Chapter 18 – Other
Aspects of Aqueous
Equilibria
Objectives:
1. Apply the common ion effect.
2. Describe the control of pH in aqueous solutions
with buffers.
3. Evaluate the pH in the course of acid-base
titrations.
4. Apply equilibrium concepts to the solubility of
ionic compounds.
The Common Ion Effect
• Add Lactate ion – effect?
• Le Chatelier’s Principle: Lactic acid will
__________________________.
• The ionization of an acid or a base is limited by
the presence of its ______________________.
Preparation of Buffer
As shown by an
universal indicator:
Acetic acid
-Acidic solution
Sodium acetate
-Basic solution
Mixing equal amounts
of acetic acid and
sodium acetate:
- Solution with lower
hydronium ion
concentration than
acetic acid.
Assume you have a 0.30 M solution of formic acid (HCO2H) and
have added enough sodium formate (NaHCO2) to make the solution
0.10 M in the salt. Calculate the pH of the formic acid solution
before and after adding sodium formate.
HCO2H (aq) + H2O (l)
H3O+ (aq) + HCO2-(aq)
Buffer
• A buffer causes solutions to resist a change
in pH when a strong acid or base is added.
• Two requirements:
-
• Buffer is usually prepared from a conjugate
acid-base pair:
-
Common Buffers
Buffer
• Acetic acid / acetate ion buffer:
• Acetic acid, the weak acid, is needed to consume any
added hydroxide ion:
CH3CO2H(aq) + OH- CH3CO2-(aq) + H2O (l) K = 1.8 x 109
• K is very large – any OH- added will be consumed
• Acetate ion, the conjugate base, will consume any added
hydronium ion:
CH3CO2- + H3O+
CH3CO2H (aq) + H2O (l) K = 5.6 x 104
• K is also large, because H3O+ is a strong acid.
What is the pH of a buffer solution composed of 0.50 M
formic acid (HCO2H) and 0.70 M sodium formate
(NaHCO2)?
HCO2H (aq) + H2O (l)
H3O+ (aq) + HCO2-(aq)
Expressions for Buffer Solutions
• Let’s rearrange:
Ka = [H3O+][HCO2-] = 1.8 x10-4
[HCO2H]
+
[H3O+] = [HCO2H]
[HCO2-]
x Ka
[H3O+] = [acid]
x Ka
[conjugate base]
pH =
[H3O ] is given by the ratio of the acid and
conjugate base concentrations multiplied by
the acid ionization constant. This is true
for all buffers from weak acid and its
conjugate base.
Apply –log to each side of the equation:
pKa + log [conjugate base]
[acid]
Henderson-Hasselbalch Equation
pH =
pKa + log [conjugate base]
[acid]
• The resulting pH of the buffer is determined
primarily by ____________________(or _____)
and it is adjusted by varying the _______________
ratio (relative number of moles).
• Diluting a buffer will _____________its pH.
• When the concentrations of acid and conjugate base
are the same, the log of ________ = _________
• If acid = conjugate base ; pH ________
• If conjugate base > acid ; pH _________
• If acid > conjugate base ; pH _________
What is the pH of a buffer solution composed of 0.50 M
formic acid (HCO2H) and 0.70 M sodium formate
(NaHCO2)?
pH =
pKa + log [conjugate base]
[acid]
HCO2H (aq) + H2O (l)
H3O+ (aq) + HCO2-(aq)
Buffer
• To be useful:
– pH control: choose a weak acid with ____
___________________________
– Buffer capacity: Concentration of buffer
should be high enough to ____________
______________________. Buffers are
usually prepared as ________________
solutions.
Describe how to prepare a buffer solution from
NaH2PO4 and Na2HPO4 to have a pH of 7.5.
Benzoic acid (C6H5CO2H, 2.0 g) and sodium benzoate (C6H5CO2-, 2.0 g)
are dissolved in enough water to make 1.0 L of solution. Calculate pH
of the solution using the Henderson-Hasselbalch equation.
Calculate the pH of 0.5 L of buffer solution composed
of 0.50 M formic acid and 0.70 M sodium formate
before and after adding 10.0 mL of 1.0 M HCl.
1) Find the amount of acid formed when HCl reacts with conjugate base.
