What is the solubility of calcium sulfate (CaSO4) in 0.30 M aqueous sodium sulfate (Na2SO4)? (Ksp of calcium sulfate = 2.0 10–5 M)
1.7 10–5 M
6.7 10–5 M
1.4 10–3 M
1.5 10–3 M
To determine the solubility of calcium sulfate (CaSO4) in 0.30 M aqueous sodium sulfate (Na2SO4), we need to consider the common ion effect.
The common ion effect states that if a solution already contains an ion that is also a component of an insoluble salt, the solubility of that salt will decrease.
In this case, calcium sulfate (CaSO4) is the insoluble salt and sodium sulfate (Na2SO4) is the soluble salt that contains the common ion sulfate (SO4^2-). We are given that the solubility product constant (Ksp) for calcium sulfate is 2.0 x 10^–5 M.
To solve this problem, we need to determine the concentration of calcium ions (Ca^2+) in the presence of the common ion sulfate (SO4^2-) from the sodium sulfate solution.
Using the Ksp expression for calcium sulfate: Ksp = [Ca^2+][SO4^2-] = (2.0 x 10^–5 M)
Since the concentration of sulfate ions from the sodium sulfate solution is 0.30 M, we can assume that the concentration of sulfate ions from the dissociation of calcium sulfate is much smaller than 0.30 M.
Let's assume x is the solubility (concentration) of calcium sulfate in the presence of 0.30 M sodium sulfate. Since calcium sulfate dissociates into one calcium ion and one sulfate ion, the concentration of calcium ions ([Ca^2+]) is equal to x.
Substituting the values into the Ksp expression, we get:
(2.0 x 10^–5) = (x)(0.30)
Simplifying the equation, we have:
x = (2.0 x 10^–5) / (0.30)
Calculating the value, we find that x ≈ 6.7 x 10^–5 M.
Therefore, the solubility of calcium sulfate (CaSO4) in 0.30 M aqueous sodium sulfate (Na2SO4) is approximately 6.7 x 10^–5 M.
So, the correct answer is: 6.7 x 10^–5 M.