Chapter 15�Additional Aspects of Aqueous Equilibria

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John D. Bookstaver

St. Charles Community College

Cottleville, MO

Aqueous

Equilibria

Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO₄ in water:

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq)

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Aqueous

Equilibria

Solubility Products

The equilibrium constant expression for this equilibrium is

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K_sp = [Ba²⁺] [SO₄²⁻]

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where the equilibrium constant, K_sp, is called the solubility product.

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Aqueous

Equilibria

Solubility Products

• K_sp is not the same as solubility.
• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

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Aqueous

Equilibria

The Common-Ion Effect

• Consider a solution of acetic acid:
• CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq)
• If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

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Aqueous

Equilibria

The Common-Ion Effect

“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

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Aqueous

Equilibria

The Common-Ion Effect

Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

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K_a for HF is 6.8  10⁻⁴.

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Aqueous

Equilibria

The Common-Ion Effect

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HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F⁻(aq)

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Because HCl, a strong acid, is also present, the initial [H₃O⁺] is not 0, but rather 0.10 M.

Aqueous

Equilibria

The Common-Ion Effect

= x

​

1.4  10⁻³ = x

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Aqueous

Equilibria

The Common-Ion Effect

• Therefore, [F⁻] = x = 1.4  10⁻³

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[H₃O⁺] = 0.10 + x = 0.10 + 1.4  10⁻³ = 0.10 M

​

• So, pH = −log (0.10)

pH = 1.00

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Aqueous

Equilibria

Factors Affecting Solubility

• The Common-Ion Effect
• If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.
• BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq)

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Aqueous

Equilibria

Factors Affecting Solubility

• pH
• If a substance has a basic anion, it will be more soluble in an acidic solution.
• Substances with acidic cations are more soluble in basic solutions.

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Aqueous

Equilibria

Factors Affecting Solubility

• Complex Ions
• Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

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Aqueous

Equilibria

Factors Affecting Solubility

• Complex Ions
• The formation of these complex ions increases the solubility of these salts.

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Aqueous

Equilibria

Factors Affecting Solubility

• Amphoterism
• Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
• Examples of such cations are Al³⁺, Zn²⁺, and Sn²⁺.

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Aqueous

Equilibria

Will a Precipitate Form?

• In a solution,
• If Q = K_sp, the system is at equilibrium and the solution is saturated.
• If Q < K_sp, more solid can dissolve until Q = K_sp.
• If Q > K_sp, the salt will precipitate until Q = K_sp.

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Aqueous

Equilibria

A solution of 750.0mL of 4.00x10^-3M Ce(NO₃)₃ and 300.0mL of 2.00x10^-2M KIO₃ is made. Will Ce(IO₃)₃ precipitate?

• K_sp Ce(IO₃)₃ = 1.9x10^-10
• Should know: Potassium & Nitrate are spectators
• Calculate [Ce³⁺] & [IO₃^-] after mixing
• [Ce³⁺] = 2.86x10^-3M
• [IO₃^-] = 5.71x10^-3M
• Net Equation: Ce(IO₃)_{3 (s)} ⇌ Ce³⁺_(aq) + 3IO₃^-_(aq)
• Q = [Ce³⁺]₀[IO₃^-]₀³
• Is Q > K?

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Aqueous

Equilibria

100.0mL of .0500M Pb(NO₃)₂ and 200.0mL of .100M NaI

• Eq: Pb(NO₃)_2(aq) + 2NaI_(aq) ⇌ PbI_2(s) + 2Na(NO₃)_2(aq)
• Calculate the equilibrium concentrations of Pb²⁺ and I^- in equilibrium with PbI₂
• Pb²⁺ & I^- will completely react, then redissolve to reach equilibrium
• Equilibrium rxn: PbI₂ ⇌ Pb²⁺ + 2I^-
• PbI₂ K_sp = 1.4x10^-8

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Aqueous

Equilibria

Stoichiometry

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Aqueous

Equilibria

Equilibrium

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Aqueous

Equilibria

• K_sp = [Pb²⁺][I^-]²
• 1.4x10^-8 = (x)(.0333-2x)²
• Since x is going to be small compared to .0333, we can ignore it
• 1.4x10^-8 = .00111x
• X = 1.3x10^-5
• [Pb²⁺] = 1.3x10^-5M
• [I^-] = .0333M

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Aqueous

Equilibria

Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.

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Aqueous

Equilibria

How to separate Group 1 Ions

• Suppose you have a solution that contained Ag⁺, Pb²⁺ & Hg₂²⁺ ions among others
• You added dilute HCl and form precipitates of AgCl, PbCl₂ and Hg₂Cl₂
• You have decanted the remaining solution and now only have these precipitates?
• How can you separate them?

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Aqueous

Equilibria

• Adding more water will not work (K_sp is too small)
• Adding more acid will not work (Cl^- is not attracted to H⁺ in sln)
• May take many tricks
• PbCl₂ is very sensitive to temperature
• Increasing temp will dissolve it
• AgCl forms a soluble complex with NH₃
• NH₃ will also form a white complex ppt with Hg⁺(oxidation) and cause remaining Hg₂²⁺ to reduce to Hg(l)

© 2009, Prentice-Hall, Inc.

Aqueous

Equilibria