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4-4-2011

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Choosing a Buffer

• Pick a buffer whose acid has a pKa near the pH that you want

• Must consider pH and capacity

Buffer Capacity: Not all buffers resist changes in pH by the same extent. The effectiveness of a buffer to resist changes in pH is called the buffer capacity.

Buffer capacity is the measure of the ability of a buffer to absorb acid or base without significant change in pH.

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Buffer capacity increases as the concentration of acid and conjugate base increases.

1. Larger volumes of buffer solutions have a larger buffer capacity than smaller volumes with the same concentration.

2. Buffer solutions of higher concentrations have a larger buffer capacity than a buffer solution of the same volume with smaller concentrations.

3. Buffers with weak acid/conjugate base ratio close to 1 have larger capacities

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- Calculate the pH of a solution which is 0.40 M in acetic acid (HOAc) and 0.20 M in sodium acetate (NaOAc). Ka=1.7x10-5

HOAc H+ + OAc-

Initial M: 0.40 0 0.20Change -x +x +xEquil M: 0.40-x x 0.20+x

0.20 >> x1.7x10-5 = [H+][OAc-] = [x][0.20] x = [H+] = 3.4x10-5 M

[HOAc] [0.40]

You should observe that 0.2>>x

pH = - log 3.4x10-5 = 4.47

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What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

0.30 – x 0.30

0.52 + x 0.52

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Ka = {0.52 * x}/0.30

1.7*10-4 = 0.52*x/0.3

X = 9.4*10-5

Again 0.3 is really much greater than x

X = [H+] = 9.4*10-5 M

pH = 4.02

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Calculate the pH of a solution that is 0.025 mol/L HCN and 0.010 mol/L NaCN. (Ka of HCN = 4.9 x 10-10)

HCN H+ + CN-

Even if you do not know it is a buffer you can still work it out easily:

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Ka = {x*(0.01+x)}/(0.025 – x), assume 0.01>>x

However if you observe it is a buffer you could work it without including the “x” as it is always very small.

4.9X10-10 = {x*(0.01)}/(0.025)X = [H+] = 1.2*10-9 M, and pH = 8.91

(all in mol/L) HCN ( aq ) + H2

O )l( H3O

+( aq ) + CN

-( aq)

Initial conc. 0. 025 N/A 0. 0 0. 010

Conc. change -x N/A +x +x

Equil. conc. 0. 025 – x N/A +x 0.010 + x

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A diprotic acid, H2A, has Ka1 = 1.1*10-3 and Ka2 = 2.5*10-6. To make a buffer at pH = 5.8, which combination would you choose, H2A/HA- or HA-/A2-? What is the ratio between the two buffer components you chose that will give the required pH?

The weak acid/conjugate base system should have a pKa as close to the required pH as possible.

pKa1 = 2.96, while pKa2 = 5.6It is clear that the second equilibrium should

be used (HA-/A2-).

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The ratio between HA- and A2- can easily be found from the equilibrium constant expression where:

HA- H+ + A2-

Ka2 = {[H+][A2-]}/[HA-]

Ka2/[H+] = [A2-]/[HA-]

[H+] = 10-5.8 = 1.6*10-6 M

{2.5*10-6/1.6*10-6} = [A2-]/[HA-]

[A2-]/[HA-] = 1.56

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Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? Kb = 1.8*10-5

NH4+ (aq) H+ (aq) + NH3 (aq)

First, get initial pH before addition of any base

Ka = {[H+][NH3]}/[NH4+]

)10-14/1.8*10-5( = {[H+] * 0.30}/0.36

[H+] = 6.7*10-10

pHinitial = 9.18

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Now calculate the pH after addition of the base:

Addition of the base will decrease NH4+ and will

increase NH3 by the same amount due to a 1:1 stoichiometry, which is always the case.

mmol of NH4+

= 0.36*80 – 0.05*20 = 27.8

mmol NH3 = 0.30*80 + 0.05 * 20 = 25

Can use mmoles instead of molarity, since both ammonia and ammonium are present in same solution

NH4+ (aq) H+ (aq) + NH3 (aq)

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NH4+ (aq) H+ (aq) + NH3 (aq)

Ka = {[H+][NH3]}/[NH4+]

)10-14/1.8*10-5( = {[H+] * 25}/27.8

[H+] = 6.2*10-10

pHfinal = 9.21

pH = pHfinal – pHinitial

pH = 9.21 – 9.18 = + 0.03

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A 0.10 M acetic / 0.10 M acetate mixture has a pH of 4.74 and is a buffer solution! What happens if we add 0.01 mol of NaOH (strong base) to 1.00 L of the acetic acid – acetate buffer solution?

