1

Indicators for Acid-Base Titrations (Sec. 9-6)

transition range needs to match the endpoint pH as closely as possible in order to minimize titration error

2

Acid-Base indicators are themselves weak acids…..

e.g. phenolthalein

H2In = HIn- = In2-

3

Ch 10: Acid-Base TitrationsTitration of 0.10 M HCl by 0.10 M NaOH

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100

mL OH-

pH

phenolthalein 8.0-9.6

Automated titrators determine the endpoint electronically by numerically calculating the 2nd derivative

2nd Derivative

-300

-200

-100

0

100

200

300

49.5 49.6 49.7 49.8 49.9 50 50.1 50.2 50.3 50.4 50.5

mL base

d2 p

H/d

mL

2

Acid-Base Titrations Curves - pH (or pOH) as a function of mL of titrant added

mL base

analyte = strong acid

titrant = strong base

mL acid

analyte = strong base

titrant = strong acid

1

2

3

4

4

I. Strong Acid-Strong Base Titration Curves (Sec. 10-1)

equivalence pt. volume:

50 mL of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the titration curve for the analysis.

1 Initial pH

2 pH before the equivalence pt.

3 pH at the equivalence pt.

5

4 pH after the equivalence pt.

mL base

[H+] = CHA so pH = -log CHA

Strong Acid - Strong Base Titration (both monoprotic)(analyte) (titrant)

Eq. Pt. pH = 7

[H+] = MaVa - MbVb

Vtotal

[OH-] = Mb(Vb beyond eq.pt.)Vtotal

6

Titration of 0.10 M HCl by 0.10 M NaOH

0

2

4

6

8

10

12

14

0 10 20 30 40 50 60 70 80 90 100

mL OH-

pH

methyl red 4.2-6.2

phenolthalein 8.0-9.6

Titration Error

Titration of 0.10 M HCl by 0.10 M NaOH

Expanded View of Equivalence Point

0

2

4

6

8

10

12

14

49 49.1 49.2 49.3 49.4 49.5 49.6 49.7 49.8 49.9 50 50.1 50.2 50.3 50.4 50.5 50.6 50.7 50.8 50.9 51

mL OH-

pH

phenolthalein 8.0-9.6

0.02 mL/50 mL =0.04% error!

7

II. Weak Acid-Strong Base Titration Curve (Sec. 10-2)

HA = H+ + A-

[HA]

]][A[H Ka

50 mL of a 0.100 M soln of the weak acid HA, Ka = 1.0 x 10-5, is titrated with 0.100 M NaOH. Calculate the titration curve for the analysis.

equivalence pt. volume:

1 Initial pH

8

2 pH before the equivalence pt.

4 pH after the equivalence pt. = same as SA-SB titration

3 pH at the equivalence pt.

9

mL base

Weak Acid - Strong Base Titration (both monoprotic)(analyte) (titrant)

Eq. Pt. Hydrolysis of the conjugate base

[OH-] = Mb(Vb beyond eq.pt.)Vtotal

HAHAa CwhenCKH x ][

mol acid

saltmolpKapH

log

Buffer region

1/2 eq. pt. pH = pKa

10

Ch 11: Titrations in Diprotic Systems

Biological Applications - Amino Acids (Sec. 11-1)

low pH high pH

R = (CH3)2CHCH2 -

Finding the pH in Diprotic Systems (Sec. 11-2)

The strength of H2L+ as an acid is much, much greater than HL -

Ka1 = 10-2.328 = 4.7 x 10-3

Ka2 = 10-9.744 = 1.8 x 10-10

So assume the pH depends only on H2L+ and ignore the

contribution of H+ from HL.

1. The acidic form H2L+

11

Calculate the pH of 0.050M H2L+

2. The basic form L-

Ka1 = 10-2.328 = 4.7 x 10-3 Ka2 = 10-9.744 = 1.8 x 10-10

Strengths of conjugate bases:

for L- Kb1 = Kw/Ka2 = 1.01 x 10-14/1.8 x 10-10 = 5.61 x 10-5

for HL Kb2 = Kw/Ka1 = 1.01 x 10-14/4.7 x 10-3 = 2.1 x 10-12

Since the second conj. base HL is so weak, we'll assume all the OH- comes from the L- form.

12

Example: Calculate the pH of a 0.050M solution of sodium leucinate

The Intermediate FormThe pH of a Zwitterion Solution - Leucine (HL form)

13

14

[H+]2 = Ka1 Ka2

-log [H+]2 = - log Ka1 - log Ka2

2 pH = pKa1 + pKa2

HLa1

a1wHLa2a1

CK

KKCKK][H

assume:

KwKa1 << Ka1Ka2CHL

Ka1 << CHL

a2a1

HL

HLa2a1 KK][H so C

CKK][H

2

pKpKpH a2a1

pH of a solution of a diprotic zwitterion

Example:pH of the Intermediate Form of a Diprotic Acid

Potassium hydrogen phthalate, KHP, is a salt of the intermediate form of phthalic acid. Calculate the pH of 0.10M KHP and 0.010M KHP.

15

Titration Curve for the Amino Acid Leucine

Titration of 10 mL of 0.100 M Leucine

with 0.100 M NaOH

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35

mL NaOH

pH

equivalence pt. volumes (Ve1 & Ve2) =

pts B and D: 1st and 2nd half eq. pt's =

pt A: init. pH (H2L+ treat as monoprotic weak acid) =

16

pt C: 1st eq. pt (HL) =

pt E: 2nd eq. pt (L-) =

17

Example p. 233:Titration of Sodium Carbonate (soda ash)

Calculate the titration curve for the titration of 50.0 mL of 0.020 M Na2CO3 with 0.100 M HCl.

equivalence pt. volumes (Ve1 & Ve2) =

18

pt C: 1st eq. pt (HCO3-) =

pt E: 2nd eq. pt (H2CO3 treat as monoprotic weak acid) =

pts B and D: 1st and 2nd half eq. pt's =

pt A: init. pH (CO32- treat as monoprotic weak base) =

pt E: 2nd eq. pt (H2CO3 treat as monoprotic weak acid) =

19

Buffers of Polyprotic Acids and Bases

Fractional Composition Diagram H3PO4

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 2 4 6 8 10 12 14

pH

alp

ha

H3PO4 HPO42- PO43-H2PO4-