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Version 2012 Updated on 042312 Copyright © All rights reserved
Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Women’s University
Chapter 12
Principles of Neutralization Titration
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Standard solutions for Acid-Base Titrations
Analyte Titrant vs Standard solution
N V = N’V’
The standard reagents used in acid-base titrations are always strong acids or strong bases, most commonly HCl, HClO4, H2SO4, NaOH, KOH. Weak acids and bases are never used as standard reagents because they react incompletely with analyte.
N= unknown V= measure
N’= known
V’= known
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Primary standards for standardizing
Acids Potassium acid phthalate
Sulfamic acid ( H2NSO3H)
HCl
Potassium hydrogen iodate
Bases TRIS(hydroxymethylaminomethane)
Sodium carbonate
Borax (= sodium tetraborate )
HgO
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General goal and procedure of acid-base titration
1) Standardization 2) Determination 3) Titration curve
4) Interpret titration curve ,
understand what is
happening during titration
Primary standard solutionknown normality(N)
known volume(V)
Acid or base standard solution
unknown normality(N’)unknown volume(V’)
Acid or base standard solution
known normality(N’)unknown volume(V’)
Analyte Sample solution
unknown Ns
known Vs
Volume of NaOH added (ml)
0 5 10 15 20 25
pH
0
2
4
6
8
10
12
14
Plots of pH versus volume of titrant
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Finding the end point
1) Indicator
2) Potentiometry : Titration curve ( pH or mV vs Va )
1st derivative titration curve
2nd derivative titration curve
Gran plot
3) Conductometry
4) Spectrometry
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Ex. Thymol blue
H2In HIn– In2 –
Red Yellow Blue
H2In H+ + HIn– pKa1 = 1.7
pH = pKa1 + log [HIn– ]/[H2In]
if [HIn– ]/[H2In] 1:10 pH = 0.70 Red color
[HIn– ]/[H2In] = 1:1 pH = pKa1 = 1.7 Orange color
[HIn– ]/[H2In] 10:1 pH = 2.70 Yellow color
HIn– H+ + In2 – pKa2 = 8.9
Transition range
pKa11
Indicator
An acid-base indicator is itself an acid or base whose different protonated species have different colors.
The approximate pH transition range of most acid-type indicator is roughly pKa 1
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Choosing an indicator
The calculated pH titration curve for the titration of 10.00ml of 0.1000M Na2CO3
with 0.1000M HCl.
Phenolphthalein
acid form :
colorless
Phenolphthalein
base form :
pink
transition range(pH):
8.0~9.0
Volume of 0.1000M HCl (ml)
0 5 10 15 20 25 30 350.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
PP 8.0~9.0
MR 4.8~6.0
MO 3.1~4.4
pH
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Calculated titration curve for the reaction of 100 mL of 0.0100M base (pKb = 5.00) with 0.0500 M HCl.
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Indicator color as a function of pH (pKa=5.0)
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Volume of 0.09270N NaOH (ml)
0 5 10 15 20 25 30
pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
Fig. Experimental titration curve.
0.06860N KHP 25.00ml vs 0.09270N NaOH
Primary standard KHP
204.22g/1000ml=1.0000N
0.35g/25ml = xN
x = 0.06860N
End point=18.50ml
0.06860N×25.00ml= x N×18.50ml
x=0.09270N
Titration of
25.00ml of
KHP with
NaOH
Volume of 0.09270N NaOH (ml)
0 5 10 15 20 25 30
pH
/V
a
0.00
1.00
2.00
3.00
4.00
5.00
Fig. The 1st derivative experimental titration curve.
0.06860N KHP 25.00ml vs 0.09270N NaOH
Experiments. Standardization of 0.1000N NaOH
Volume of 0.09270N NaOH (ml)
0 5 10 15 20 25 30
p
H/
Va)/ V
a
-6.00
-4.00
-2.00
0.00
2.00
4.00
6.00
Fig. The 2nd derivative experimental titration curve.
