? 10 3 . 85 6 K K pH [ H ] w w [ OH ] - - PDF document

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CEE 680 Lecture #16 2/21/2020 Print version Updated: 21 February 2020 Lecture #16 Buffers & Titrations (Benjamin, Chapter 5) (Stumm & Morgan, Chapt.1 3 ) David Reckhow CEE 680 #16 1 Titrations of Acids and Bases Weak acid

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SLIDE 1

CEE 680 Lecture #16 2/21/2020 1

Lecture #16 Buffers & Titrations (Benjamin, Chapter 5)

(Stumm & Morgan, Chapt.1‐3 )

David Reckhow CEE 680 #16 1

Updated: 21 February 2020

Print version

Titrations of Acids and Bases

3 4 5 6 7 8 0.2 0.4 0.6 0.8 1 f value pH David Reckhow CEE 680 #16 2

 Weak acid with a strong base

3HC C O O H 3HC C O O

NaOH H2O

Na

HCl NaCl

HAc NaAc

HAc NaOH

85 . 3 3 7 . 4

10 10 10 ] [

   

  

T aC

K H

 

85 . 7 3 3 . 9 14

10 10 10 10 ] [ ] [

     

   

T b w w

C K K OH K H

?

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SLIDE 2

CEE 680 Lecture #16 2/21/2020 2

Defining the Titration Curve

 A titration is complete when the equivalents of

titrant (t) added equals the equivalents of sample (s)

  • riginally present

 equt = equs  VtNt = VsNs

 we can define the extent of a base titration as:  At any point from the start of the titration, we have a

mixed solution of the acid and conjugate base

 We must use the ENE in place of the PBE

David Reckhow CEE 680 #16 3 s B s s B B

moles equ M V N V f  

Defining the Titration Curve (cont.)

 The ENE is:

 for this problem (titration of HAc with NaOH):

 [Na+] + [H+] = [Ac‐] + [OH‐]

 and in general, for a base titration:

 CB  [Na+] = [A‐] + [OH‐] ‐ [H+]

 and combining with the definition for f:

David Reckhow CEE 680 #16 4 T T T B s B s s B B

C H OH C H OH A C C moles equ M V N V f ] [ ] [ ] [ ] [ ] [

1     

         

Amount of base added at any point during the titration in equivalents/liter Amount of acid originally present in moles/liter (which is the same as the total of acid + conjugate base present throughout)

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SLIDE 3

CEE 680 Lecture #16 2/21/2020 3

Alpha & f curves

0.2 0.4 0.6 0.8 1 2 4 6 8 10 12 14 pH alpha

David Reckhow CEE 680 #16 5 3 4 5 6 7 8 9 10 11 0.2 0.4 0.6 0.8 1 f value pH

f1 1 0

Log C and f curves

 Titration of

10‐2 M HAc

 compare to

Stumm & Morgan’s Figure 3.3

David Reckhow CEE 680 #16 6

pH

1 2 3 4 5 6 7 8 9 10 11 12 13 14

g

0.0 0.2 0.4 0.6 0.8 1.0

pH 3.35 pH 4.7 pH 8.35 Starting Point Mid-point End Point

pH

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Log C

  • 7
  • 6
  • 5
  • 4
  • 3
  • 2
  • 1

H+

OH-

HAc Ac-

f

0.0 0.2 0.4 0.6 0.8 1.0

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SLIDE 4

CEE 680 Lecture #16 2/21/2020 4

Reverse Titration (acid)

 The reverse titration is the addition of a strong acid

(e.g., HCl) to the fully titrated acetic acid (e.g., NaAc). This re‐forms the original HAc and produces NaCl too.

 we can define the extent of an acid titration as:  As with the forward titration, we have a mixed

solution of the acid and conjugate base

 We must use the ENE in place of the PBE

David Reckhow CEE 680 #16 7 s A s s A A

moles equ M V N V g  

Reverse titration (cont.)

 The ENE is:

 for this problem (titration of NaAc with HCl):

 [Na+] + [H+] = [Ac‐] + [OH‐] + [Cl‐]

 and for an acid titration of a pure base (Na form):

 CT  [HA] + [A‐] = [Na+]

 and combining with the definition for g:

David Reckhow CEE 680 #16 8 T T T A s A s s A A

C OH H C OH H HA C C moles equ M V N V g ] [ ] [ ] [ ] [ ] [

   

         

Amount of acid added at any point during the titration in equivalents/liter Amount of base originally present in moles/liter (which is the same as the total of acid + conjugate base present throughout)

[Cl-] = [Na+] - [Ac-] + [H+] - [OH-] CA  [Cl-] = [HA] + [H+] - [OH-]

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SLIDE 5

CEE 680 Lecture #16 2/21/2020 5

 For a monoprotic acid/base:

 f + g equals 1 throughout a titration

David Reckhow CEE 680 #16 9

1 ] [ ] [ ] [ ] [

1 1

         

  

   

