Download Unit 8 Chemical Equilibrium 8B

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Unit 8 Chemical Equilibrium
8B-7 Solubility Equilibria
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Recall that for a solute to be dissolved in a solvent requires three steps:
1. Attractive forces between solute particles are broken
2. Attractive forces between some solvent particles are broken
3. Attractive forces between solute and solvent particles are formed
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Each of these processes exists as an equilibrium system.
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The solubility of the solute depends on the equilibrium position of these
processes, especially steps 1 and 3.
If the equilibria lie on the side of solute-solute bonds breaking and solute-solvent
bonds forming, then the solute is soluble in the solvent.
If either of these equilibria lie in opposition to these processes, the solute will be
slightly, sparingly, or even in-soluble.
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The solubility equilibrium for an aqueous solution of a salt is written as:
MX(s)
Ma+ + Xa-

And so the equilibrium constant for this solubility, known as the solubility
product constant, is:
Solutions reach a state of equilibrium when the solvent is saturated with solute
ions.

𝐾𝑠𝑝 =
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[𝑀𝑎+] [𝑋𝑎−]
1
= [𝑀𝑎 +][𝑋𝑎−]
The Ksp value is very large for very soluble compounds and very small for very
insoluble compounds.
Ksp values are also found in Appendix D, page 1063.
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Example 1 – write the solubility product constant for ferric hydroxide, Fe(OH)3.
Example 2 – A saturated solution of calcium phosphate, Ca3(PO4)2, has ion
concentrations of [Ca2+] = 4.53 x 10–7 M and [PO43–] = 3.02 x 10–7 M at 25 oC.
Determine the Ksp value of calcium phosphate at this temperature.

The equilibrium constant, Ksp, is related to the molar or mass solubility values; in
fact, we can convert between them using an “ICE” chart.
Example 3 – Determine the molar concentrations of the ions in a saturated solution
of calcium fluoride.
Example 4 – Determine the solubility of calcium fluoride in grams per liter.

As with other systems leading towards an equilibrium, the position of a solution
system relative to its solubility product equilibrium can be determined by
calculating the “reaction quotient” or trial ion product (Q) of a solution.
o If Q = Ksp, then the reaction is at equilibrium (saturated solution)
o If Q < Keq, then the reaction must proceed forward to reach equilibrium
(unsaturated solution)
o If Q > Keq, then the reaction must proceed in reverse to reach eq’m
(supersaturated solution)
Predicting the formation of precipitates
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We can make a qualitative prediction of whether two ionic solutions will form a
precipitate upon mixing by referring to a solubility chart.
However, a more quantitative approach is available using the Ksp values.
First, when the solutions are mixed, identify all of the ions involved.
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Use the solubility chart to identify which ions are likely to form a precipitate (if
any).
Determine the concentrations of these ions.
State the balanced equation of the possible precipitation rxn, and determine the
trial ion product, Q, value.
Compare the Q value with the known Ksp value for that precipitate.
Example 5 – predict whether a precipitate forms from the mixing of 200.0 mL of
0.0075 M calcium chloride and 300.0 mL of 0.0100 M potassium hydroxide.
FACTORS AFFECTING SOLUBILITY
1. Common Ion Effect
 As predicted by the Le Chatelier Principle, the addition of a soluble ion in a
solubility equilibrium will result in the equilibrium shifting left, to form more
precipitate.
Example 6 – Compare the molar solubility of lead (II) sulfate in distilled water and in a
solution containing 0.005 M sodium sulfate.
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2. pH
basic salts (i.e. salt of strong base and weak acid) will be more soluble under low
pH conditions.
This is because the acid of the solution will consume the basic anion and drive the
equilibrium to the right.
Acidification of the oceans is leading to increased dissolving of carbonate corals.
Amphoteric salts, such as the hydroxides or oxides of Al3+, Zn2+, Sn2+, or Cr3+, will
have increased solubility under both very low or very high pH conditions.
3. Formation of Complex Ions
 Many transition metal ions form complexes with soluble molecules that contain
lone pairs (similar to hydration, but with other molecules than water).
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 The formation of these complexes effectively removes the metal ions from the
equilibrium, so the equilibrium shifts right, to increase the solubility and produce
more metal ions.
Assigned Questions: BLBMW: p. 744 #49 – 51; 53 – 65 (odd), 67, 71
SCH4UA
Unit 8 Chemical Equilibrium
8B-7 Solubility Equilibria
Example 1 – write the solubility product constant for ferric hydroxide, Fe(OH)3.
Example 2 – A saturated solution of calcium phosphate, Ca3(PO4)2, has ion
concentrations of [Ca2+] = 2.25 x 10–6 M and [PO43–] = 1.50 x 10–6 M at 25 oC.
Determine the Ksp value of calcium phosphate at this temperature.
Example 3 – Determine the molar concentrations of the ions in a saturated solution
of calcium fluoride.
Example 4 – Determine the solubility of calcium fluoride in grams per liter.
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Example 5 – predict whether a precipitate forms from the mixing of 200.0 mL of
0.0075 M calcium chloride and 300.0 mL of 0.0100 M potassium hydroxide.
Example 6 – Compare the molar solubility of lead (II) sulfate in distilled water and in a
solution containing 0.005 M sodium sulfate.
Assigned Questions: BLBMW: p. 744 #49 – 51; 53 – 65 (odd), 67, 71