CHM 122
Chapter 15 - Applications of Aqueous Equilibria
Neutralization: occurs when aqueous solutions of an acid and a base are mixed in the proper proportion during
titration, which a reaction takes place causing acidic and basic properties of the solutions to disappear. The
hydrogen ion, which is responsible for the acidic properties, has reacted with the hydroxide ion, which is
responsible for the basic properties thus producing neutral water.
General neutralization
Acid + Base  Water + Salt
E.g
HCl(aq) + NaOH
 H2O(l) + NaCl
Types of neutralization
Strong acid and strong base
Molecular equation:
Net-ionic equation :
NaOH(aq) 
HCl(aq) +
+
H aq) +
OH(aq) 
-
H2O(l) + NaCl(aq)
HF(aq) +
NaOH(aq)
Net-ionic equation:
HF(aq) +
-
pH = 7.00
H2O(l)
Weak acid and strong base
Molecular equation:
Kn = 1/Kw
 H2O(l)
OH(aq) 
+
H2O(l) +
NaF(aq)
Kn = Ka x (1/Kw)
pH > 7.00
F-(aq)
Weak base and strong acid
Molecular equation:
NH3(aq) +
HCl(aq)  NH4Cl(aq)
Net-ionic equation:
NH3(aq) +
H+(aq)
 NH4+(aq)
Kn = Kb x (1/Kw)
pH < 7.00
Weak acid and weak base
Molecular equation: HCN(aq) +
Net-ionic equation:
HCN(aq) +
NH3(aq)
NH3(aq)
 NH4CN(aq)
 NH4+(aq) + CN-(aq)
Dang1
Kn = Ka x Kb x (1/Kw)
pH < ???
The Common-Ion Effect
The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the
ions already involved in the equilibrium. In other words, adding a common ion to solution decreases dissociation
of acid or base.
Example: A 1.0 M solution of HF at a pH of 1.58 Ka(HF) = 6.8 x 10-4. How does the pH of the solution change
when 42.0 g of NaF is added to the solution? Assume the volume remains constant at 2.0 L.
First consider what happens to equilibrium when F- is added?
HF (aq) + H2O(l)
H3O+ (aq) + F- (aq)
- Equilibrium will shift to the left.
- H3O+ will decrease, (increase pH).
15.3 Buffer Solutions
A solution that resists changes in pH when a small amount of acid or base is added. The best buffer systems
consist of either
a) a weak acid and a salt containing its conjugate base (e.g. HC2H3O2 and NaC2H3O2);
b) a weak base and a salt containing its conjugate acid (e.g. NH3 and NH4Cl).
Why?
While a weak acid will partially ionize to produce its conjugate base, it will not produce enough conjugate base to be
considered a buffer. (The same problem occurs for a weak base and its conjugate acid).
For these buffer systems:
1) The acid component of the buffer can neutralize added base and the base component of the buffer can neutralize
added acid.
2) Since they are a conjugate acid-base pair, the acid and base in the buffer don’t react with one another.
Buffers are important in Biochemistry because many of the enzymes that make your body run are designed to work
at one particular pH, if the solution doesn’t have the right pH things go wrong 
Organisms (and humans) have built-in buffers to protect them against changes in pH.
Human blood is maintained by a combination of CO3-2, PO4-3 and protein buffers.
Blood: (pH 7.4)
Death = 7.0 <pH > 7.8 = Death
Buffer Problems Involving Addition of Acid or Base
Stoichiometric calculation for the acid-base neutralization reaction; assume the neutralization, reaction goes to
completion:

Add a small amount of strong base (-OH) to a buffer solution
◦
Acid (HA) component of solution neutralizes the added base
Dang2
Buffer Solutions

Add a small amount of strong acid (H3O+) to a buffer solution
◦
Base (A-) component of solution neutralizes the added acid
◦ For each of the neutralization reaction, write an equation and make a change to see how much acid and base are
left
◦Since buffers contain a weak acid/base pair, you can set up the problem using either a weak
◦acid ionization reaction and K or a weak base ionization reaction and K .
a
b
Example
pH of human blood (pH = 7.4) controlled by conjugated acid-base pairs (H2CO3/HCO3-). Write
an equation for this buffer mixture then neutralization equation for the following effects
◦
With addition of HCl
◦
With addition of NaOH
Example

Calculate the pH of 100.0 mL DI water

Calculate the new pH after adding 1.0 mL of 0.10M HCl to the above water solution. Is water a buffer
solution? Why?
Example

50.0 mL of 0.100 M HCl was added to a .100L buffer consisting of 0.025 moles of sodium acetate and
0.030 moles of acetic acid. What is the pH of the buffer before and after the addition of the acid? Ka
of acetic acid is 1.7 x 10-5. Assume the volume is constant
Dang3
Example
a.
Calculate the pH of the buffer that results from mixing 60.0mL of 0.250M HCHO2 and 15.0
mL of 0.500M NaCHO2.
Ka = 1.7 x 10-4
b. Calculate the pH after addition of 8.0 mL of 0.150 M NaOH.
Dang4
Assume volume is additive
c. Calculate the pH after addition of 16.0 mL of 0.150 M NaOH.
Assume volume is additive.
Buffer Capacity
is the amount of acid or base the buffer can neutralize before there is a significant change in pH. The buffer
capacity is a measure of the effectiveness of a buffer.





A measure of amount of acid or base that the solution can absorb without a significant change in
pH.
Depends on how many moles of weak acid and conjugated base are present.
Buffer capacity is greater when larger amounts of HA and A- are present.
pH will stay relatively constant as long as [HA] and [A-] are greater than the amount of acid or base added.
Buffers work best when [HA] and [A-] are approximately equal. For buffers to be effective, the ratio of
Base:Acid should be between 1:10 to 10:1.
 For an equal volume of solution: the more concentrated the solution, the greater buffer capacity
 For an equal concentration: the greater the volume, the greater the buffer capacity
Example

The following pictures represent solutions that contain a weak acid HA and/ or its sodium salt NaA.
(Na+ ions and solvent water molecules have been omitted for clarity
Dang5

Which of the solutions are buffer
solution?

Which solution has the greatest
buffer capacity?
Example

What is the maximum amount of acid that can be added to a buffer made by the mixing of 0.35
moles of sodium hydrogen carbonate with 0.50 moles of sodium carbonate? How much base can be
added before the pH will begin to show a significant change?
15.4 The Henderson-Hasselbalch Equation
Example: Calculate the pH of a buffer solution that is 0.50 M in benzoic acid (HC7H5O2) and 0.150
M in sodium benzoate (NaC7H5O2). Ka = 6.5 x 10-5
Example: How would you prepare a NaHCO3-Na2CO3 buffer solution that has pH = 10.40
10-11
Example
Ka2 = 4.7 x
You prepare a buffer solution of .323 M NH3 and (NH4)2SO4. What molarity of (NH4)2SO4 is
necessary to have a pH of 8.6? (pKb NH3= 4.74)
Dang6