unit 19 acid base equilibria: titrations
DESCRIPTION
CHM 1046 : General Chemistry and Qualitative Analysis. Unit 19 Acid Base Equilibria: Titrations. Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL. Textbook Reference : Chapter 19 (sec. 5-8) Module 9. Soln -Unknown Concentration ( M ): Acid. Titration. - PowerPoint PPT PresentationTRANSCRIPT
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AqueousEquilibria
Unit 19Acid Base Equilibria:
Titrations
Dr. Jorge L. Alonso
Miami-Dade College – Kendall Campus
Miami, FL
CHM 1046: General Chemistry and Qualitative Analysis
Textbook Reference:
•Chapter 19 (sec. 5-8)
•Module 9
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AqueousEquilibria
Titration A volumetric
technique in which one can determine the concentration of a solute in a solution of unknown concentration, by making it react with another solution of known concentration (standard).
•Standard-of known Conc.(M)
MolesB = M x V
MolesA = M x V
{*TitrationMovie}
•Soln-Unknown Concentration (M): Acid
If: MolesA = MolesB
Then: (M x V)A = (M x V)B
• Known Volume (V)
•Measure Vol to reach end pt.
L L
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AqueousEquilibria
Determining the Concentration of Solutions by Titration
Example:
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
Neutralization: equivalence point
# mol1(acid) = # mol2(base)
(Standard)
xa HA (aq) + xb MOH (aq) MA (aq) + H2O (l)
1 mol1(acid) = 1 mol2(base)
xa xb
H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + H2O(l)
A known concentration of base (or acid) is slowly added to a solution of acid (or base) until neutralization occurs.
NaOH1SOH? 42
NaOH2
SOH1 42
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AqueousEquilibria
Titration Calculations: Stoichiometry using Molarities
Neutralization: 1 moles(acid) = 1 moles(base)
Since moles = MV = moles x Liter Liter
BxAx
MA x VA MB x VB=
xA HN + xB MOH MN + HOH
B or Ax = coefficients from balanced equations Where
==ηA
AxAx
MWg#
xA xB
=ηB
Bx Bx
MWg#
=
* Equation Useful for determining Molarities and Volumes at the Equivalence Point of a Titration *
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AqueousEquilibria
Solution Stoichiometry Problems: Molarity
A
BBA V 2
V1M M
H2SO4 + 2 NaOH 2HOH + Na2SO4
BxAxMA x VA = MB x VB
1 2
Problem: A volume of 16.3 mL of a 0.30M NaOH solution was used to titrate 25.00 mL H2SO4. What is the concentration of H2SO4 in the solution of unknown concentration?
mL 25.00 2
mL 3.160.30M 1 = 0.098 M H2SO4
Titration of a Strong Acid with a Strong Base
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AqueousEquilibria
Titration Graph: pH vs. Volume
Excess acid
Excess base
Acid = Base
Methyl Red Indicator
Phenolphthalein Indicator
{Titration2}
SA
SB
pHmeter
mL of NaOH
pH
0 1.2
10 1.4
20 1.6
30 1.7
40 1.9
50 7.0
60 12.0
70 12.2
80 12.3
Titration Data:
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AqueousEquilibria
Titrations: The Strength of Acids & Bases
Strong Base with Weak Acid
Strong Base with Strong Acid
Weak Base with Weak Acid
Weak Base with Strong AcidSA
SB
SB
WA
SA
WB
WA
WB
Phenolphthalein 8-10
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AqueousEquilibria
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
(2) Titration of a WA with a SB
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AqueousEquilibria
Selecting Appropriate Indicators
• an indicator is chosen so that it will change color at a pH just beyond the equivalence point (mid point of the steep vertical portion of the graph). The first point at which the indicator permanently changes color marks the end of the titration and is called the indicator end-point. Dropping a perpendicular from the indicator end-point to the x-axis is a very close estimation of the equivalence point.
Phenolph
Phenolph
Meth Red Meth Red
?????
