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.
Larger volumes of buffer solutions have a larger buffer capacity than smaller volumes with the same concentration.
Buffer solutions of higher concentrations have a larger buffer capacity than a buffer solution of the same volume with smaller concentrations.
Buffers with weak acid/conjugate base ratio close to 1 have larger capacities 3

4. You should observe that 0.2>>x
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:
Change -x +x +x
Equil M: x x x
0.20 >> x
1.7x10-5 = [H+][OAc-] = [x][0.20] x = [H+] = 3.4x10-5 M
[HOAc] [0.40]
You should observe that 0.2>>x
pH = - log 3.4x =

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
0.52 -x +x +x x x x
0.30 – x  0.30
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
Calculate the pH of a solution that is mol/L HCN and mol/L NaCN. (Ka of HCN = 4.9 x 10-10)
HCN D H+ + CN-
Even if you do not know it is a buffer you can still work it out easily: 7

8. + -
(all in mol/L)
HCN
(aq) + H
O (l) D H O
(aq) CN
(aq) 2 3
Initial conc. 0.
025
N/A 0. 0. 10
Conc. change - x
N/A +x +x
Equil. conc. 0.
025 – x
N/A +x x
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

9. A diprotic acid, H2A, has Ka1 = 1. 1. 10-3 and Ka2 = 2. 5. 10-6
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 = 5.6
It 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- D H+ + A2-
Ka2 = {[H+][A2-]}/[HA-]
Ka2/[H+] = [A2-]/[HA-]
[H+] = = 1.6*10-6 M
{2.5*10-6/1.6*10-6} = [A2-]/[HA-]
[A2-]/[HA-] = 1.56

11. First, get initial pH before addition of any base
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 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* * 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. DpH = pHfinal – pHinitial DpH = 9.21 – 9.18 = + 0.03
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
DpH = pHfinal – pHinitial
DpH = 9.21 – 9.18 =

14. A 0. 10 M acetic / 0. 10 M acetate mixture has a pH of 4
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 mo
les ) CH
COOH (aq) + OH - -
(aq) g H
O (l) + CH
COO
(aq) 3 2 3
Initial
0.10
0.01
N/A 0. 10
Change - x - x
N/A + x
Final
(where x
= 0.01
0.1 – x
0.01 –
x =
N/A
0.10 + x
due to
limiting OH - )
= 0.09
0.00
= 0.11 14

15. (all in mol/L) CH 3
COOH (aq) H 2
O (l) D H O +
(aq) + CH
COO -
(aq)
Initial conc. 0. 09
N/A
0.0
0.1 1
Conc. change x +x
Equil. conc. –
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
pH = - log [H3O+]
pH = - log 1.5 x 10-5
pH = 4.82 15

16. New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M
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 16

17. Note we’ve made the assumption that x << 0.09 M!
(all in mol/L) CH
COOH (aq) H
O (l) D + - H O
(aq) + CH
COO
(aq) 3 2 3 3
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
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
Note we’ve made the assumption that x << 0.09 M!
pH = - log [H3O+]
pH = - log 2.2 x 10-5
pH = 4.66 17

18. Assume 0.25>>x, which is always safe to assume for buffers
Calculate the pH of a L buffer solution that is 0.25 mol/L in HF and 0.50 mol/L in NaF.
(all in mol/L) H F
(aq) + 2
O (l) D 3 O -
Initial conc.
0.25
N/A
0.0 0. 5
Conc. change x +x
Equil. conc. –
0 + x
Assume 0.25>>x, which is always safe to assume for buffers 18

19. M 10 x 1.8 0.50 0.25 3.5 ] [F [HF] K O [H = pH = - log [H3O+]4 a 3 - + =
pH = - log [H3O+]
pH = - log 1.8 x 10-4
pH = 3.76
No. mol HF = 0.25*0.1 = 0.025
No. mol F- = 0.50*0.1 = 0.050 19

20. Use moles: New HF = 0.027 mol and new F- = 0.048 mol
a) What is the change in pH on addition of mol of HNO3? It is easier here to use moles not molarity
(all in mo
les ) F -
(aq) + H O +
(aq) g H
O (l) + H F
(aq) 3 2
Initial 0.
050
0.0 2
N/A 0.
025
Change - x - x
N/A + x
Final
(where x
= 0.0 2 0. 50 – x
0.0 2 – x
N/A 0. 25 + x
due to
limiting H O + )
= 0. 48
= 0.00
= 0. 27 3
Use moles: New HF = mol and new F- = mol
(all in mol/L) HF
(aq)
+ H
O (l) D H O +
(aq) + F -
(aq) 2 3
Initial conc. 0. 02 7
N/A
0.00 0.
048
Conc. change - x
N/A +x +x
Equil. conc. 0. 02 7 – x
N/A +x 0.
048
+ x )
(0.027
(x)(0.048 10 x
3.5
[HF] ]
][F O [H K 4 3 a = - + 20

21. Notice we’ve made the assumption that x << 0.027 M.
[HF]
0.027 [H O + ] = K =
3.5 x 10 - 4 = -
2.0 x 10 4 M 3 a [F - ]
0.048
Notice we’ve made the assumption that
x << M.
pH = - log [H3O+]
pH = - log 2.0 x 10-4
pH = 3.71 21

22. Use moles: New HF = 0.021 mol and new F- = 0.054 mol
b) What is the change in pH on addition of mol of KOH?
(all in mo
les )
HF (aq) + OH -
(aq) g H
O (l) + F -
(aq) 2
Initial 0.
025
0.004
N/A
0.050
Change - x - x
N/A + x
Final
(where x
= 0.0 4 0.
025 – x
0.04 – x
N/A
0.050 + x
due to
limiting OH - )
= 0.
021
= 0.00
= 0.
054
Use moles: New HF = mol and new F- = mol
(all in mol/L) HF
(aq)
+ H2O
(l) D H + O
(aq) + F -
- (aq) 3
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 [H O +
][F - ]
(x)(0.054 ) - 4 K =
3.5x10 - 4 = 3 so
3.5x10 = a
[HF]
(0.021 ) 22

23. Notice we’ve made the assumption that
[HF]
0.21 + - x - 4 [H O ] = K =
3.5 x 10 4 =
1.4 10 M 3 a - [F ]
0.54
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 23