0.06860N KHP 25.00ml vs 0.09270N NaOH
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Calculation of concentration (Finding of NaOH concentration)
HOOCC6H4COOK (mw 204.44) NaOH (mw 40.01) 1 Eq wt.
204.22 g 1 N × 1000 mL
20.422 mg 0.1 N × 1 mL
a g x N’ × Ve mL
x N’ = (a g × 1000 mL) / (204.22 g × Ve mL)
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Volume of 0.0927N NaOH(m l)
0 10 20 30 40 50 60
pH
2.00
4.00
6.00
8.00
10.00
12.00
14.00
Fig . The 1st derivative experimen tal titration cur ve. 0.1094N Venigar 25.00ml vs 0.0927N NaOH
Volume of 0.0927N NaOH(m l)
0 10 20 30 40 50 60
( p
H/
Va)/ V
a
-4.00
-3.00
-2.00
-1.00
0.00
1.00
2.00
3.00
4.00
Fig. The 2nd deriv ative experimen tal titration cur ve. 0.1094N Venigar 25.00ml vs 0.0927N NaOH
Standardization of NaOH
0.09270N
Titation of
25.00ml of Venigar
with NaOH
End point = 29.50ml
x 25ml = 0.09270N 29.50ml
Venigar : x = 0.1094N
Vo lume of 0.09 270N NaOH (ml)
0 5 10 15 20 25 30
pH
/V
a
0.00
1.00
2.00
3.00
4.00
5.00
Fig. The 1st derivative ex per imental titration cur ve.
0. 06860N PHP 25. 00ml vs 0.0927N Na OH
Experiments. Determination of acetic acid in venigar
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Gran plot
HA = H+ + A–
Ka = [H+] fH+ [A–]fA– / [HA]fHA
HA vs NaOH
[A–] = (moles of OH– delivered) / (total volume) = VbFb / (Vb + Va)
[HA] = (original moles of HA – moles of OH –) / (total volume)
= (VaFa – VbFb ) / (Vb + Va)
Ka = [H+] f H+ VbFb fA– / (VaFa – VbFb ) fHA
[H+] fH+ Vb = (fHA / fA– ) Ka{(VaFa – VbFb )/ Fb}
10–pH Vb = (fHA / fA– ) Ka(Ve – Vb)
= –(fHA / fA– ) Ka Vb + (fHA / fA– ) KaVe 10
–pH V
bVb(l)
Slope =–(fHA / fA– ) Ka
Ve
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Gran plot for the first equivalence point. This plot gives an estimate of Ve that differs from that in Figure 12.7 by 0.2μL (88.4 versus 88.2μL). The last 10-20% of volume prior to Ve is normally used form a Gran plot.
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Conductometric end-point detection
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Calculated titration curve for the reaction of
50.00 mL of 0.02000 M NaOH with 0.1000 M HCl.
Titration of strong acid with strong base
Ve mL×0.1000M = 50.00 mL × 0.02000 M
Ve = 10.00 mL
0.02000M NaOH
50.00 mL
0.1000MHCl
Neutralization :
HCl + NaOH NaCl + H2O
A
Va(mL)
0 2 4 6 8 10 12 14 16 18 20
0
2
4
6
8
10
12
14
B
C
D
pH
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A. Before the beginning of titration : Va = 0.00 mL.
