If solubility of Ca3 (PO4)2 is ‘s’ moles per liter at 25 oC, calculate its solubility productat that temperature.
Question
If solubility of Ca3 (PO4)2 is ‘s’ moles per liter at 25 oC, calculate its solubility productat that temperature.
Solution
The solubility product (Ksp) of a compound is the product of the molar concentrations of its constituent ions, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.
The dissolution of Ca3(PO4)2 in water can be represented by the following equation:
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43-(aq)
If the solubility of Ca3(PO4)2 is 's' moles per liter, then the concentration of Ca2+ ions in the solution is 3s and the concentration of PO43- ions is 2s.
The solubility product (Ksp) is then given by:
Ksp = [Ca2+]^3 * [PO43-]^2
Substituting the concentrations into the equation gives:
Ksp = (3s)^3 * (2s)^2 = 108s^5
So, the solubility product of Ca3(PO4)2 at 25°C is 108s^5.
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