1. Chapter 16 Lecture 2 Titrations
Strong Acid—Strong Base Titrations
Titration Basics
Titration = addition of a measurable volume of a known solution (titrant) to an unknown solution until it is just consumed
Use the stoichiometry of the reaction of the known and unknown to calculate the concentration of the unknown solution
A pH curve shows the change in pH versus volume of titrant as the titration proceeds
pH meter can be used to monitor
pH during the titration
An acid-base indicator can be used
to signal reaching the equivalence point

2. The unit mmol
Titrations often involve small amounts of a compound. Rather than working with small numbers of moles, the millimole unit can be used.
Since 1 ml = 1 x 10-3 L, Molarity can be expressed either as:
Titration of a strong acid with a strong base
Net reaction = H OH H2O
Example: HNO3(aq) NaOH(aq) Na NO H2O
Let’s titrate 50.0 ml of M HNO3 with M NaOH.
0.0 ml of NaOH added: pH = -log(0.2 M) = 0.699

3. 10.0 ml of NaOH added:
H OH H2O
Initial 10 mmol 1 mmol
Equilibrium 9 mmol 0 mmol
[H+] = 9 mmol / 60 ml = 0.15 M pH = 0.82
20.0 ml NaOH added:
Initial 10 mmol 2 mmol
Equilibrium 8 mmol 0 mmol
[H+] = 8 mmol / 70 ml = 0.11 M pH = 0.94
50.0 ml NaOH added:
Initial 10 mmol 5 mmol
Equilibrium 5 mmol 0 mmol
[H+] = 5 mmol / 100 ml = 0.05 M pH = 1.30

4. e) 100.0 ml of NaOH added: Equivalence Point
H OH H2O
Initial 10 mmol 10 mmol
Equilibrium 0 mmol 0 mmol
[H+] = 1 x 10-7 M from water pH = 7.00
f) ml NaOH added:
Initial 10 mmol 15 mmol
Equilibrium 0 mmol 5 mmol
[OH-] = 5 mmol / 200 ml = M pOH = 1.60 pH = 12.40
g) ml NaOH added:
Initial 10 mmol 20 mmol
Equilibrium 0 mmol 10 mmol
[OH-] = 10 mmol / 250 ml = 0.04 M pOH = 1.40 pH = 12.60

5. Important points:
pH increases slowly far from the equivalence point
pH changes quickly near the equivalence point
The equivalence point of a strong acid—strong base titration = 7.00
5) The titration of a strong base with a strong acid is almost identical

6. Titration of a Weak Acid with a Strong Base
Addition of a strong base to a weak acid forms a Buffer Solution
HA OH A H2O
If not enough base has been added to complete the reaction: HA/A- buffer
Problems can be worked either as equilibrium or buffer problems
Either way, you must consider the stoichiometry of the reaction first, then consider the equilibrium or buffer problem
Example: Titrate 50.0 ml of 0.10 M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10 M NaOH
0.0 ml NaOH added: typical weak acid problem
HA H A-
Initial
Equilibrium 0.1 – x x x
2) 10.0 ml NaOH added: stoichiometry first, then buffer problem
OH HA A H2O
Initial 1 mmol 5 mmol
Equilibrium mmol 1 mmol
pH = pKa log([A-]/[HA]) = log(1/4) = 4.14

7. 3) 25.0 ml NaOH added: stoichiometry first, then buffer problem
OH HA A H2O
Initial 2.5 mmol 5 mmol
Equilibrium mmol 2.5 mmol
pH = log(2.5/2.5) = 4.74
When HA = A-, we are half-way to the equivalence point. pH = pKa
Titrations are a very precise way of finding pKa’s of acids and bases.
4) 40.0 ml NaOH added: stoichiometry first, then buffer problem
OH HA A H2O
Initial 4 mmol 5 mmol
Equilibrium mmol 4 mmol
pH = log(4/1) = 5.35
5) 50.0 ml NaOH added: stoichiometry first, then weak base problem
Initial mmol mmol
Equilibrium mmol mmol
We are now at the equivalence point

8. A H2O HA OH-
Initial M
Equilibrium x x x
The major species at the equilibrium point of a weak acid titration, is its conjugate base. Thus, the pH at the equivalence point > 7.00.
(The major species at the equilibrium point of a weak base titration, is its conjugate acid. Thus, the pH at the equivalence point < 7.00)
60.0 ml NaOH added: stoichiometry first, then strong base problem
OH HA A H2O
Initial 6 mmol 5 mmol
Equilibrium 1 mmol mmol
pOH = -log(1mmol/110ml) = -log(9.1 x 10-3 M) = 2.04 pH = 11.96

9. 7) 75.0 ml NaOH added: stoichiometry first, then strong base problem
OH HA A H2O
Initial 7.5 mmol 5 mmol
Equilibrium 2.5 mmol mmol
pOH = -log(2.5mmol/125ml) = -log(0.02 M) = 1.70 pH = 12.30
Important Points
pH increases more rapidly at the start than for a strong acid
pH levels off near pKa due to HA/A- buffering effect
Curve is steepest near equivalence point. Equivalence Point > 7.0
Curve is similar to strong acid—strong base after eq. pt. where OH- is major

10. Example: Titrate 50.0 ml 0.1 M HCN (Ka = 6.2 x 10-10) with 0.10 M NaOH
pH after 8 ml NaOH added
pH at halfway point
pH at equivalence point
Conclusions about weak acid titrations
The same amount of base is required
to reach the equivalence point of any acid.
It is not the Ka that matters, it is how much
acid is there.
2) pH at the equivalence point varies
depending on the Kb of the conjugate base
of the weak acid. The weaker the acid, the
higher the pH at the eq. pt.
3) Ka effects the shape of the titration
curve, but only before the eq. pt.

