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

Acid-Base indicators are themselves weak acids…..

e.g. phenolthalein

H2In = HIn- = In2-

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

pH

analyte = strong acid

titrant = strong base

mL acid

pH

analyte = strong base

titrant = strong acid

1

23

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.

4 pH after the equivalence pt.

mL base

pH

[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

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!

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

2 pH before the equivalence pt.

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

3 pH at the equivalence pt.

mL base

pH

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

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+

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.

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

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

[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 pf 0.10M KHP and 0.010M KHP.

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) =

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

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

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) =

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) =

Buffers of Polyprotic Acids and BasesFractional 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-