Buffers and Titrations, Part 2
Ch. 21
Suggested HW: 14, 18, 30, 42, 44
Weak Acid/Base Chemistry Recap
• For a weak acid dissociation
HA aq + H2 O L ↔ H3 O+ aq + A− (aq)
H3 O+ [A− ]
Ka =
[HA]
• For the protonation of the conjugate base:
A− aq + H2 O L ↔ OH − aq + HA (aq)
OH − [HA]
Kb =
[A− ]
• Therefore:
H3 O+ [A− ] OH − [HA]
+ OH − = K = 𝟏𝟎−𝟏𝟒
Ka x Kb =
x
=
H
O
3
w
[HA]
[A− ]
Buffer Recap
• For a buffer (HA/A-)
pH = pKa + log
base
acid
where pKa = -log Ka
• As the base concentration goes up, the logarithm of the ratio of base
to acid also goes up, resulting in an increase in pH of the buffer.
The opposite is then true for the addition of acid.
Buffer Capacity and pH Range
• Two important characteristics of a buffer are its capacity and pH
range.
▫ Buffer capacity is the amount of acid or base that a buffer can
neutralize before the pH begins to change vastly.
 ex. A buffer of 0.1M HF/ 0.1M NaF has the same pH as 1.0 M HF/
1.0 M NaF, but less buffer capacity
• pH range is the range over which a buffer works effectively. In real
practice, the maximum buffer range is reached when one component is
10 times more concentrated than the other.
 𝑝𝐻 = 𝑝𝑘𝑎 + log
𝐵
𝐴
 for B:A ratio of 10:1 , 𝑝𝐻 = 𝑝𝐾𝑎 + 1
 for B:A ratio of 1: 10, 𝑝𝐻 = 𝑝𝐾𝑎 − 1
max buffer range =
pKa ± 1
• When designing a buffer, the pKa of the acid should be as close as
possible to the desired pH
Example: Selecting and Preparing a Buffer
• You work for a major pharmaceutical company that has developed a
new drug to combat sickle cell anemia. You have a fresh batch of
live cells to test this drug. The cells must be maintained in 100 mL
of a buffered solution at pH=7.4. The buffering capacity should be
high, so it is recommended that the concentration of the acid
component be 0.5M. Assuming that you have access to any acid and
base (non-ionic forms as well as sodium salts), what conjugate
acid/base pair would you use to create your buffer? What masses of
each would you dissolve into solution? Use the pka table on the next
slide.
Example contd.
• Potential options for a pH=7.4 buffer would be (acid/base):
▫ HClO/NaClO
 pKa =7.40 (buffer range of 6.40 to 8.40)
▫ NaH2PO4/Na2HPO4
 pKa = 7.21 (buffer range of 6.21 to 8.21)
Example contd.
• Using the hypochloric acid/ hypochlorite buffer with [HClO]=0.5 M :
[𝑁𝑎𝐶𝑙𝑂]
7.4 = 7.4 + log
[𝐻𝐶𝑙𝑂]
[𝑁𝑎𝐶𝑙𝑂]
0 = log
[𝐻𝐶𝑙𝑂]
100 =
[𝑁𝑎𝐶𝑙𝑂]
[𝐻𝐶𝑙𝑂]
[𝑁𝑎𝐶𝑙𝑂]
1=
[𝐻𝐶𝑙𝑂]
𝐻𝐶𝑙𝑂 = 0.5 M
𝑁𝑎𝐶𝑙𝑂 = 0.5 M
0.5 mol H2 CO3 52.45 g 𝐻𝐶𝑙𝑂
Acid: 0.100 L x
x
= 𝟐. 𝟔𝟐 𝐠
L
mol 𝐻𝐶𝑙𝑂
0.5 mol Na𝐶𝑙𝑂 74.45 g NaHCO3
Base: 0.100 L x
x
= 𝟑. 𝟕𝟐 𝐠
L
mol NaHCO3
Dissolve in
100 mL of
solution.