Select appropriate indicator for following:
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AqueousEquilibria
SA
SB SA
SB
(1) Titration of a SA with a SB (or SB with a SA)
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AqueousEquilibria
Acid-Base Neutralization Equations(1) Strong Acids and Bases are represented in completely dissociated state:
as H+ and OH-
(2) Weak Acids and Bases are represented in undissociated state: as HA and B (or MOH)
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AqueousEquilibria
(H2CO3 + K+)
(H2CO3 + Ca2+ + C2H3O2- )
(H2CO3 + Zn2+ + SO42-)
(H2C O3 + Zn2+)
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AqueousEquilibria
HA + OH- H2O + A-
]HA[
]A][H[
AK
(2) Acid Base Neutralization: when not at the end-point or using WA or WB
(3) WA or WB Equilibrium problems
HA ↔ H+ + A-
Use: Mole ICEnd Table Use: [ICE] Table
II 0.0038 η 0.0030 η 0 η
C - 0.0030 η - 0.0030 η + 0.0030 η
End 0.0008 η 0 η 0.0030 η
Mole HA + MOH → MA + H20(.15M)(.025L) (.10M)(.030L)
II 0.061 0 0
C -x +x +x
E 0.061 - x x x
HA ↔ H+ A-
][
][log
HA
ApKpH a
Equations and Tables used in solving A-B Titration Problems
(1) Acid Base Neutralization: when you reach the end-point using SA or SB
B
B
A
A MVMV
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AqueousEquilibria
Example: 25 mL of 0.15M HCl with 0.10M NaOH.
(1) Titration of a SA with a SB
B
B
A
A MVMV
mL5.37
B
B
A
A
BA
BAB M
MVV
M
LMVB 10.0 )1(
)1( 025.15.0
II 0.0038 η 0.0030 η 0
C - 0.0030 η - 0.0030 η + 0.0030 η
End 0.0008 η 0 0.0030 η
(1) What volume of NaOH is required to reach the equivalence point?
Mole HX + MOH → MX + H20
(2) What is the pH of the initial strong acid? (strong acid problem)
(3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (excess SA problem)
In strong acid [HA]=[H+], so pH=-log [H+]
(.15M)(.025L) (.10M)(.030L) )L030.L025(.
)0008.0(
L]H[]HX[
pH = -log (1.4 x 10-2) = 1.9Salt of SA & SB: will not Hydrolyze
= -log (0.15) = 0.82
M104.1]H[ 2
*what is used for neutralization Rx?*
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AqueousEquilibria
Example: 25 mL of 0.15M HCl with 0.1M NaOH.
(4) What is the pH at the equivalence point?
(5) What is the pH after the equivalence point? Lets say after 40. mL of NaOH. (excess SB problem)
II 0.0038 η 0.0040 η 0 η
C - 0.0038 η - 0.0038 η + 0.0038 η
End 0 η 0.0002 η 0.0038 η
mole HX + MOH → MX + H20
(.15M)(.025L) (.10M)(.040L)
Salt of SA & SB: will not Hydrolyze
Only salt + water present and salt will not hydrolyze water since it is derived from SA & SB. So pH = 7
MLL
OH 3101.3)040.025(.
0002.0][
pOH = -log(3.1x10-3) = 2.5
pH + pOH = pKw pH = pKw - pOH
pH = 14 – 2.5 = 11.5
(1) Titration of a SA with a SB
*what is used for neutralization Rx?*
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AqueousEquilibria
• The pH >7 at the equivalence point.
• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
(2) Titration of a WA with a SB
HA ↔ H+ + A-
SB (OH-)
HA + OH- H2O + A-
]HA[
]A][H[
AK
][
][log
HA
ApKpH a
*what is used for equilibrium Rx?*
*what is used for neutralization Rx?*
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AqueousEquilibria
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH.
B
B
A
A MVMV
mL5.37
B
B
A
A
BA
BAB M
MVV
M
LMVB 10.0 )1(
)1( 025.015.0
(2) What is the initial pH of the acetic acid? (Before titration, WA Equilibrium problem)
II 0.15 0 0
C -x +x +x
E 0.15 - x x x
HA ↔ H+ A-
)15.0(
))((108.1 5
x
xx
][
]][[
HA
AHK a
(1) What volume of NaOH is required to reach the equivalence point?
0016.0][
107.2)15.0)(108.1( 652
Hx
x 8.2)0016.0log( pH
(2) Titration of a WA with a SB
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AqueousEquilibria
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH.