T T

C OH H C H OH g f

pH Buffers & Buffer Intensity

 Definitions

 Buffer: a solution that resists large pH changes when

a base or acid is added

 commonly a mixture of an acid and its conjugate base

 Buffer Intensity: the amount of strong acid or strong

base required to cause a small shift in pH

 Significance

 Natural Waters

 wide range  poorly buffered waters are susceptible to acid precipitation

David Reckhow CEE 680 #16 10

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SLIDE 6

CEE 680 Lecture #16 2/21/2020 6

 Engineered Processes

 certain treatments need large pH shifts (e.g., softening)  others need to resist large shifts (e.g., biotreatment)

 Laboratory

 buffers needed to calibrate pH meters  used in experimentation to maintain constant pH. This

simplifies data analysis and interpretation

David Reckhow CEE 680 #16 11

Making a Buffer

Acid & conjugate base

 best to have a reservoir of each so

there is resistance to change in both directions

 Mirror questions

 Given a desired pH,

what should the buffer composition be?

 Given an acid/conjugate

base mixture, what will the pH be?

David Reckhow CEE 680 #16 12

HA A- HA A-

Base Acid

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SLIDE 7

CEE 680 Lecture #16 2/21/2020 7

Acetic Acid System: Alpha & f curves

0.2 0.4 0.6 0.8 1 2 4 6 8 10 12 14 pH alpha

David Reckhow CEE 680 #16 13 3 4 5 6 7 8 9 10 11 0.2 0.4 0.6 0.8 1 f value pH

Base addition

1 0

Acid addition

Buffers: Acetic Acid & Sodium Acetate Example

 1. List all species present

 H+, OH‐, HAc, Ac‐, Na+

 2. List all independent equations

 equilibria

 Ka = [H+][Ac‐]/[HAc] = 10‐4.77  Kw = [H+][OH‐] = 10‐14

 mass balances

 CHAc + CNaAc = [HAc]+[Ac‐]

 electroneutrality: (positive charges) = (negative charges)

 Note: we can’t use the PBE because we’re adding an acid and its

conjugate base

 [Na+] + [H+] = [OH‐] + [Ac‐]

David Reckhow CEE 680 #16 14

1 2 3 4 Five total

CNaAc = [Na+]

5

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SLIDE 8

CEE 680 Lecture #16 2/21/2020 8

Simplified HAc/NaAc Example

 3. Use simplified ENE & solve for Ac‐ and HAc

 [Na+] + [H+] = [OH‐] + [Ac‐]  [Na+]  [Ac‐]  CNaAc  [Ac‐]

 4. Plug back in to Ka equation and solve for H+

 Ka = [H+][Ac‐]/[HAc]  Ka = [H+] CNaAc/ CHAc  [H+]= KaCHAc/CNaAc  pH = pKa + log(CNaAc/CHAc)

 or more generally

 pH = pKa + log(CA/CHA)

David Reckhow CEE 680 #16 15

4

Kw = [H+][OH-] [OH-] = Kw/[H+]

2 3 1 3+4+5 1+3+4+5

Assumes [Na+]>>[H+], and [Ac-]>>[OH-] CNaAc = [Na+]

5

CHAc + CNaAc = [HAc]+[Ac-] CHAc + CNaAc = [HAc]+CNaAc CHAc = [HAc]

4+5

Henderson‐Hasselbalch Equation

 Classic H‐H equation

 Just a re‐arrangement of equilibrium equation

 Always correct

 Empirical H‐H

 Assumes buffer salts swamp H+ and OH‐

David Reckhow CEE 680 #16 16

HA A a

C C pK pH log   ] [ ] [ log HA A pK pH

a 

 

Lawrence Henderson was a biochemist, born 3 Jun 1878 in Lynn MA, established the fatigue lab at Harvard

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SLIDE 9

CEE 680 Lecture #16 2/21/2020 9

Simplified HAc/NaAc Example (cont.)

 Solution #1

 CNaAc (= CA) = 10 mM  CHAc (= CHA) = 10 mM

 Solution #2

 CNaAc (= CA) = 20 mM  CHAc (= CHA) = 2 mM

David Reckhow CEE 680 #16 17

7 . 4 10 10 log 7 . 4 log     

HA A a

C C pK pH 7 . 5 2 20 log 7 . 4 log     

HA A a

C C pK pH

Observations

  • 1. pH = pKa, when equal amounts of acid and conjugate base are added
  • 2. pH is independent of CT (eventually at low CT this breaks down)

Exact Solutions: Summary

Monoprotic

Acids:

 [H+]3 + {Ka}[H+]2 ‐ {Kw + KaC}[H+] ‐ KWKa = 0

Bases:

 [H+]3 + {C+Ka}[H+]2 ‐

{Kw}[H+] ‐ KWKa = 0

Mixed Acid/Bases (i.e., buffers):

 [H+]3 + {CA+Ka}[H+]2 ‐ {Kw + KaCHA}[H+] ‐ KWKa = 0

David Reckhow CEE 680 #16 18

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SLIDE 10

CEE 680 Lecture #16 2/21/2020 10

To next lecture

David Reckhow CEE 680 #16 19