11. Titration of a Weak Base with a Strong Acid
Example: Find the Ka of an unknown acid. 2 mmol of the acid in 100 ml of water was titrated with 0.05 M NaOH. After 20 ml NaOH was added, the pH = (Do stoichiometry, then buffer problem)
Titration of a Weak Base with a Strong Acid
Similar problem to the titration of a weak acid with a strong base
Determine major species from the stoichiometry
Calculate pH from weak acid, buffer, or weak base accordingly
Example: Titrate 100 ml of 0.10 M NH3 (Kb = 1.8 x 10-5) with 0.1 M HCl.
No HCl added: weak base pH calculation
HCl added, but before the equivalence point:
HCl will react to completion with NH3
Major species: NH3, NH4+
Buffer problem
At the equivalence point
All NH3 has been converted to NH4+
Weak acid problem, find Ka from Kb/KW
Past the equivalence point:
Excess HCl, is a strong acid
Weak acid NH4+ won’t effect pH
Strong acid problem

12. Titrations of Polyprotic Acids and Bases
Multiple Inflection Points = Multiple Equivalence Points will be seen
The volume required to reach each equivalence point will be the same
H2SO H HSO Ka1 = 1.6 x 10-2
HSO H SO Ka2 = 6.4 x 10-8

13. Acid-Base Indicators Finding the equivalence point of a titration
Use a pH meter
Plot pH versus titrant volume
Center vertical region = equivalence point
Use an Acid-Base Indicator
Acid-Base Indicator = molecule that changes color based on pH
Choose an indicator that changes color at the equivalence point
End Point = when the indicator changes color. If you have chosen the wrong indicator, the end point will be different than the eq. pt.
Indicators are often Weak Acids that lose a proton (causing the color change) when [OH-] reaches a certain concentration
HIn OH In H2O

14. Examine the behavior of a Hypothetical Indicator (Ka = 1 x 10-8)
HIn H In-
red blue
From the Ka equation, we can develop a formula showing how color behaves depending on the pH
The ratio of HIn and In- will determine the color
The ratio is a function of pH and a constant
We should be able to predict the color at different pH’s
If we add a few drops of HIn to a pH = 1.0 solution: a)
HIn is the dominant species so the color will be red
If we add OH- to the solution, eventually [In-] = [HIn]
The color will be equal parts red and blue = purple

15. Usually, about 10% of the HIn must react before our eyes can detect a color change.
Example: What is the pH at the color change of Bromthymol Blue (Ka = 1 x 10-7, HIn = yellow, In- = blue) starting in a strongly acidic solution?
We can use the Henderson-Hasselbalch equation on Indicators as well
pH = pKa log([In-]/[HIn])
pH = pKa log(1/10) for a color change
log(1/10) = -1
pH for color change starting in acid is always pKa – 1 for any Indicator
For a basic solution titrated with acid, [In-]/[HIn] = 10/1 for color change
Log(10/1) = +1, pH for color change will equal pKa + 1
Useful range for a pH Indicator is always pKa +/- 1

16. 4) The useful range and colors of common Indicators:

17. Choosing an Indicator
We always want the end point (color change) to coincide with the eq. pt.
Strong Acid—Strong Base titrations have a large pH change at the equivalence point, so you get sharp color change.
What indicator would work best for HCl/NaOH titration?
Eq. Pt. = 7.00
We will be going from acidic to basic, so we want [In-]/[HIn] = 1/10
pH = pKa log(1/10) = pKa – 1
Since we know the pH at eq. pt. = 7.00, pKa of the indicator we need should be 8.00 (Ka = 1 x 10-8)
If we were going from basic to acidic, we would want pKa = 6.00
How crucial is titrating exactly to the equivalence point?
NaOH = 99.9 ml pH = 5.3
NaOH = ml pH = 7.0
NaOH = ml pH = 8.7
Wide range of Indicators would tell us the eq. pt. to within 0.1 ml

18. Indicators for weak acid titrations
The weaker the acid, the smaller the pH change at the eq. pt.
We have less flexibility in choosing our indicator
0.1 M HC2H3O2 titrated with 0.1 M NaOH has pH = 8.7 at eq. pt.
Phenolphthalein ok (8.0—10.0 endpoints)
Thymol Blue ok (8.0—9.2 endpoints)
Methyl Red in not ok (4.2—6.2 endpoints)
HCl titration
HC2H3O2 titration