2) Calculate [H3O+] for the buffer
3) Convert [H3O+] to pH
HCO2H (aq) + H2O (l)
H3O+ (aq) + HCO2-(aq)
Biological Buffer Systems
• HPO42- / H2PO4• HCO3-/H2CO3
• Read book p. 862 ed 6. (p. 822 ed 7)
Titration and pH
Titration of a Strong Acid
with a Strong Base
• Initial pH (depending on strong
acid concentration) – very acidic
• As NaOH is added, pH increases
very slowly until the equivalence
point.
• Equivalence point – same number
of moles of acid and base
• H3O+ + OHH2O
• = neutral pH = _______
• As more NaOH is added, pH
becomes basic.
The pH of the equivalence point in an acidbase titration is the mid-point in the vertical
portion of the pH vs volume of titrant curve.
Strong-acid/strong-base titration : pH =____
Titration of a Weak Acid with
a Strong Base
• The initial pH depends on
the Ka of the weak acid and
the [acid].
• As NaOH is added the pH
increases and the conjugate
base of the acid is form.
• A buffer is generated!
• At halfway point of the
titration the
[acid]=[conjugate base]
• pH = __________
• At the equivalence point the
solution contains only the
conjugate base salt since all
acid and base reacted
producing the salt.
• At eq. pt: The pH depends on ______
of the conjugate base and [conjugate
base].
What is the pH of the solution when 35.0 mL of 0.1 M
NaOH has been added to 100 mL of 0.1 M acetic acid?
1) Find the [acid] remaining and the [conjugate base] formed after adding NaOH.
2) Find pH for the buffer generated.
CH3CO2H (aq) + H2O (l)
CH3CO2H (aq) + OH- (aq)
H3O+ (aq) + CH3CO2-(aq)
H2O (aq) + CH3CO2- (aq)
Titration of a Weak Base with a
Strong Acid
• The initial pH depends on the
Kb of the weak base and the
[base].
• As HCl is added the pH
decreases and the conjugate
acid of the base is form.
• A buffer is generated!
• At halfway point of the
titration the
[base]=[conjugate acid]
• pH = _________________
• At the equivalence point the
solution contains only the
conjugate acid salt since all
base and acid reacted
producing the salt.
• The solution is __________.
Titration of a Weak Diprotic
Acid with a Strong Base
• Curve shows two
inflection points.
• First produces a
weak base
• Second produces a
stronger base.
If you require 36.78 mL of 0.0105 M HCl to reach the equivalence
point in the titration of 25.0 mL of aqueous ammonia. What was the
concentration of NH3 in the original ammonia solution? What is the
pH of the solution at the equivalence point?
NH3 + H3O+
NH4+ + H2O
Indicators
• An organic compound that is itself a __________________.
• The acid form of the compound has one color and the conjugate
base another.
Indicators
• Indicator needs to be chosen so that it changes color
at a pH closer to the anticipated equivalence point.
Solubility of Salts
•
•
•
•
Precipitation reaction:
CaCl2(aq) + Na2CO3 (aq)
CaCO3(s) + 2 NaCl (aq)
Insoluble salt in solvent:
AgBr(s)
Ag+ + BrKsp – Solubility product constant
Is an equilibrium constant
K = [Ag+][Br-] = Ksp
-When the product of the
Ksp =
concentrations is larger than Ksp
the salt will precipitate!
Solubility – quantity present in
some volume of a saturated
solution (gr/L, etc).
Insoluble Salts
Identify as soluble or insoluble:
Pb(NO3)2
Fe(OH)3
ZnCl2
CuS
ZnSO4
(NH4)2CO3
Calculate the Ksp value for CaF2 if the concentration of
calcium ions in solution (solubility) is 2.4 x 10-4 mol/L
CaF2
Ca2+ + 2 F-
Knowing the Ksp for MgF2 is 5.2 x 10-11, what is
the solubility in moles/L?
MgF2
Mg2+ + 2 F-
Common Ion Effect
Adding an ion “common” to an equilibrium
causes the equilibrium to
___________________________.
Insoluble Salts
Separation of Ions
What amount of Cl- is required to precipitate
the Hg2Cl2 from a Hg22+ solution of 0.01 M
concentration.
A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to
precipitate red Ag2CrO4 and yellow PbCrO4. Which
precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution: The substance whose Ksp is first exceeded
precipitates first.
Remember
• Go over all the contents of your
textbook.
• Practice with examples and with
problems at the end of the chapter.
• Practice with OWL tutor.
• Work on your OWL assignment for
Chapter 18.