Chapter 16 Lecture 2 Titrations

I. Strong Acid—Strong Base TitrationsA. Titration Basics

1) Titration = addition of a measurable volume of a known solution (titrant) to an unknown solution until it is just consumed

2) Use the stoichiometry of the reaction of the known and unknown to calculate the concentration of the unknown solution

3) A pH curve shows the change in pH versus volume of titrant as the titration proceeds

a) pH meter can be used to monitor

pH during the titration

b) An acid-base indicator can be used

to signal reaching the equivalence point

4) The unit mmol

a) Titrations often involve small amounts of a compound. Rather than working with small numbers of moles, the millimole unit can be used.

b)

c) Since 1 ml = 1 x 10-3 L, Molarity can be expressed either as:

B. Titration of a strong acid with a strong base

1. Net reaction = H+ + OH- H2O

2. Example: HNO3(aq) + NaOH(aq) Na+ + NO3- + H2O

3. Let’s titrate 50.0 ml of 0.200 M HNO3 with 0.100 M NaOH.

a) 0.0 ml of NaOH added: pH = -log(0.2 M) = 0.699

mol 10 x 1mol1000

1 mmol 1 3-

ml

mmol

ml 1000

mmol 1000

L

molM

b) 10.0 ml of NaOH added:

H+ + OH- H2OInitial 10 mmol 1 mmol -------Equilibrium 9 mmol 0 mmol -------

[H+] = 9 mmol / 60 ml = 0.15 M pH = 0.82

c) 20.0 ml NaOH added:

H+ + OH- H2OInitial 10 mmol 2 mmol -------Equilibrium 8 mmol 0 mmol -------

[H+] = 8 mmol / 70 ml = 0.11 M pH = 0.94

d) 50.0 ml NaOH added:

H+ + OH- H2OInitial 10 mmol 5 mmol -------Equilibrium 5 mmol 0 mmol -------

[H+] = 5 mmol / 100 ml = 0.05 M pH = 1.30

e) 100.0 ml of NaOH added: Equivalence Point

H+ + OH- H2OInitial 10 mmol 10 mmol -------Equilibrium 0 mmol 0 mmol -------

[H+] = 1 x 10-7 M from water pH = 7.00

f) 150.0 ml NaOH added:

H+ + OH- H2OInitial 10 mmol 15 mmol -------Equilibrium 0 mmol 5 mmol -------

[OH-] = 5 mmol / 200 ml = 0.025 M pOH = 1.60 pH = 12.40

g) 200.0 ml NaOH added:

H+ + OH- H2OInitial 10 mmol 20 mmol -------Equilibrium 0 mmol 10 mmol -------

[OH-] = 10 mmol / 250 ml = 0.04 M pOH = 1.40 pH = 12.60

4) Important points:

a) pH increases slowly far from the equivalence point

b) pH changes quickly near the equivalence point

c) 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

II. Titration of a Weak Acid with a Strong BaseA. Addition of a strong base to a weak acid forms a Buffer Solution

1) HA + OH- A- + H2O

2) If not enough base has been added to complete the reaction: HA/A- buffer

3) Problems can be worked either as equilibrium or buffer problems

4) Either way, you must consider the stoichiometry of the reaction first, then consider the equilibrium or buffer problem

B. Example: Titrate 50.0 ml of 0.10 M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10 M NaOH

1) 0.0 ml NaOH added: typical weak acid problem

HA H+ + A-

Initial 0.1 0 0

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 0 ------

Equilibrium 0 4 mmol 1 mmol ------

pH = pKa + log([A-]/[HA]) = 4.74 + log(1/4) = 4.14

2.87pH

0.00134x0.1

x10 x 1.8

25

3) 25.0 ml NaOH added: stoichiometry first, then buffer problem

OH- + HA A- + H2OInitial 2.5 mmol 5 mmol 0 ------Equilibrium 0 2.5 mmol 2.5 mmol ------

pH = 4.74 + 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- + H2OInitial 4 mmol 5 mmol 0 ------Equilibrium 0 1 mmol 4 mmol ------

pH = 4.74 + log(4/1) = 5.35

5) 50.0 ml NaOH added: stoichiometry first, then weak base problem

OH- + HA A- + H2OInitial 5 mmol 5 mmol 0 ------Equilibrium 0 0 mmol 5 mmol ------

We are now at the equivalence point

A- + H2O HA + OH-

Initial 0.05 M ----- 0 0

Equilibrium 0.05 - 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)

6) 60.0 ml NaOH added: stoichiometry first, then strong base problem

OH- + HA A- + H2O

Initial 6 mmol 5 mmol 0 ------

Equilibrium 1 mmol 0 5 mmol ------

pOH = -log(1mmol/110ml) = -log(9.1 x 10-3 M) = 2.04 pH = 11.96

ml 100

mmol 5

8.72pH5.28pOH10 x 5.27][OHx

0.05

x10 x 5.56

10 x 1.8

10 x 1

K

KK

6

210

5

14

a

Wb

7) 75.0 ml NaOH added: stoichiometry first, then strong base problem

OH- + HA A- + H2O

Initial 7.5 mmol 5 mmol 0 ------

Equilibrium 2.5 mmol 0 5 mmol ------

pOH = -log(2.5mmol/125ml) = -log(0.02 M) = 1.70 pH = 12.30

C. Important Points

1) pH increases more rapidly at the start than for a strong acid

2) pH levels off near pKa due to HA/A- buffering effect

3) Curve is steepest near equivalence point. Equivalence Point > 7.0

4) Curve is similar to strong acid—strong base after eq. pt. where OH- is major

D. Example: Titrate 50.0 ml 0.1 M HCN (Ka = 6.2 x 10-10) with 0.10 M NaOH

1) pH after 8 ml NaOH added

2) pH at halfway point

3) pH at equivalence point

E. Conclusions about weak acid titrations

1) 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.

