Chapter 16: Applications of Aqueous Equilibria

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Chapter 16 Applications of Aqueous Equilibria

• Renee Y. Becker
• Valencia Community College

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Acid-base Neutralization Reaction

• Acid-base neutralization reaction
• 1.      Products are water and a salt
• 2.      Four types
• a)      Strong Acid-Strong Base
• b)     Weak Acid-Strong Base
• c)      Strong Acid-Weak Base
• d)     Weak Acid-Weak Base

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Strong Acid-Strong Base

• HCl(aq) NaOH(aq) ? H2O(l) NaCl(aq)
• 1. Because HCl is a strong acid and NaOH is a
  strong base they are both strong electrolytes,
  they dissociate nearly 100
• 2.      Complete ionic equation
• H Cl- Na OH- ? H2O(l) Na Cl-
• 3.      Net ionic equation
• H3O OH- ? 2 H2O(l)

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Strong Acid-Strong Base

• 4.      Kn 1 / H3O OH- 1 / Kw 1 / 1 x
  10-14 1 x 1014
• a)      Kn is the equilibrium constant with
  respect to neutralization
• 1 x 1014 is a large and means that for a
  strong acid-strong base reaction proceeds
  essentially 100 to completion
• 5.      pH 7

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Weak Acid-Strong Base

• HF(aq) NaOH(aq) ? H2O(l) NaF
• 1. Because HF is a weak acid-weak electrolyte it
  does not dissociate well and will not be ionized
  on the reactant side
• 2. Complete ionic equation
• HF Na OH- ? H2O Na F-
• 3. Net Ionic equation
• HF OH- ? H2O F-

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Weak Acid-Strong Base

• 4. To obtain the equilibrium constant, Kn we
  need to multiply known equilibrium constants
  for reactions that add to give the net ionic
  equation for the neutralization
• HF(aq) H2O(l) ? H3O(aq) F-(aq) Ka
  3.5 x 10-4
• H3O(aq) OH- ? 2 H2O(l) 1/Kw 1 x 1014
• Net HF OH- ? H2O(l) F- Kn Ka
  (1/Kw)
• 3.5 x 10-4 (1 x 1014) 3.5 x 1010

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Weak Acid-Strong Base

• For any weak acid-strong base reaction
• 1. Kn Ka (1/Kw)
• 2.  100 completion because of OH- strong
  affinity for protons
• 3. The pH will be gt 7, due to basicity of
  conjugate base

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Strong Acid-Weak Base

• NH3(aq) HCl(aq) ? NH4(aq) Cl-
• 1. A strong acid is completely dissociated into
  H3O and A- ions
• 2. Neutralization reaction is a proton transfer
• 3. Net ionic equation
• H3O NH3 ? H2O NH4

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Strong Acid-Weak Base

• Just as before we can obtain the equilibrium
  constant for the neutralization reaction by
  multiplying known equilibrium constants for
  reactions that add to give the net ionic equation
  
• NH3 H2O ? NH4 OH- Kb 1.8 x 10-5
• H3O OH- ? 2 H2O 1/Kw 1 x 1014
•  
• Net H3O NH3 ? H2O NH4 Kn
  Kb(1/Kw)
• 1.8 x 109

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Strong Acid-Weak Base

• For any strong acid-weak base reaction
• 1.  Kn Kb(1/Kw)
• 2.  100 completion because H3O is a powerful
  proton donor
• 3. The pH will be lt 7, due to the conjugate
  acid

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Weak Acid-Weak Base

• CH3CO2H NH3 ? NH4 CH3CO2-
• 1. Both acid and base are largely undissociated
• 2.  Neutralization reaction is a proton transfer
  from the weak acid to the weak base
• 3. The equilibrium constant can be obtained by
  adding equations for the acid dissociation, the
  base protonation and the reverse of the
  dissociation of water
•      

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Weak Acid-Weak Base

• CH3CO2H H2O ? H3O CH3CO2- Ka 1.8 x
  10-5
• NH3 H2O ? NH4 OH- Kb 1.8 x 10-5
• H3O OH- ? 2 H2O 1/Kw 1 x 1014
• Net CH3CO2H NH3 ? NH4 CH3CO2-
• Kn Ka(Kb)(1/Kw) 3.2 x 104
• For any weak acid-weak base reaction
• 1.      Kn Ka(Kb)(1/Kw)
• Kn is smaller so the reaction does not go to
  completion

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Example 1 The Common-Ion Effect

• Calculate the pH of a solution prepared by
  dissolving .10 mol acetic acid and .10 mol sodium
  acetate in water, and then diluting the solution
  to a volume of 1.00 L

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Buffer Solutions

• 1.  A weak acid and its conjugate base
• 2.  Resist drastic changes in pH
• 3.   If a small amount of OH- is added, the pH
  increases but not by much because the acid
  component of the buffer neutralizes the OH-
• 4. If a small amount of H3O is added the pH
  decreases but not by much because the conjugate
  base in the buffer neutralizes the added H3O

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Buffer Solutions

• 5.  Examples
• a) CH3CO2H CH3CO2-
• b) HF F-
• c) NH4 NH3
• d) H2PO4- HPO42-
• 6. Very important in biological systems (blood
  is a buffer) (H2CO3 HCO3-)

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Example 2

• Addition of OH- to a buffer, we add 0.01 mol
  solid NaOH to 1.00 L of a 0.10 M acetic acid-0.10
  M sodium acetate solution. What is the pH?
• Because solutions involving a strong acid or
  base go to nearly 100 completion we must
  account for the neutralization before we can
  calculate H3O

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Example 3

• Addition of H3O to a buffer, we add 0.01 mol HCl
  to 1.00 L of a 0.10 M acetic acid-0.10 M sodium
  acetate solution. What is the pH?