(3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (WA Buffer problem)
II 0.0038 η 0.0030 η 0 η
C - 0.0030 η - 0.0030 η + 0.0030 η
End 0.0008 η 0 η 0.0030 η
mole HA + MOH → MA + H20
(.15M)(.025L) (.10M)(.030L)
Salt of WA & SB: WILL Hydrolyze H2O
][
][log
HA
ApKpH a
]log[]log[ HAApKa
2.5)8.1()3.1()7.4()015.0log()055.0log()108.1log( 5 pH
MLL
HA 015.0030.0025.0
)0008.0(][
M
LLA 055.0
030.0025.0
)0030.0(][
(2) Titration of a WA with a SB
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AqueousEquilibria
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.1M NaOH.
(4) What is the pH at the equivalence point? This happens @ 37.5 mL (Hydrolysis of Salt derived from a WA & SB)
II 0.0038 η 0.0038 η 0 η
C - 0.0038 η - 0.0038 η + 0.0038 η
End 0 η 0 η 0.0038 η
mole HA + MOH → MA + H20
(.15M)(.025L) (.10M)(.038L)
Salt of WA & : will Hydrolyze water
Salt is NaC2H3O2
Na+ = derived from SB (NaOH), will not hydrolyze.C2H3O2
- = derived from WA (acetic acid) it WILL hydrolyze water.
(2) Titration of a WA with a SB
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AqueousEquilibria
Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.1M NaOH.
(4) What is the pH at the equivalence point? This happens @ 37.5 mL (Hydrolysis of Salt derived from a WA & SB)
II 0.061 0 0
C -x +x +x
E 0.061 - x x x
C2H3O2- + H2O ↔ HC2H3O2 + OH-
[C2H3O2-] =.
)038.025(.
0038.0
LL
M066.0
)061.0(
))((106.5 10
x
xx
][
]][[
A
OHHAK b
6
11102
101.6][
107.3)066.0)(106.5(
OHx
x
2.5)101.6log( 6 pOH
baw KKK
105
14
106.5108.1
100.1
a
wb K
KK
II 0.0038 η 0.0038 η 0 η
C - 0.0038 η - 0.0038 η + 0.0038 η
End 0 η 0 η 0.0038 η
mole HA + MOH → MA + H20
(.15M)(.025L) (.10M)(.038L)
pH + pOH = pKw pH = pKw - pOH
pH = 14 – 5.2 = 8.8
Salt is NaC2H3O2
(2) Titration of a WA with a SB
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AqueousEquilibria
2005B Q1
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AqueousEquilibria
Mole ICEnd Table:
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AqueousEquilibria
• The pH at the equivalence point in these titrations is < 7.
• Methyl red is the indicator of choice.
MOH ↔ M+ + OH-
SA (H3O+)
H3O+ + MOH 2H2O + M+
(3) Titration of a WB with a SA
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AqueousEquilibria
Calculation of pH
B + H+ H2O + A- ][
]][[KB B
OHBH
(1) Acid Base Neutralization (2) WB Equilibrium problem
B+ H2O ↔ BH+ + OH-
Use: Mole ICEnd Table Use: [ICE] Table
II 0.0038 η 0.0030 η 0 η
C - 0.0030 η - 0.0030 η + 0.0030 η
End 0.0008 η 0 η 0.0030 η
Mole B + H+ → A- + H20(.15M)(.025L) (.10M)(.030L)
II 0.061 0 0
C -x +x +x
E 0.061 - x x x
B + H2O ↔
BH+ OH-
][
][log
B
BHpKpOH B
(3) Titration of a WB with a SA:
Weak base and strong acid
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AqueousEquilibria
2007A Q1 Titration: weak acid with strong base
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AqueousEquilibria
Use: Mole ICEnd Table Mole HA + OH- → A- + H20
(.40 M)(.025L) (.40M)(.015L)
II 0.010 η 0.0060 η 0 η
C - 0.0060 η - 0.0060 η + 0.0060 η
End 0.0040 η 0 η 0.0060 η
lnso
solu
LM
L)025.0015.0(
)006.0(
= 0.15 M F-
lnso
solu
LM
(e) What’s pH?
L)025.0015.0(
)004.0(
= 0.10 M HF
For F-:
For HF:
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AqueousEquilibria
Use: [ICE] Table
HA H+ ↔ A-
II 0.10 0 0.15
C -x +x +x
E 0.10 - x x 0.15 + x
0.15 M F-
0.10 M HF
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AqueousEquilibria
Titrations of Polyprotic Acids
In these cases there is an equivalence point for each dissociation.