NaOH Na+ + OH–,
0.02000 M [OH–] = 0.02000 M
[H+] = 1.00×10–14/ 2.000×10–2
pH = 12.30
B. Before the equivalence point : 0 < Va < Ve
remaining NaOH
V a Ve
Initial NaOH amount =F × Vi
added HCl amount = used NaOH amount = F× Va
Remaining NaOH amount =
[OH–] = {(Ve – Va )/ Ve}FNaOH {Vi /(Vi + Va)}
Ex. Va = 3.00 mL
[OH–] = {(10.00 – 3.00)/10.00}(0.02000){50.00 /(50.00 +3.00)} = 0.0132 M
[H+] = 1.00×10–14/ 0.0132 = 7.58×10–13
pH = 12.12
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C. At the equivalence point : Va = Ve
H2O = H+ + OH–,
[H+] [OH–] = 1.00×10–14
[H+] = 1.00×10–7 pH = 7.00
D. After the equivalence point : Va > Ve
excess HCl
V aVe
added HCl amount = FHCl× Va
initial NaOH amount = used HCl = F × Vi
Excess HCl amount = FHCl(Va–Vi)
= FHCl (Va–Ve)
[H+] = FHCl {(Va– Ve) /(Vi + Va)}
Ex. Va = 10.50 mL
[H+] = (0.1000) {10.50 –10.00) /(50.00 + 110.50)} = 8.26×10–4 pH = 3.08
ViNaOH
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Titration curves for NaOH with HCl.
A. 50.0 mL of 0.0500 M NaOH with 0.1000 M HCl.
B. 50.00 mL of 0.00500 M NaOH with 0.0100 M HCl.
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Titration curves for HCl with NaOH.
A. 50.0 mL of 0.0500 M HCl
with 0.1000 M NaOH.
B. 50.00 mL of 0.000500 M HCl
with 0.00100 M NaOH.
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Changes in pH during the totration of strong acid with strong base.
A: 50 mL of 0.0500 M HCl vs 0.1000 M NaOH B: 50 mL of 0.000500 M HCl vs 0.001000 M NaOH
Va of NaOH (mL)
0 5 10 15 20 25 30 35
0.001.002.003.004.005.006.007.008.009.00
10.0011.0012.0013.0014.00
A
B
The effects of titrant and analyte concentrations on neutralization titration curves.
For the titration with diluted concentrations, the change in pH in equivalence point region is markedly less than concentrated solution
pH
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Weak acid titrated with strong base
1) Titration reaction :
CH3COOH + NaOH CH3COO– Na+ + H2O
strong + weak complete reaction
K = 1/Kb = 1.76 ×109
2) Calculation of equivalence point :
[CH3COOH] Vi mL = [NaOH] Ve mL
0.1000 M×5.00 mL = 0.1000 M × Ve mL
Ve= 5.00 mL
3) Titration curve :
Changes in pH during the titration of a weak acid with a strong base.A: 5.00 mL of 0.1000M HOAC vs 0.1000 M NaOHB: 5.00 ml of 0.001000M HOAC vs 0.001000M NaOH
Va of NaOH(mL)
0 1 2 3 4 5 6 7 8
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
pH
A
B
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Titration curves for the titration of acetic acid with NaOH.
A. 0.1000 M acetic acid
with 0.1000M NaOH.
B. 0.001000 M acetic acid
with 0.00100 M NaOH.
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A. Before adding any titrant : Va = 0.00
HAC = H+ + AC–
[H+] = KaF = 1.75 ×10–5 ×0.1000
= 1.323 × 10–3
pH= 2.88
0.1000M HAC
25.00ml
0.1000MNaOH
Titration curve.
0.1000 M HAC vs 0.1000 M NaOH.
Va of 0.1000 M NaOH (mL)
0 1 2 3 4 5 6 7pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
A
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B. Before equivalence point : 0 < Va < Ve
HAC + NaOH H2O + Na+AC–
[H+] = Ka[HAC] / [AC–]
initial HAC amount = F×Ve
– added NaOH amount
= used amount HAC= F×Va
Remaining HAC amount =F(Vi–Va)
[HAC] = {(F×Ve) – (F×Va)}/(Vi +Va)
[AC–] = (F×Va) / (Vi +Va)
pH = pKa + log [AC–] /[HAC]
at Va = Ve/2 pH= pKa = 4.76 Titration curve.
0.1000 M HAC vs 0.1000 M NaOH.
Va of 0.1000 M NaOH (mL)
0 1 2 3 4 5 6 7pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
B
Ve/2
Vi
VeVa
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Titration curve.