F. 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 = 6.00. (Do stoichiometry, then buffer problem)

III. Titration of a Weak Base with a Strong AcidA. Similar problem to the titration of a weak acid with a strong base

1) Determine major species from the stoichiometry2) Calculate pH from weak acid, buffer, or weak base accordingly

B. Example: Titrate 100 ml of 0.10 M NH3 (Kb = 1.8 x 10-5) with 0.1 M HCl.1) No HCl added: weak base pH calculation2) HCl added, but before the equivalence point:

a) HCl will react to completion with NH3

b) Major species: NH3, NH4+

c) Buffer problem3) At the equivalence point

a) All NH3 has been converted to NH4+

b) Weak acid problem, find Ka from Kb/KW 4) Past the equivalence point:

a) Excess HCl, is a strong acid

b) Weak acid NH4+ won’t effect pH

c) Strong acid problem

Titrations of Polyprotic Acids and Bases

1. Multiple Inflection Points = Multiple Equivalence Points will be seen

2. The volume required to reach each equivalence point will be the same

H2SO3 H+ + HSO3- Ka1 = 1.6 x 10-2

HSO3- H+ + SO3

2- Ka2 = 6.4 x 10-8

IV. Acid-Base IndicatorsA. Finding the equivalence point of a titration

1) Use a pH meter

a) Plot pH versus titrant volume

b) Center vertical region = equivalence point

2) Use an Acid-Base Indicator

a) Acid-Base Indicator = molecule that changes color based on pH

b) Choose an indicator that changes color at the equivalence point

c) End Point = when the indicator changes color. If you have chosen the wrong indicator, the end point will be different than the eq. pt.

d) Indicators are often Weak Acids that lose a proton (causing the color change) when [OH-] reaches a certain concentration

HIn + OH- In- + H2O

B. Examine the behavior of a Hypothetical Indicator (Ka = 1 x 10-8)

1) HIn H+ + In-

red blue

2) From the Ka equation, we can develop a formula showing how color behaves depending on the pH

a) The ratio of HIn and In- will determine the color

b) The ratio is a function of pH and a constant

c) We should be able to predict the color at different pH’s

3) If we add a few drops of HIn to a pH = 1.0 solution:

a)

b) HIn is the dominant species so the color will be red

4) If we add OH- to the solution, eventually [In-] = [HIn]

The color will be equal parts red and blue = purple

[HIn]

][In

][H

K

[HIn]

]][In[HK a

a

[HIn]

][In

1

10 x 1

1.0

10 x 1

][H

K -7-8a

5) Usually, about 10% of the HIn must react before our eyes can detect a color change.

6) 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?

C. We can use the Henderson-Hasselbalch equation on Indicators as well

1) pH = pKa + log([In-]/[HIn])

2) pH = pKa + log(1/10) for a color change

a) log(1/10) = -1

b) pH for color change starting in acid is always pKa – 1 for any Indicator

3) For a basic solution titrated with acid, [In-]/[HIn] = 10/1 for color change

a) Log(10/1) = +1, pH for color change will equal pKa + 1

b) Useful range for a pH Indicator is always pKa +/- 1

10

1

[HIn]

][In changeColor

-

6.00pH1x10][H10K10

1

[HIn]

][In

][H

K 6a

a

4) The useful range and colors of common Indicators:

D. Choosing an Indicator

1) We always want the end point (color change) to coincide with the eq. pt.

2) Strong Acid—Strong Base titrations have a large pH change at the equivalence point, so you get sharp color change.

a) What indicator would work best for HCl/NaOH titration?

b) Eq. Pt. = 7.00

c) We will be going from acidic to basic, so we want [In-]/[HIn] = 1/10

d) pH = pKa + log(1/10) = pKa – 1

e) 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)

f) If we were going from basic to acidic, we would want pKa = 6.00

3) How crucial is titrating exactly to the equivalence point?

a) NaOH = 99.9 ml pH = 5.3

b) NaOH = 100.0 ml pH = 7.0

c) NaOH = 100.1 ml pH = 8.7

d) Wide range of Indicators would tell us the eq. pt. to within 0.1 ml

4) Indicators for weak acid titrations

a) The weaker the acid, the smaller the pH change at the eq. pt.

b) We have less flexibility in choosing our indicator

c) 0.1 M HC2H3O2 titrated with 0.1 M NaOH has pH = 8.7 at eq. pt.

i. Phenolphthalein ok (8.0—10.0 endpoints)

ii. Thymol Blue ok (8.0—9.2 endpoints)

iii. Methyl Red in not ok (4.2—6.2 endpoints)

HCl titration HC2H3O2 titration