Unit 19 Acid Base Equilibria: Titrations

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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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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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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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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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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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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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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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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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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SA

SB SA

SB

(1) Titration of a SA with a SB (or SB with a SA)

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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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(H2CO3 + K+)

(H2CO3 + Ca2+ + C2H3O2- )

(H2CO3 + Zn2+ + SO42-)

(H2C O3 + Zn2+)

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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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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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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


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• 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.


HA ↔ H+ + A-

SB (OH-)

HA + OH- H2O + A-

]HA[

]A][H[

AK

][

][log

HA

ApKpH a

*what is used for equilibrium Rx?*


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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


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(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(][


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(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


(.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.


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(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


(.15M)(.025L) (.10M)(.038L)

pH + pOH = pKw pH = pKw - pOH

pH = 14 – 5.2 = 8.8

Salt is NaC2H3O2


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2005B Q1

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Mole ICEnd Table:

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• 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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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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2007A Q1 Titration: weak acid with strong base

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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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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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Titrations of Polyprotic Acids

In these cases there is an equivalence point for each dissociation.

Ka1

Ka2

Ka3

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2005B Q1 Titration: weak acid strong base

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2005B Q1

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2000 QA Titration: weak base strong acid

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2001 Q3Titration: weak acid strong base

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Answers 2001 Q3

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2002A Q1 Titration: weak acid strong base

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2003A Q1 Titration: weak base strong acid

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2006B Q1 Titration: weak acid strong base

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Expressing Concentrations of

Solutions: Molarity (& Normality*)

* For MDC students only!

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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=

=


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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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For titrations:

MM-g

solute g #

MM-g

mole 1 solute g # moles

AAA

MB x VBAx Bx =

Since


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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


=Ax

B

MW

g #

Bx=

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2006 (B)

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2007 (A)

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Unit 19 Acid Base Equilibria: Titrations

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