Chap'15: Applications of Aqueous Equilibria

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Chap.15 Applications of Aqueous Equilibria

• Overview
• Buffer Solutions
• Acid-Base Reactions, Titrations
• Solubility Equilibria

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• The Common-Ion Effect
• Consider a solution of HF (Ka 7.2 x 104)
• HF(aq) ? H(aq) F(aq)
• Now add some solid NaF to this solution.
• NaF(s) ? Na(aq) F(aq)
• How will this affect the HF equilibrium?
• F is a common ion. According to Le Chateliers
  Principle, the HF equilibrium will be shifted to
  the left (reactant) side. Thus, H decreases.

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• Now consider an aqueous NH3 solution.
• NH3(aq) H2O(l) ? NH4(aq) OH(aq)
• Add some solid NH4Cl
• NH4Cl(s) ? NH4(aq) Cl(aq)
• The increase in NH4 will shift this
  equilibrium to the left, decreasing OH.
• Conclusion The addition of a common ion will
  repress the dissociation of a weak acid or weak
  base.

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• Buffer Solution contains a weak acid plus its
  salt (conjugate base) or a weak base plus its
  salt (conjugate acid).
• With both an acidic and a basic species present,
  buffer solutions can resist changes in pH.
• Ex 1. A solution contains 0.10 mol HC2H3O2 and
  0.20 mol of NaC2H3O2 in 1.0 L total volume. What
  is the pH of this solution?
• Always consider complete reactions before
  equilibria.

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• NaC2H3O2(aq) ? Na(aq) C2H3O2(aq)
• 0
  0.20 M 0.20 M
• 
  neutral
• HC2H3O2(aq) ? H(aq)
  C2H3O2(aq)
• I. 0.10 M 0
  0.20 M
• C. -x x
  x
• E. 0.10x x
  0.20x
• Now set up the mass-action expression.
• Note that there are now two contributions to the
  acetate ion (the salt provides 0.20 M and the
  acid provides x M).

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• By our usual approximation, (0.10x) 0.10.
• But then we can also say that (0.20 x) 0.20.
• Solving, x H 9.0 x 106 M
• and pH 5.05

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• In buffer solutions, the initial concentrations
  of both the weak acid (or base) and the salt are
  generally gtgt x.
• Then,
• Now, we can derive another useful relationship
  from this.

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• Taking the logarithm of both sides
• And rearranging
• This is the Henderson-Hasselbalch equation for
  buffer solutions.

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• In our example, pKa log(1.8 x 105) 4.74
• Then, pH 4.74 log (0.20 / 0.10)
• 4.74 0.30 5.04
• Ex 2 A solution is 0.15 M in NH3 and 0.25 M in
  NH4Cl. What is the pH of this solution?
• First, the salt dissociates
• NH4Cl ? NH4 Cl
• 0 0.25 M
  0.25 M
• 
  neutral ion

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• Now, the weak base establishes equilibrium
• NH3 H2O ? NH4
  OH
• I. 0.15
  0.25 0
• C. -x
  x x
• E. 0.15-x
  0.25x x
• x OH 1.1 x 105 M
• pOH 4.96 pH 9.04

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• In order to use the Henderson-Hasselbalch
  equation, we need the Ka for the acidic component
  of the buffer. In this case, it is NH4.
• Since salts do not appear in weak acid tables, we
  need to calculate Ka.
• Ka for NH4 Kw / Kb for NH3
• (1.00 x 1014) / (1.8 x 105) 5.6 x
  1010
• Then, pKa log (5.6 x 1010) 9.25

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• Now, pH pKa log CBo / CAo
• 9.25 log (0.15 / 0.25)
  9.03
• The I.C.E. method and the Henderson-Hasselbalch
  equation give the same result.
• Next, we will see how to prepare a buffer
  solution of a specified pH.

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Preparation of Buffer Solutions

• ex. How much solid NaC2H3O2 must be added to 100
  mL of 0.10 M HC2H3O2 to make a buffer of pH 3.65?
  Assume that there is no volume change.
• We will use the Henderson-Hasselbalch equation.
• pH pKa log CBo / CAo

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• We have seen before that the pKa for acetic acid
  is 4.74.
• We know that the initial concentration of the
  acid is 0.10 M.
• 3.65 4.74 log CBo / (0.10)
• log CBo / (0.10) 3.65 4.74
  1.09
• CBo / (0.10) 10 1.09 8.13
  x 102
• CBo (0.10) (8.13 x 102) 8.13 x
  103 M
• (8.13 x 103 mol/L) (0.100 L) (82.0 g/mol)
  0.067 g of
• 
  NaC2H3O2

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Summary of the Four Types of pH Problems

• 1) Strong acid only or strong base only
  complete ionization or dissociation. H or
  OH is given by the initial acid or base
  concentration.
• Weak acid only or weak base only the substance
  is in equilibrium with its ions. The standard
  I.C.E. setup is used.
• Salt of weak acid or salt of weak base only.
• a) Salt dissociates completely.
• b) One of the ions acts as a weak acid or
  base. The K value must be calculated.

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• Buffer Solutions contain weak acid or base and
  its salt.
• a) The salt dissociates completely.
• b) The weak acid or base is in equilibrium.
  Use the I.C.E. approach, where the initial
  concentration of the conjugate is due to the
  salt.