Chapter 16: Aqueous Ionic Equilibria I
Buffers
Chem 102
Dr. Curtis
Buffers
• a buffer is a solution that resists a change in pH when
an acid or base is added to it
• A buffer contains one of the following scenarios:
• Significant amounts of a weak acid and its
conjugate base
• or
• Significant amounts of a weak base and its
conjugate acid
2
Identifying a buffers
• Which solution is a buffer?
• 0.100 M HNO2 and 0.100 M HCl
• 0.100 M HNO3 and 0.100 M NaNO3
• 0.100 M HNO2 and 0.100 M NaCl
• 0.100 M HNO2 and 0.100 M NaNO2
3
Acid-Base reactions
• The key to buffers is to understand that:
• Strong acids react fully with weak bases (to form H2O)
• Strong bases react fully with weak acids (to form H2O)
• Example
HA(aq) + OH (aq) ! H2 O(`) + A (aq)
• if you had 0.5 moles of acetic acid and react with 0.1
moles of NaOH, you will get 0.1 moles of C2H3O2-(aq)
• the weak acid partially neutralizes the NaOH added
4
Acid-Base reactions
• The key to buffers is to understand that:
• Strong acids react fully with weak bases (to form H2O)
• Strong bases react fully with weak acids (to form H2O)
• Example
A (aq) + H3 O+ (aq) ! HA(aq) + H2 O(`)
• if you had 0.5 moles of acetate and react with 0.1 moles
of HCl, you will get 0.1 moles of acetic acid
• the weak base partially neutralizes the strong acid
added
5
Buffers have both
• If you have a solution with both a weak acid and its
conjugate base, it will neutralize either a strong base or
a strong acid
HC2 H3 O2 (aq) + C2 H3 O2 (aq)
HC2 H3 O2 (aq) + NaOH(aq) ! H2 O(`) + NaC2 H3 O2 (aq)
NaC2 H3 O2 (aq) + HCl(aq) ! HC2 H3 O2 (aq) + NaCl(aq)
6
BAA table
• Use moles (not molarity)
• Tracks the stoichiometry of the reaction
• Not an ICE table
7
Calculate the pH of a buffer (before perturbation)
• pH of a solution that is 0.100 M HC2H3O2 and 0.100 M
NaC2H3O2 (Ka = 1.8 x 10-5)
CH3COOH(aq)
H2O(l)
H3O+(aq)
CH3COO-(aq)
I
0.100 M
-
~0 M
0.100 M
C
-x
-
+x
+x
E
0.100 - x
-
x
0.100 + x
⇌
8
Henderson-Hasselbalch Equation
HA(aq) + H2 O(`) )* H3 O+ (aq) + A (aq)
[H3 O+ ][A ]
[HA]
+
Ka =
) [H3 O ] = Ka
[HA]
[A ]
• [H3O+] is proportional to the ratio of [HA] / [A-]
• Take the log of both sides to get the HendersonHasselbalch equation:
✓
◆
[A ]
pH = pKa + log
[HA]
9
Henderson-Hasselbalch Equation
pH = pKa + log
✓
[A ]
[HA]
◆
• Assumes x is small compared to initial acid
concentration
• [HA] should be 100 to 1000 times larger than Ka
• Note: log(1) = 1, so if [base] = [acid], pH = pKa for a
buffer
• Note: The base and acid are in the same solution, so
you can use moles instead of molarity (the volume
cancels)
10
Example
• Calculate the pH of a solution that is 0.195 M HC2H3O2
and 0.125 M NaC2H3O2 (Ka = 1.8 x 10-5).
