indicators for acid-base titrations (sec. 9-6)
TRANSCRIPT
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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
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Acid-Base indicators are themselves weak acids…..
e.g. phenolthalein
H2In = HIn- = In2-
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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
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2
3
4
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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.
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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
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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!
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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
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2 pH before the equivalence pt.
4 pH after the equivalence pt. = same as SA-SB titration
3 pH at the equivalence pt.
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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
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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+
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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.
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Example: Calculate the pH of a 0.050M solution of sodium leucinate
The Intermediate FormThe pH of a Zwitterion Solution - Leucine (HL form)
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[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.
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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) =
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pt C: 1st eq. pt (HL) =
pt E: 2nd eq. pt (L-) =
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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) =
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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) =
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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-