0.1000 M HAC vs 0.1000 M NaOH.
C. Equivalence point Va = Ve
HAC + NaOH H2O + Na+AC–
initial 1 1
final 0 0 1
AC– = HAC + OH–
[OH–] = KbF’ = KwF’ /Ka
F’=( F×Vi) / (Vi +Va) Va of 0.1000 M NaOH (mL)
0 1 2 3 4 5 6 7
pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
C
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Titration curve.
0.1000 M HAC vs 0.1000 M NaOH.
D. After equivalence point Va> Ve
added NaOH amount = F×Va
initial HAC amount
= used NaOH amount = F×Ve
Excess NaOH amount = F×(Va –Ve)
[OH–] = FNaOH ×(Va– Vi) / (Vi +Va)
Va of 0.1000 M NaOH (mL)
0 1 2 3 4 5 6 7
pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
D
Vi
Ve
Va
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The effect of acid strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 mL of 0.1000 M acid with 0.1000 M base.
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Weak base titrated with strong acid
1) Titration reaction :
NH3 + HCl NH4+Cl–
2) Equivalence point :
NH3 HCl
0.1100 M×20.00 mL = 0.1000 M×Ve mL
Ve = 22.00 mL
3) Titration curve :
A. Initial Va = 0
NH3 + H2O = NH4+ + OH–
Kb = Kw / Ka = [NH4+][OH– ]/[NH3]
1.79×10–5 = [OH– ]2 / 0.1100–[OH– ] [OH– ]= KbF =1.4×10–3
pH = 11.15
The calculated titration curve forthe titration of 20.00 mL of 0.1100 Mammonia with 0.1000 M HCl.
0.1000M HCl volume (mL)
0 10 20 30 40 50
pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
A
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B. Before the equivalence point 0<Va< Ve
Major constituents: NH3 + NH4+Cl–
[OH]= Kb [NH3]/ [NH4+]
initial NH3 amount = F×Ve
– added HCl amount
= used amount NH3 = F×Va
Remaining NH3 amount =F(Vi–Va)
[NH3] = {(F×Ve) – (F×Va)}/(Vi +Va)
[NH4+] = (F×Va) / (Vi +Va)
pH = pKb + log [NH4+] / [NH3]
at Va = Ve/2 pOH= pKb = 4.74
pH = 9.26
0.1000M HCl volume (ml)
0 10 20 30 40 50
pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
Fig. 12-4. The calculated titration curve for the titration of 20.00 ml of 0.1100M ammonia with 0.1000M HCl.
B
Ve/2
Vi
VeVa
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0.1000M HCl volume (ml)
0 10 20 30 40 50
pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
Fig. 12-4. The calculated titration curve for the titration of 20.00 ml of 0.1100M ammonia with 0.1000M HCl.
C. Equivalence point Va = Ve
NH3 + HCl NH4+Cl–
initial 1 1
final 0 0 1
NH4+ = NH3 + H+
[H+] = KaF’
F’=( F×Vi) / (Vi +Va)
C
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D. After equivalence point Va> Ve
added HCl amount = F×Va
initial NH3 amount
= used HCl amount = F×Ve
Excess HCl amount = F×(Va –Ve)
[H+] = FHCl ×(Va– Ve) / (Vi +Va)
0.1000M HCl volume (ml)
0 10 20 30 40 50
pH
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
Fig. 12-4. The calculated titration curve for the titration of 20.00 ml of 0.1100M ammonia with 0.1000M HCl.
D
Vi
Ve Va
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The effect of base strength (dissociation constant) on titration curves. Each curve represents the titration of 50.00 ml of 0.1000 M base with 0.1000 M HCl.
