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CH 17: Solubility and Complex-Ion EquilibriaPowerPoint Presentation

CH 17: Solubility and Complex-Ion Equilibria

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Solubility Equilibria

Solubility Product Constant, Ksp

Same as Kc, Kp, Kw, Ka, & Kb Prod / reactant Coefficients are exponents, omit solids and pure liquids

CaF2(s) Ca2+(aq) + 2 F-(aq)

Ksp = [Ca2+][F-]2

Example1: Ksp Expressions

Write Ksp expressions for the following

a) Mg(OH)2

- SrCO3
- Ca3(AsO4)2
- Fe(OH)3

Example 2:

The solubility of silver bromate, AgBrO3, in water is 0.0072 g/L. Calculate Ksp

Example 3:

Calculate Ksp for copper(II)iodate, Cu(IO3)2. The solubillity of copper(II)iodate in water is 0.13 g/100mL

Precipitation of Ionic Compounds

Ion Product (IP)

Same as Ksp but at some time, t, snapshot like Qc, reaction quotient

CaF2(s) Ca2+ + 2 F-

IP = [Ca2+][F-]2

If IP > Ksp solution is supersaturated and precipitation will occur

If IP = Ksp the solution is saturated and equilibrium exists

If IP< Ksp the solution is unsaturated and ppt will not occur

Example 4:

Will a precipitate form on mixing equal volumes of the following solutions?

- 3.0 x 10-3 M BaCl2 and 2.0 x 10-3 M Na2CO3 (Ksp = 2.6 x 10-9 for BaCO3)
- 1.0 x 10-5 M Ba(NO3)2 and 4.0 x 10-5 M Na2CO3

Example 4:

- From each of the following ion concentrations in a solution, predict whether a ppt will form in the solution
A) [Ba2+] = 0.020 M [F-] = 0.015 M

B) [Pb2+] = 0.035 M [Cl-] = 0.15 M

Example 5:

- The following solutions are mixed:
1 L of a 0.00010 M NaOH

1 L of a 0.0020 M MgSO4

Is a ppt expected, explain

Measuring Ksp and Calculating Solubility from Ksp

Example 6: A saturated solution of Ca3(PO4)2 has [Ca2+] = 2.01 x 10-8 M and

[PO43-] = 1.6 x 10-5 M.

Calculate Ksp for Ca3(PO4)2

Factors that Affect Solubility

1. The Common ion effect

MgF2(s) Mg2+(aq) + 2 F-(aq)

If we try dissolve this in a aqueous solution of NaF the equilibrium will shift to the left. This will make MgF2 less soluble

2. Formation of Complex ions

Complex ion: An ion that contains a metal cation bonded to one or more small molecules or ions, NH3, CN- or OH-

AgCl(s) Ag+ + Cl-

Ag+ + 2 NH3 Ag(NH3)2+

Ammonia shifts the equilibrium to the right by tying up Ag+ ion in the form of a complex ion

Factors that Affect Solubility

3. The pH of the solution

a) An ionic compound that contains a basic anion becomes more soluble as the acidity of the solution increases

CaCO3(s) Ca2+ + CO32-

H3O+ + CO32- HCO3- + H2O

Net: CaCO3(s) + H3O+ Ca2+ + HCO3- + H2O

Solubility of calcium carbonate increases as the pH decreases because the CO32- ions combine with protons to give HCO3- ions. As CO32- ions are removed from the solution the equilibrium shifts to the right to replenish the carbonate

PH has no effect on the solubility of salts that contain anions of strong acids because these anions are not protonated by H3O+

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