This reaction goes to completion since OH- is a strong base and keeps occurring until we run out of the limiting reagent OH-

)all in moles( CH 3 COOH (aq) + OH- ) aq ( H2 O (l) + CH 3COO-

) aq(

Initial 0.10 0.01 N/A 0. 10

Change -x - x N/A +x

Final )where x =0.01

due to limiting OH- (

0.10 – x

=0.09

0.01 – x =

0.00

N/A 0.10 + x

=0.11

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With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check, after calculations are done!), we find

x)(0.09

x)(x)(0.1110 x 1.8

COOH][CH

]COO][CHO[HK 5

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33a

pH = - log [H3O+]pH = - log 1.5 x 10-5

pH = 4.82

)all in mol/L( CH3

COOH (aq) + H2

O (l) H3O

+) aq + (CH

3COO

-) aq(

Initial conc. 0. 09 N/A 0.0 0.11

Conc. change -x N/A +x +x

Equil. conc. 0. 09 – x N/A +x 0.11 + x

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What happens if we add 0.01 mol of HCl (strong acid) to 1.00 L of the acetic acid – acetate buffer solution?

CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)

This reaction goes to completion and keeps occurring until we run out of the limiting reagent H3O+

New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M

)all in moles( CH3COO- )aq( + H3O+ )aq( → H2O )l( + CH3COOH )aq(

Initial 0.10 0.01 N/A 0.10 Change -x -x N/A +x Final )where x = 0.01 due limiting H3O

+( 0.10 – x = 0.09

0.01 – x = 0.00

N/A 0.10 + x = 0.11

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With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check, after calculations are done!), we find

x)(0.11

x)(x)(0.0910 x 1.8

COOH][CH

]COO][CHO[HK 5

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33a

Note we’ve made the assumption that x << 0.09 M!pH = - log [H3O+]pH = - log 2.2 x 10-5

pH = 4.66

)all in mol/L( CH3COOH (aq) + H 2 O (l) H

3O

+) aq + (CH

3COO-

) aq(

Initial conc. 0.11 N/A 0.0 0. 09

Conc. change -x N/A +x +x

Equil. conc. 0.11 – x N/ A +x 0. 09 + x

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Calculate the pH of a 0.100 L buffer solution that is 0.25 mol/L in HF and 0.50 mol/L in NaF.

x)(0.25

x)(x)(0.5010 x 3.5

[HF]

]][FO[HK 43

a

)all in mol/L( HF ) aq ( + H2

O (l) H3

O+

) aq ( + F-

) aq(

Initial conc. 0.25 N/A 0.0 0. 50

Conc. change -x N/A +x +x

Equil. conc. 0.25 – x N/A +x 0. 5 0 + x

Assume 0.25>>x, which is always safe to assume for buffers

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pH = - log [H3O+]pH = - log 1.8 x 10-4

pH = 3.76No. mol HF = 0.25*0.1 = 0.025No. mol F- = 0.50*0.1 = 0.050

M 10 x 1.80.50

0.2510 x 3.5

[]F

]HF[10 x 3.5

[]F

]HF[K[O]H

44

4a3

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a) What is the change in pH on addition of 0.002 mol of HNO3? It is easier here to use moles not molarity

Use moles: New HF = 0.027 mol and new F- = 0.048 mol

)all in moles( F - ) aq ( +H3 O+ ) aq ( H 2 O (l) + HF ) aq(

Initial 0. 050 0.002 N/A 0. 025

Change -x - x N/A +x

Final )where x =0.0 02

due to limiting H3 O+ (

0. 050 – x

=0. 048

0.002 – x

=0.00

N/A 0. 025 + x

=0. 027

)all in mol/L( HF ) aq ( +H2

O (l) H3O

+) aq + ( F

-) aq(

Initial conc. 0. 027 N/A 0.00 0. 048

Conc. change -x N/A +x +x

Equil. conc. 0. 027 – x N/A +x 0. 048 +x

)(0.027

)(x()0.04810 x 3.5

]HF[

[][FO]HK

43a

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Notice we’ve made the assumption that x << 0.027 M.

pH = - log [H3O+]pH = - log 2.0 x 10-4

pH = 3.71

M 10 x 2.00.0480.02710 x 3.5

[]F

]HF[K[O]H 44

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b) What is the change in pH on addition of 0.004 mol of KOH?

(all in moles ) HF )aq( +OH - ( aq ) H2O )l( + F - ( aq) Initial 0. 025 0.004 N/A 0.050 Change -x - x N/A +x Final (where x =0.0 4 due to limiting OH-)

0. 025 – x =0. 021

0.04 – x =0.00

N/A 0.050 + x =0. 054

)(0.021

)(x()0.0543.5x10 so

]HF[

[][FO]H3.5x10K

434a

(all in mol/L) HF ( aq ) +H2O ( l ) H 3O+

( aq ) + F- - )aq(

Initial conc. 0.021 N/A 0.00 0.054 Conc. change -x N/A +x +x Equil. conc. 0.021 – x N/A +x 0.054 +x

Use moles: New HF = 0.021 mol and new F- = 0.054 mol

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Notice we’ve made the assumption that x << 0.21 M. We should check this!

pH = - log [H3O+]pH = - log 1.4 x 10-4

pH = 3.87

M 10x 1.40.54

0.2110 x 3.5

[]F

]HF[K[O]H

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4a3