think this question is relevant for Professor Bindell's class at UCF.
The Ksp of CaSO4 is 4.93 x 10-5. Calculate the solubility (in g/L) of CaSO4(s) in 0.300 M Na2SO4 (aq) at 25 °C.
For this problem, we’re being asked to calculate the solubility (in g/L) of CaSO4(s) in 0.300 M Na2SO4(aq). Since the compounds are ionic compounds, they form ions when dissociating in water. The dissociation of CaSO4 and Na2SO4 in water are as follows:
The sulfate ion, SO42–, has a charge of –2. Calcium is in Group 2A so it’s charge is +2:
CaSO4(s) ⇌ Ca2+(aq) + SO42–(aq)
The sulfate ion, SO42–, has a charge of –2. Sodium is in Group 1A so it’s charge is +1:
Na2SO4(s) → 2 Na+(aq) + SO42–(aq)
Notice that there is a common ion present, SO42–. The common ion effect states that the solubility of a salt is lower in the presence of a common ion.
We can construct an ICE table for the dissociation of CaSO4. Remember that solids are ignored in the ICE table.