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Titration of Weak Base with Strong Acid
[H+] = KaF’
F’=( F×Vi) / (Vi +Va)
[OH–] = KbF’ = KwF’ /Ka
F’=( F×Vi) / (Vi +Va)
Equivalence point
After equivalence point
(Va>Ve)
pH = pKb + log[NH4+]/[NH3]pH = pKa + log [A–] /[HA]
Before the equivalence point (0<V a<V e)
[OH-] = KbF =1.4×10–3 [H+] = KaFInitial
B + H2O → BH+ + OH- HA + OH- → H2O + A-Titration reaction
Weak Base with Strong BaseWeak Acid with Strong Base
(Vi + Va)}(Va– Ve){FHCl=[H+]
(Vi + Va)
}(Va– Ve){FNaOH=[OH-]
Comparison of Weak Acid/ Base with Strong Base/Acid
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Determining the pK values for amino acids
Amino acids contain both an acidic and a basic group.
alanine
All naturally occurring amino acids are left-handed (L) form.
The amine group behaves as a base, while the carboxyl group acts as an acid.
Aspartic acid
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Curves for the titration of 20.00ml of 0.1000M alanine with 0.1000 M NaOH and 0.1000M HCl. Note that the zwitterion is present before any acid or base has been added. Adding acid protonates the carboxylate group with a pKa of 2.35. Adding base reacts with the protonated amine group with a pKa of 9.89.
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Plots of relative amounts of acetic acid and acetate ion during a titration.
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Amino acidin aceticacid +
0.020M HClO4
in acetic acid
0.010M CH3COOKin acetic acid
Acid base titration in non-aqueous media
HOOCCHNH2 + HClO4 HOOCCHNH3+ClO4
–
R CH3COOH R
(solvent)
HClO4 + CH3COOK CH3COOH + K+ClO4–
CH3COOH(solvent)
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Titrants used in non-aqueous titrimetry
Acidic titrants Perchloric acid
p- Toluenesulfonic acid
2,4-Dinitrobenzenesulfonic acid
Basic titrants Tetrabutylammonium hydroxide
Sodium acetate
Potassium methoxide
Sodium aminoethoxide
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Selected solvents for non-aqueous titration
Solvent Autoprotolysis constant Dielectric
(pKHS) constant
Amphiprotic Glacial acetic acid 14.45 6.1
Ethylenediamine 15.3 12.9
Methanol 16.7 32.6
Aprotic or basic Dimethylformamide 36.7
Benzene 2.3
Methyl isobutylketone 13.1
Pyridine 12.3
Dioxane 2.2
n-Hexane 1.9
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Acid and base strengths that are not distinguished in aqueous solution may be distinguishable in non-aqueous solvents.
Ex. Perchloric acid is a stronger acid than hydrochloric acid in acetic acid solvent,
neither acid is completely dissociated.
HClO4 + CH3COOH = ClO4– + CH3COOH2
+ K = 1.3×10–5
strong acid strong base weak base weak acid
HCl + CH3COOH = Cl– + CH3COOH2+ K = 5.8×10–8
Differentiate acidity or basicity of different acids or bases
differentiating solvent for acids …… acetic acid, isobutyl ketone
differentiating sovent for bases …… ammonia, pyridine
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Hammett acidity function, Ho, for aqueous solutions of acids.
Acid Name Ho
H2SO4 (100%) Sulfuric acid –11.93
H2SO4 SO3 Fuming sulfuric acid –14.14
H SO3F Fluorosulfuric acid –15.07
H SO3F+10%SbF5 Super acid –18.94
H SO3F+7%SbF53SO3 –19.35
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Titration of a mixture of acids with tetrabutylammonium hydroxide in methyl isobutyl ketone solvent shows that the order of acid strength is HClO4 > HCl > 2-hydroxybenzoic acid > acetic acid > hydroxybenzene. Measurements were made with a glass electrode and a platimum reference electrode. The ordinate is proportional to pH, with increasing pH as the potential becomes more positive.
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Summary
Acid-Base titration
Titration curve
Indicator
Detection of end-point
Potentiometry
Gran plot
Conductometry
Non-aqueous titration
Hammett acidity function, Ho