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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 Titrations
Titration of 0.10 M HCl by 0.10 M NaOH
14
12
10
phenolthalein 8.0-9.6
pH
8
6
4
2
0
0
10
20
30
40
50
60
70
80
90
100
mL OH-
2nd Derivative
300
Automated titrators determine the
endpoint electronically by numerically
calculating the 2nd derivative
d2 pH/d mL2
200
100
0
49.5 49.6 49.7 49.8 49.9
-100
50
50.1 50.2 50.3 50.4 50.5
-200
-300
mL base
Acid-Base Titrations Curves - pH (or pOH)
as a function of mL of titrant added
4
2
analyte = strong acid
titrant = strong base
3
1
mL base 
analyte = strong base
titrant = strong acid
mL acid 
I. Strong Acid-Strong Base
Titration Curves (Sec. 10-1)
50 mL of 0.100 M HCl 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.
3
pH at the equivalence pt.
4
pH after the equivalence pt.
Strong Acid - Strong Base Titration (both monoprotic)
(analyte)
(titrant)
[H+] = MaVa - MbVb
Vtotal
[OH-] = Mb(Vb beyond eq.pt.)
Vtotal
Eq. Pt. pH
=7
[H+] = CHA so pH = -log CHA
mL base 
Titration of 0.10 M HCl by 0.10 M NaOH
14
12
10
phenolthalein 8.0-9.6
pH
8
6
methyl red 4.2-6.2
4
2
0
0
10
20
30
40
50
mL OH-
60
70
80
90
100
Titration Error
Titration of 0.10 M HCl by 0.10 M NaOH
Expanded View of Equivalence Point
14
12
10
phenolthalein 8.0-9.6
pH
8
6
0.02 mL/50 mL
=0.04% error!
4
2
0
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
mL OH-
51
II. Weak Acid-Strong Base
Titration Curve (Sec. 10-2)
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.
HA = H+ + A-
[H ][A  ]
Ka 
[HA]
equivalence pt. volume:
1
Initial pH
2
pH before the equivalence pt.
3
pH at the equivalence pt.
4
pH after the equivalence pt. =
same as SA-SB titration
Weak Acid - Strong Base Titration (both monoprotic)
(analyte)
(titrant)
pH  pKa  log
mol salt
mol acid
[OH-] = Mb(Vb beyond eq.pt.)
Vtotal
Buffer region
Eq. Pt. Hydrolysis of the
conjugate base
1/2 eq. pt. pH = pKa
[ H ]  K C when x  C

a
HA
mL base 
HA
Ch 11: Titrations in Diprotic Systems
Biological Applications - Amino Acids (Sec. 11-1)
R = (CH3)2CHCH2 -
low pH
high pH
Finding the pH in Diprotic Systems (Sec. 11-2)
1. The acidic form H2L+
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.
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 Form
The pH of a Zwitterion Solution - Leucine (HL form)
assume:
K a1K a2CHL  K wK a1
[H ] 
K a1  CHL

K a1K a2CHL
[H ] 
CHL

KwKa1 << Ka1Ka2CHL
Ka1 << CHL
so [H]  K a1K a2
[H+]2 = Ka1 Ka2
-log [H+]2 = - log Ka1 - log Ka2
2 pH = pKa1 + pKa2
pK a1  pK a2
pH 
2
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.
Titration Curve for the Amino Acid Leucine
Titration of 10 mL of 0.100 M Leucine
with 0.100 M NaOH
14
12
10
pH
8
6
4
2
0
0
5
10
15
20
mL NaOH
25
30
35
equivalence pt. volumes (Ve1 & Ve2) =
pt A: init. pH (H2L+ treat as monoprotic weak acid) =
pts B and D: 1st and 2nd half eq. pt's =
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 A: init. pH (CO32- treat as monoprotic weak base) =
pts B and D: 1st and 2nd half eq. pt's =
pt C: 1st eq. pt (HCO3-) =
pt E: 2nd eq. pt (H2CO3 treat as monoprotic weak acid) =
pt E: 2nd eq. pt (H2CO3 treat as monoprotic weak acid) =
Buffers of Polyprotic Acids and Bases
Fractional Composition Diagram H 3PO4
H3PO4
HPO42-
H2PO4-
PO43-
1.00
0.90
0.80
alpha
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0
2
4
6
8
pH
10
12
14