• Solve with an ICE chart and with HendersonHasselbalch equation
11
Acid/Base reactions in a buffer
• Two-step process
• Understand the reaction stoichiometry first
• Then establish a new equilibrium
• BAA table (reaction table, or BRA chart)
• Before addition, Addition, After addition
• The key is that a strong acid will react with the weak
base in a buffer or
• a strong base will react with the weak acid in a buffer
12
Perturbation of a buffer
• Two-step process
• Understand the reaction stoichiometry first
• Then establish a new equilibrium
• If you add acid to a buffer, the pH must decrease slightly
• If you add base to a buffer, the pH must increase slightly
• Adding strong acid removes some of the conjugate base
• Adding strong base removes some of the weak acid
• Then a new equilibrium is established
13
Perturbation of a buffer
• Two-step process
• Understand the reaction stoichiometry first
• Then establish a new equilibrium
• Example: Say you have a buffer made with the generic
weak acid HA and its conjugate
• pKa = 4.82 (Ka = 1.5 x 10-5)
• 1.0 L total volume
• 0.100 M HA
• 0.100 M A14
Calculate the pH of the buffer
• pKa = 4.82 (Ka = 1.5 x 10-5)
• 1.0 L total volume
• 0.100 M HA (0.100 moles)
• 0.100 M A- (0.100 moles)
pH = pKa + log
✓
[A ]
[HA]
✓
◆
= pKa + 0 = 4.82
pH = pKa + log
0.100M
0.100M
◆
15
Perturb the buffer
• Add 0.025 moles of strong acid (assume no volume
change)
• It will react completely
A-(aq)
Before
addition
0.100
moles
HA(aq)
H2O(l)
0.100
moles
0.025
moles
Addition
After
addition
H+(aq) →
0.075
moles
0
moles
0.125
moles
16
Then establish a new equilibrium
HA(aq)
H2O(l) ⇌
H3O+(aq)
A-(aq)
I
0.125 M
-
~0 M
0.075 M
C
-x
-
+x
+x
E
0.125 - x
-
x
0.075 + x
x(0.075 + x)
0.075x
Ka = 1.5 ⇥ 10 =
⇡
0.125
✓ 0.125◆ x
0.125
+
x = [H3 O ] = Ka
= 2.5 ⇥ 10 5 M
0.075
5
pH = 4.60
17
Then establish a new equilibrium
HA(aq)
H2O(l) ⇌
H3O+(aq)
A-(aq)
I
0.125 M
-
~0 M
0.075 M
C
-x
-
+x
+x
E
0.125 - x
-
x
0.075 + x
• Or use Henderson-Hasselbalch equation
pH = pKa + log
✓
[A ]
[HA]
◆
= pKa + log
✓
0.075 + x
0.125 x
◆
⇡ pKa + log
✓
0.075
0.125
◆
pH = 4.60
18
Phew! That was tough!
• Think about our answer
• We had a buffer and we added acid to it
• Expect pH to drop a little bit
• 4.82 to 4.60
• Or log of a number less than one must be negative, so
pH < pKa if you have more base than acid
• Remember: volume changes can usually be ignored
because the new volume cancels (can use moles in
equations)
19
Adding base to a buffer
• What if we added 0.025 moles of strong base to our
original buffer?
• It will react with the weak acid, producing more of the
conjugate base
• Establish new equilibrium
• Answer: pH will increase slightly
• pH = 5.04
20
21
Buffers from a weak base and conjugate acid
• Henderson-Hasselbalch equation still applies!
• Example: 0.05 M NH3 and 0.04 M NH4Cl
• For NH3: pKb = 4.75 (pKa for NH4+ = 9.25)
pH = pKa + log
pH = pKa + log
✓
[N H3 ]
[N H4+ ]
◆
✓
[N H3 ]
[N H4+ ]
◆
= 9.25 + log
✓
0.05 M
0.04 M
◆
pH = 9.35
22
Buffer effectiveness
• Buffers resist pH change best when [acid] = [base]
• e.g. 0.1 M HA with 0.1 M A- would be a “better” buffer
than 0.1 M HA with 0.05 M A• [acid] and [base] must be within factor of 10 to have an
effective buffer
• pH = pKa + log(10) through pH = pKa + log(0.1)
• pH = pKa ± 1
23
Buffer effectiveness
• Buffers best resist change to pH when [acid] and [base]
are relatively high
• Example 0.1 M HA with 0.1 M A- would be a “better”
buffer than 0.001 M HA with 0.001 M A• acid or base can be used up more quickly the less
there is
24
Buffer capacity
• Amount of acid or base you can add to a buffer without
causing a large change in pH
• Buffer capacity increases when [acid] is close to
[base]
• Buffer capacity increases when [acid] and [base] are
high
25
Making a buffer
• Make a weak acid solution with a salt of the conjugate
base
• 0.100 M HNO2 and 0.100 M NaNO2
• Or: Start with a weak acid and add enough strong base
to create the conjugate base in solution
• Add 100 mL of 0.1 M NaOH to 100 mL of 0.2 M HNO2
• Or: Start with a weak base and add enough strong acid
to create the conjugate acid in solution
26
BAA chart - use moles
HNO2(aq)
OH-(aq)
Before
0.02 moles
addition
Addition
After
0.01 moles
addition
→
NO2-(aq)
H2O(l)
0
0.01
moles
0.01
moles
27
Making a buffer
• Add 100 mL of 0.1 M NaOH to 100 mL of 0.2 M HNO2
• Start: (0.100 L)(0.2 mol/L) = 0.02 mol HNO2
• Add: (0.100 L)(0.1 mol/L) = 0.01 mol OH- added
• Reaction goes to completion:
HNO2 (aq) + OH (aq)
! NO2 (aq) + H2 O(`)
• End up with 0.01 mol HNO2 and 0.01 mol NO2- in a
volume of 200 mL
• Equal concentrations of a weak acid and its conjugate
base
28
Making a buffer with a specific pH
• Which acid and its conjugate base (as a sodium salt)
would you choose to create a buffer with pH = 7.35?
• What would be the ideal molar ratio of acid to salt?
• HClO2 (pKa = 1.95)
• HNO2 (pKa = 3.34)
• HCOOH (pKa = 3.74)
• HClO (pKa = 7.54)
29