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Given solubility of Ba^2+ is 2.66×10^-8mol/l in saturated solution of Ba3(PO4)2.Calculate the Ksp value for barium phosphate
Accepted Answer
The Ksp for barium phosphate (Ba3(PO4)2) can be calculated from the solubility of Ba^2+ in a saturated solution. Here's how:
1. Write the balanced dissolution equilibrium:
Ba3(PO4)2(s) ⇌ 3Ba^2+(aq) + 2PO4^3-(aq)
2. Set up an ICE table:
| | Ba^2+ | PO4^3- |
|--------------|-------|--------|
| Initial (I) | 0 | 0 |
| Change (C) | +3s | +2s |
| Equilibrium (E) | 3s | 2s |
3. Relate solubility to equilibrium concentrations:
The given solubility of Ba^2+ is 2.66×10^-8 mol/L. Since the stoichiometry is 3:2 for Ba^2+ to PO4^3-, the solubility of PO4^3- is (2/3)
2.66×10^-8 mol/L = 1.77×10^-8 mol/L
4. Substitute into Ksp expression:
Ksp = [Ba^2+]^3 [PO4^3-]^2
Ksp = (3s)^3 (2s)^2 = 108s^5
Ksp = 108
(2.66×10^-8)^5 = 6.25×10^-38
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