Final answer:
The maximum solubility of CaSO4 in a 0.500 M Na2SO4 solution at 25°C can be calculated using the solubility product constant and is found to be 0.0134 g/L.
Step-by-step explanation:
To calculate the solubility of calcium sulfate (CaSO4) in the presence of a sodium sulfate (Na2SO4) solution, we must consider the common ion effect. The presence of sulfate ions from Na2SO4 will affect the solubility of CaSO4. The solubility product (Ksp) for CaSO4 is given as 4.93×10⁻⁵.
In a 0.500 M Na2SO4 solution, the concentration of sulfate ions [SO4²⁻] is already present at 0.500 M. Therefore, adding CaSO4 to this solution, it will not significantly increase the sulfate concentration. In fact, we must find the maximum concentration of calcium ions [Ca²⁻] that can be present in the solution before reaching the solubility product threshold.
Since the [SO4²⁻] from Na2SO4 already exceeds the saturation level, the reaction quotient Q can be expressed as Q = [Ca²⁻][SO4²⁻]. To prevent precipitation, Q must be less than Ksp. As such, we calculate [Ca²⁻] as follows:
Ksp = [Ca²⁻][SO4²⁻] 4.93×10⁻⁵ = [Ca²⁻](0.500 M) [Ca²⁻] = 9.86×10⁻µ M
The molar mass of CaSO4 is approximately 136.14 g/mol, so the solubility in g/L can be found by multiplying the molarity of [Ca²⁻] by the molar mass of CaSO4:
Solubility in g/L = 9.86×10⁻µ M × 136.14 g/mol = 0.0134 g/L (to three significant figures).
This calculated value is the maximum solubility of CaSO4 in a 0.500 M Na2SO4 at 25°C without causing precipitation.