Enduring Understanding 6.C.3: Solubility

• The dissolution of a substance in a solvent is a reversible process, and therefore has an equilibrium constant, Ksp associated with it.
• Solubility can be understood in terms of chemical equilibrium.
• MX(s) ⇆ M+(aq) + X-(aq)    Ksp = [M+] [X-]
• This can be used to determine if a precipitate forms when two solutions are mixed, or what the concentration of a species in solution will be.
• Sodium, potassium, ammonium, and nitrate salts are almost always highly soluble in water.

• Sample Problem 1: If 100 mL of a 0.010M CaCl2 solution is mixed with 400 mL of a 0.020M Na3PO4 solution, (final volume 500 mL), will a precipitate form? The Ksp of Ca3(PO4)2 is 1.0 x 10-26
• 3 CaCl2(aq) + 2 Na3PO4(aq) ⇆ Ca3(PO4)2(s) + 6 NaCl(aq)
• The resulting ionic equation is:
• Ca3(PO4)2(s) ⇆ 3 Ca2+(aq) + 2 PO43-(aq)
• So the Ksp will be given by:
• Ksp = [Ca2+]3[PO43-]2 = 1.0 x 10-26
• The concentration of Ca2+ will be: (0.010M) x (100 mL/500 mL)= 0.0020M
• The concentration of PO43- will be: (0.020M) x (400 mL/500 mL)= 0.016M
• The solubility product Qsp will be:
• Qsp = [Ca2+]3[PO43-]2= 0.00203 x 0.0162 = 2.0 x 10-12
• Since Qsp >> Ksp , a precipitate of Ca3(PO4)2(s) will form.
• The solubility of a poorly soluble salt decreases in the presence of a common ion.
• For example, in the system MX(s) ⇆ M+(aq) + X-(aq), if extra X- was added, for example as NaX, the solubility of MX(s) would be lower.
• Sample Problem 2: The molar solubility of of Ca3(PO4)2(s) in water is 2.5 x 10-6 mol/L. What is the molar solubility of Ca3(PO4)2(s) in an 0.1M solution of sodium phosphate?
• Ca3(PO4)2(s) ⇆ 3 Ca2+(aq) + 2 PO43-(aq)    Ksp = [Ca2+]3[PO43-]2 = 1.0 x 10-26
• x moles of Ca3(PO4)2(s) dissolve to form 3x moles of Ca2+ and 2x moles of PO43-
• In pure water, we could substitute x this into the Ksp to give:
• [Ca2+]3[PO43-]2 = (3x)3(2x)2 = 1.0 x 10-26
• But, this is an 0.1M sodium phosphate solution, so the equation becomes:
• [Ca2+]3[PO43-]2 = (3x)3(2x + 0.1)2 = 1.0 x 10-26
• Ca3(PO4)2(s) has a very low solubility. If we assume x << 0.1, this reduces to:
• [Ca2+]3[PO43-]2 = (3x)3(0.1)2 = 1.0 x 10-26
27 x 3 (0.01) = 1.0 x 10-26
x3 = 3.7 x 10-26
x = 3.3 x 10-9
• So the molar solubility of Ca3(PO4)2(s) in 0.1M sodium phosphate is 3.3 x 10-9 mol/L, which about 1/1000 of the solubility in water.

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