Ka1
Ka2
Ka3
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AqueousEquilibria
2005B Q1 Titration: weak acid strong base
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AqueousEquilibria
2005B Q1
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AqueousEquilibria
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AqueousEquilibria
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AqueousEquilibria
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AqueousEquilibria
2000 QA Titration: weak base strong acid
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AqueousEquilibria
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AqueousEquilibria
2001 Q3Titration: weak acid strong base
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AqueousEquilibria
Answers 2001 Q3
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AqueousEquilibria
2002A Q1 Titration: weak acid strong base
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AqueousEquilibria
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AqueousEquilibria
2003A Q1 Titration: weak base strong acid
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AqueousEquilibria
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AqueousEquilibria
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AqueousEquilibria
2006B Q1 Titration: weak acid strong base
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AqueousEquilibria
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AqueousEquilibria
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AqueousEquilibria
Expressing Concentrations of
Solutions: Molarity (& Normality*)
* For MDC students only!
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AqueousEquilibria
moles of soluteLiters of solution
=Molarity (M)
BxAx
B or Ax = coefficients for the acid (A) and the base (B) from the balanced neutralization equations
Where
MA x VA = MB x VB
Ax
(mol/L)A x LA (mol/LB) x LB
Bx
Ax Bx
mol molA B=
=
xA HN + xB MOH MN + HOH
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AqueousEquilibria
BxAxMA x VA MB x VB
Ax
(mol/L)A x LA (mol/LB) x LB
Bx
Ax Bx
mol molA B=
=
=
BxAxmolesA MB x VB=
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AqueousEquilibria
For titrations:
MM-g
solute g #
MM-g
mole 1 solute g # moles
AAA
MB x VBAx Bx =
Since
xA HN + xB MOH MN + HOH
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AqueousEquilibria
mol of soluteL of solution
M =
Molarity (M) vs. Normality (N)
n
MM-g EW-g
n
equiv of soluteL of solution
N =
MM-g
mole 1 Solute g # moles
EW-g
mole 1 Solute g # sequivalent
A/B = # H+ or #OH-
nRedox = #e- involved in balanced
redox equation.
Where:M = N
When n = 1
That is when using HCl, KHP NaOH
But not when using H2SO4, Ca(OH)2
Lesson for MDC students only:
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Acid g-MM (g/)
+ H20 to
HCl 36 g + 1L =
H2SO4 98 g + 1L =
H3PO4 98 g + 1L =
Molarity (M) vs. Normality (N)
Molarity (/L) Normality (eq/L)
1M
1M
1M
1N
2N
3N
=
=
=
g-EW
36/1
=36
98/2
=49
98/3
=32.7
Eq(g/gEW)
36/36
98/49
98/33
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2M H3PO4 3M Ca(OH)2 Using Molarity
1N H3PO4 1N Ca(OH)2 Using Normality
n
2H3PO4 + 3 Ca(OH)2 6 HOH + Na3PO4
N = M or M = N
NA x VA = NB x VB
n
Using Normality for titrations:
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Titration of a WA with a SB
• The pH at the equivalence point will be >7.
• Phenolphthalein is commonly used as an indicator in these titrations.
• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
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Titration: measuring the Equivalence Point (H+ = OH-)
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. The end-point of a titration is when indicator changes color.
Methyl red in base
(range R4-6Y)
Phenolphthalein in base
(range C 8-10 F)
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Titration of a SA with a SB
Excess acid
Excess base
H+ = OH-
xA HN + xB MOH MN(aq) + HOH 1. From the start of the titration
the pH goes up slowly. Just before the equivalence point, the pH increases rapidly.
2. At the equivalence point, moles H+ = moles OH-, and the solution contains only water and the salt from the cation of the base and the anion of the acid.
3. Just after the equivalence point, the pH increases rapidly. As more base is added, the increase in pH again levels off.
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TitrationSolution of unknown concentration (M1 x V1 = #mol1)
Neutralization: 1 mola = 1 molb
(equivalence point){*TitrationMovie}
Solution of known concentration (M2 x V2 = #mol2)
xa
xb
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Stoichiometric/Volumetric Calculations
A
MW
g #
ACID
BASE
BxAxMA x VA MB x VB=
xA
xB
ACID ACID
BASE BASE
xA HN + xB MOH MN + HOH
=Ax
B
MW
g #
Bx=
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2006 (B)
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2007 (A)
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