Buffers & Titrations
Chapter 18
1
Buffers
• A soln that resists change in pH when
strong acid or strong base is added
• Made from weak acid/conjugate-base
• Or
• Made from weak base/conjugate-acid
• Strong acids and strong bases don’t make
buffers!
• So how does it work?
2
How buffers work
• Acetic acid/acetate buffer system:
C2H3O2H(aq) + H2O(l)  C2H3O2-(aq) + H3O+(l)
• Add base:
C2H3O2H(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)
• Add acid:
H3O+(aq) + C2H3O2-(aq)  H2O(l) +C2H3O2H(aq)
3
Common Ion Effect: CIE
• The ionization of an acid or a base is limited by
•
the presence of its conjugate base or acid.
HAc(aq) + H2O(l)  Ac-(aq) + H3O+(aq)
Acetate ion is added in form of NaAc
– Which way will this shift the rxn?
– Would you expect a greater or lesser acidity if the CIE
was lacking?
• Let’s look at the next problem
4
Calculating pH of a buffer soln
• You have an acetic acid/acetate buffer
with a 0.700 M conc of acetic acid and a
0.600 M conc of acetate ion. What’s the
pH of the buffer? (Ka = 1.8 x 10-5)
5
Solution
HAc(aq) + H 2O(l)  Ac-(aq) + H 3O + (aq)
I 0.700M -C -x
-E 0.700 - x --
0.600
0
+x
+x
0.600+x x
[0.600+x][x]
K a = 1.8 x 10-5 
[0.700  x]
CIE: addition or subtraction of "x" from original concentrations very small

[0.600][x]
[0.700]
x=2.110-5 M
pH=-log(2.1 10-5 )=4.68
6
Let’s work on this
• Consider 100.0 mL of a buffer solution
that is 1.00M in HAc and 1.00M in NaAc.
What is the pH after addition of 25.0 mL
of 1.00M NaOH?
7
Solution
HAc(aq) + H 2O (l)  Ac- (aq) + H 3O + (aq)
I 1.00M -C -x
-E 1.00 - x --
1.00
0
+x
+x
1.00 + x x
[1.00+x][x] [1.00][x]
K a = 1.8 x 10-5 

[1.00  x]
[1.00]
x=1.8 x 10-5 M
Thus, [HAc]  1.00M  1.8 x 10-5 M =1.00M
And [Ac- ]=1.00M +1.8 x 10-5 M=1.00M
(By the way, the pH=4.74)
HAc(aq)  OH -(aq)  Ac-(aq) + H 2 O(l)
mol
1
 0.1000L acid 
 0.800M
L acid
0.1250L total
mol
1
base: 1.00
 0.0250L 
 0.200M
L
0.1250L
mol
1
acetate ion: 1.00
 0.1000L 
 0.800M
L
0.1250L
I 0.800M 0.200 0.800
-C -0.200 -0.200 +0.200 -E 0.600
0
1.000
[1.000][x]
So K a = 1.8 x 10-5 
, x = [H 3O + ]  1.1 x 10-5
[0.600]
acid: 1.00
pH = -log(1.1 x 10-5 )  4.96
8
Henderson-Hasselbalch equation
• Useful for previous problem
– Let’s take a look
[H 3O + ][A - ]
Ka =
[HA]
[HA]
Re-arrange: [H 3O ]= -  K a
[A ]
Take the negative log
+
[A - ]
pH = pKa + log(
)
[HA]
9
Titrations
• Used to determine quantity of acid or base in
unknown (analyte)
10
• When sample is neutralized (H3O+ = OH-)
– Equivalence point
• Determined by use of pH indicator
11
Different titration types
• Strong acid-strong base: equivalence pt =
7.0 (contains neutral salt)
12
Different titration types
• Weak acid-strong base: equivalence pt is
greater than 7 (basic salt)
• pH @ half-equivalence pt (halfway pt) =
pKa
13
Problem
• We start with 50.0 mL of 0.100 M HAc.
25.0 mL of 0.100 M NaOH is then added.
What is the pH of the resulting solution?
14
Solution
HAc(aq)  OH -(aq)  Ac-(aq) + H 2 O(l)
mol
1
acid: 0.100
 0.0500L 
 0.0667M
L
0.0750L
mol
1
base: 0.100
 0.0250L 
 0.0333M
L
0.0750L
I 0.0667 0.0333
0
-C -0.0333 -0.0333 +0.0333 -E 0.0334
0
0.0333
--
[A - ]
[0.0333]
-5
pH = pKa + log(
)= -log(1.8 x 10 )  log(
)  4.74
[HA]
[0.0334]
15