Titration Curves
• In a titration, one substance of
known concentration (e.g. base)
is added to another (e.g. acid). A
colored indicator or pH meter can
be used to monitor the reaction
progress.
• Using a pH meter, a titration
curve can be generated, showing
pH as a function of the volume of
titrant added
Titration Curves
• The shape of a titration curve makes it possible to determine the
equivalence point (point at which the moles of acid and base
are stoichiometrically equivalent)
• Titration curves can also be used to determine values of Ka or Kb if
weak acids/bases are being titrated
• To understand the characteristics of titration curves, we will
examine three types of acid/base titrations:
1. Strong Base into Strong Acid
2. Strong Base into Weak Acid
3. Strong Base into Polyprotic acid
Strong Acid/Strong Base Titration Curves
(e.g. 0.1M NaOH added to 0.1M HCl)
1.
The general shape of this type of
titration curve is shown to the right.
2.
The initial pH is simply the pH of the
HCl(aq) before addition of NaOH(aq)
3.
At first, the pH rises slowly due to the
presence of un-neutralized acid.
4.
At the equivalence point, pH=7 and
the moles of acid and base are
stoichiometrically equal
5.
Beyond equivalence, pH rises rapidly
due to presence of free base
Group Example
• Calculate the pH when 49 mL of 0.1M NaOH is added to 50 mL of
0.1M HCl? (ans: pH = 3)
• Calculate the pH when 51 mL of 0.1M NaOH is added to 50 mL of
0.1M HCl? (ans: pH = 11)
Weak Acid/Strong Base (e.g. 0.1M NaOH
added to 0.1M CH3COOH with Indicator)
1. As the acid is neutralized, its
conjugate base forms, resulting
in the formation of a buffer
when [B]/[A] > 1/10.
2. At the midpoint of the
buffer region, pH = pKa;
[CH3COOH] = [CH3COO-]
3. At the equivalence point, all of
the acetic acid is now acetate,
which is a weak base. Thus,
the pH at the equivalence
point is greater than 7.
Group Example
• Given that Ka of CH3COOH is 1.8 x 10-5 M, calculate the pH when 49
mL of 0.1 M NaOH is added to 50 mL of 0.1M CH3COOH.
▫ First, calculate the moles of acetic acid remaining
𝐶𝐻3 𝐶𝑂𝑂𝐻 𝑎𝑞 + 𝑂𝐻 − (𝑎𝑞) ↔ 𝐶𝐻3 𝐶𝑂𝑂− (aq) + H2O
I 0.005 mol
0.0049 mol
0
C -.0049 mol -.0049 mol
+.0049 mol
E .0001 mol
0 mol
.0049 mol
• Now, plug the concentrations into the Ka expression to find [H3O+].
The total volume is 99 mL. Refer to slide 2 for clarification.
1.8 𝑥
10−5
𝐻3 𝑂+ [𝐶𝐻3 𝐶𝑂𝑂−]
𝐻3 𝑂+ [.0495𝑀]
=
=
[𝐶𝐻3 𝐶𝑂𝑂𝐻]
[.001𝑀]
𝐻3 𝑂+ = 3.64 𝑥 10−7 M
pH = 6.44
Example
• Calculate the pH at the equivalence point from the titration of 50
mL of 0.1M acetic acid by 50 mL of 0.1M NaOH as shown on slide
13. (ans: pH= 8.72)
• Hint!!! How do you calculate Kb?
Strong Base/Polyprotic Acid Titration Curves
(e.g. 0.1M NaOH added to 0.1M H3PO4)
From Last Image
• Initial point:
▫ pH of 0.1M H3PO4
• First midpoint:
▫ [H3PO4] = [H2PO4-], pH at 1st midpoint = pKa # 1
• First equivalence point:
▫ No H3PO4 remains.
• Second midpoint:
▫ [H2PO4-] = [HPO42-], pH at 1st midpoint = pKa # 2
• Second equivalence point:
▫ No H2PO4- remains
• Third midpoint:
▫ [HPO42-] = [PO43-] , pH at 1st midpoint = pKa # 3
• Third equivalence point:
▫ Only phosphate remains