What is the solubility of Ca3(PO4)2 in 0.045 M of CaCl2
To determine the solubility of Ca3(PO4)2 in 0.045 M CaCl2, we need to consider the common ion effect. The presence of Ca2+ ions from CaCl2 can affect the solubility of Ca3(PO4)2.
1. Write and balance the chemical equation:
Ca3(PO4)2(s) ⇌ 3 Ca2+(aq) + 2 PO4^3-(aq)
2. Calculate the initial concentration of Ca2+ ions:
Since the concentration of CaCl2 is given as 0.045 M, the initial concentration of Ca2+ ions is also 0.045 M.
3. Use an ICE (Initial, Change, Equilibrium) table to analyze the common ion effect:
Let's assume the solubility of Ca3(PO4)2 is represented as "x" (in mol/L).
Ca3(PO4)2 ⇌ 3 Ca2+ + 2 PO4^3-
Initial (M) 0 0.045 0
Change (M) -x +3x +2x
Equilibrium (M) 0 0.045 + 3x 2x
4. Write the expression for the solubility product constant (Ksp) of Ca3(PO4)2:
Ksp = [Ca2+]^3 * [PO4^3-]^2
5. Set up the Ksp expression using the equilibrium concentrations:
Ksp = (0.045 + 3x)^3 * (2x)^2
6. Determine the solubility (x) using the Ksp expression and the common ion effect:
Since the Ksp expression includes a degree-5 polynomial equation, solving it can be complicated. However, you can use numerical methods or software to find the approximate value of x.
Please note that this is a simplified explanation, and the actual calculation of solubility may involve additional considerations, such as the activity coefficient.