Final answer:
The solubility of CaSO4 in a 0.250 M Na2SO4 solution is calculated by solving the Ksp expression for [Ca2+], and then converting this molarity to grams per liter using CaSO4's molar mass. The solubility is found to be 2.686 g/L, which is less than in a solution without the common ion present.
Step-by-step explanation:
To calculate the solubility of CaSO4 in the presence of 0.250 M Na2SO4, we must consider the common ion effect and the solubility product constant (Ksp) of CaSO4. The Ksp expression for CaSO4 is given by:
Ksp = [Ca2+][SO42-]
Since Na2SO4 already provides a sulfate concentration of 0.250 M, adding CaSO4 would further increase this concentration minimally, because of its limited solubility due to the common ion effect. Therefore, we can assume that [SO42-] remains approximately 0.250 M. Plugging into the Ksp expression:
4.93 × 10−5 = [Ca2+](0.250)
Now solving for [Ca2+], we get:
[Ca2+] = ÷ 4.93 × 10−5 / 0.250 = 1.972 × 10−4 M
To find the solubility in g/L, we need to convert moles to grams using the molar mass of CaSO4 (136.14 g/mol):
Solubility (g/L) = 1.972 × 10−4 mol/L × 136.14 g/mol = 2.686 g/L
This result shows the solubility of CaSO4 in a solution already containing Na2SO4 is reduced due to the common ion effect.