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Buffer capacity

• 1. Measure of the amount of acid or base that a
  solution can absorb without a significant change
  in pH
• 2.    Measure of how little the pH changes with
  the addition of a given amount of acid or base
• 3.   Depends on how many moles of weak acid and
  conjugate base are present
• 4.  The more concentrated the solution (acid
  conj. base), the greater the buffer capacity
• 5. The greater the volume (acid and conj. Base),
  the greater the buffer capacity

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Henderson-Hasselbalch Equation

• Henderson-Hasselbalch Equation
• pH pKa log base/acid
• Tells us how the pH affects the dissociation of
  a weak acid
• 2. Also tells us how to prepare a buffer
  solution with a given pH.
• a) Pick a weak acid that has a pKa close to the
  desired pH
• b) Adjust the base/acid ratio to the value
  specified by the HH equation
• The pKa of the weak acid should be within 1 pH
  unit of desired pH

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Example 4

• I want to prepare a buffer solution with a pH of
  7.00 and one of a pH of 9.00 which of the
  following pairs of weak acid-conj. bases should I
  use?
•  
• CH3CO2H CH3CO2- Ka 1.8 x 10-5 pKa
  4.74
• HF F- Ka 3.5 x 10-4 pKa 3.46
• NH4 NH3 Ka 5.56 x 10-10 pKa 9.25
• H2PO4- HPO42- Ka 6.2 x 10-8 pKa 7.21

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Example 5

• Use the HH equation to calculate the pH of a
  buffer solution prepared by mixing equal volumes
  of 0.20 M NaHCO3 and 0.10 M Na2CO3 (Ka 5.6 x
  10-11 for HCO3-)

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Example 6

• Give a recipe for preparing a NaHCO3-Na2CO3
  buffer solution that has a
• pH 10.40

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pH Titration Curves

• 1.  A plot of the pH of the solution as a
  function of the volume of the added titrant
• 2.  A solution of a known concentration of base
  or acid is added slowly from a buret to a second
  solution with an unknown concentration of acid
  or base
• 3.  Progress is monitored with a pH meter or by
  color of indicator
• 4. Equivalence point is the point at which
  stoichiometrically equivalent quantities of acid
  and base have been mixed together

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pH Titration Curves

• 5. Endpoint is when the color of the acid-base
  indicator changes
• 6. There are four important types of titration
  curves
• a)      Strong Acid-Strong Base
• b)      Weak Acid-Strong Base
• c)      Weak Base-Strong Acid
• d)      Polyprotic Acid-Strong Base
• We will only be calculating for Strong
  acid-strong base titrations but you are
  responsible to be able to recognize and label the
  titration curves for all

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Strong Acid-Strong Base
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Weak Acid-Strong Base
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Strong Acid-Weak Base
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Polyprotic Acid-Strong Base
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Strong Acid-Strong Base Titrations

• Titration of a strong acid (50 mL of a 0.02 M
  HCl) by a strong base (0.030 M NaOH)
• There are four main calculations for this type of
  titration
• 1.  Before any Base has been added
• 2.  Before the equivalence point
• 3.   At the equivalence point
• 4. After the equivalence point

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1. Before any base has been added

• 1.  Since HCl is a strong acid the initial
  concentration of H3O initial molarity 0.02
  M
• pH -log0.02 1.70

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2. Before the equivalence point

• Lets say we have added 10 mL of 0.03 M NaOH
• The added OH- ions will neutralize some of the
  H3O ions
• Moles of H3O ions MV 0.02 .05 L .001
  mol H3O
• Moles of OH- ions MV 0.03 .01L .0003
  moles
• Molarity of H3O after addition of NaOH
• Moles H3O - moles OH- /total volume,L
• M (.001 - .0003) / (.05 L .01 L) 1.2 x
  10-2 M H3O
• pH - log1.2 x 10-2 1.92

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3. At the equivalence point

• At the equivalence point the pH 7
• To find the volume of NaOH would give you the
  equivalence point use equation
• MV MV .02(50 mL) .03(x mL)
• 33.3 mL NaOH

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4. After the equivalence point

• After the equivalence point you have neutralized
  all of the acid and now you have excess base, 45
  mL NaOH
• Find the moles of acid base
• Moles of acid MV .001 Moles H3O
• Moles of base MV .03 .045 L .00135
  moles OH-

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4. After the equivalence point

• OH- (moles of base moles of acid)/ total
  volume, L
• (.00135 - .001) / (.05 L .045 L) 3.7 x
  10-3 M
• Kw H3O OH-
• H3O 1 x 10-14 / 3.7 x 10-3 2.7 x 10-12
•  
• pH -log 2.7 x 10-12 11.57

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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 Ca2F-2

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Example 7 Measuring Ksp and Calculating
Solubility from Ksp

• 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

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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

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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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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 Ca2F-2
•  
• If IP gt Ksp solution is supersaturated and
  precipitation will occur
• If IP Ksp the solution is saturated and
  equilibrium exists
• If IPlt Ksp the solution is unsaturated and ppt
  will not occur

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Example 8

• 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