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It's fairly easy to compute the solubility
of a compound in either moles/L or g/L from the Ksp value of the
compound. Simply write out the Ksp
expression, then express the equilibrium concentration of the ions in terms
of s, the molar solubility. Solve for s and you have the molar
solubility: converting to g/L is done by multiplying by the molar mass.
Example: The Ksp of PbCl2 is
1.7*10-5. What is the solubility of the compound in moles/L and
grams/L?
Solution: First, write down the reaction when lead chloride is added
to water and the Ksp expression for the reaction.
- PbCl2(s) < = > Pb+2(aq) + 2Cl-(aq)
- Ksp = [Pb+2][Cl-]2 =
1.7*10-5
For each mole of PbCl2 that dissolves, 1 mole of
Pb+2 and two moles of Cl- are formed. If we define the
solubility of PbCl2 in moles/L as s, then [Pb+2] = s at
equilibrium and [Cl-] = 2s at equilibrium. Now just solve for s.
- Ksp = [Pb+2][Cl-]2
- 1.7*10-5 = (s)(2s)2
- s = 0.016 mol/L
Thus the molar solubility of PbCl2 is
0.016 mol/L. To figure out the solubility in g/L, simply multiply the molar
solubility by the molar weight of PbCl2, 278.1 g/mol
- 0.016 mol/L * 278.1 g/mol = 4.4 g/L
Thus, the solubility of
PbCl2 is 4.4 g/L
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