1
4-4-2011
2
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.
3
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
4
- 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
5
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
6
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
7
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:
8
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
9
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-).
10
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
11
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
12
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)
13
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
14
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
15
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
3
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
16
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
17
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
3
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
18
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
19
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
20
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
21
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
a3
22
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
23
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
4
4a3