PP 23: Buffer Equilibria
Drill: Calculate the pH of 0.20 M HQ.
(KaHQ = 2.0 x 10-4)
Buffer Solutions:
• A solution that resists changes in pH
• Made from the combination of a weak acid & its salt (Acidic buffer solutions)
• Made from the combination of a weak base & its salt (Basic buffer solutions)
• A solution containing acetic acid & sodium acetate (Acidic buffer solution)
• A solution containing ammonia & ammonium chloride (Basic buffer solution)
• An acidic buffer solution works best when the acid to salt ratio is 1 : 1
• A basic buffer solution works best when the acid to base ratio is 1 : 1
• The buffering capacity of a solution works best when the pH is near the pKa or V\ the pKb
• pKa = - log Ka
• pKb = - log Kb
Buffer Equilibria: To solve buffer equilibrium problems, use the same 5 steps
•
Buffer Equilibria Demonstration: Calculate the pH of a solution containing 0.10 M HAc
& 0.10 M NaAc:
(Ka = 1.8 x 10-5)
1. Set up & balance rxns:
HAc  H+ + AcNaAC  Na+ + Ac-
2. Assign Eq. amounts:
HAc  H+ + Ac0.10 – x
x
x
NaAC  Na+ + Ac0.10 – all 0.10 0.10
In buffer solutions, x
3. Write Keq:
Ka = [H+][Ac-] / [HAc]
is always insignificant
4. Substitute:
Ka = [x][0.10 + x] / [0.10 - x]
when added to or
5. Solve for x:
Ka = (x)(0.10) / (0.10)
subtracted from
Ka = x = [H+] = 1.8 x 10-5 M
a real number.
pH = 4.74
Buffer Equilibria problem:
Calculate the pH of a solution containing 0.20 M HBz
& 0.10 M KBz:
(Ka = 6.4 x 10-5)
Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: KaHZ
Buffer Equilibria problems:
 Calculate the pH of 0.10 M NH3 in 0.20 M NH4NO3:
(Kb = 1.8 x 10-5)
 Calculate the pH of a solution containing 0.10 M HBz & 0.20 M KBz: (Ka = 6.4 x 10-5)
 Calculate the pH of a solution containing 0.50 M R-NH2 in 0.10 M R-NH3I: (Kb = 4.0 x 10-5)
Derivations from an equilibrium constant: Rxn:
HA  H+ + A-
Ka = [H+][A-] / [HA]
Divide both sides by [H+]
Ka / [H+] = [A-] / [HA]
This will give you the salt to acid ratio
More Derivations from an equilibrium constant: Rxn: HA  H+ + AKa = [H+][A-] / [HA]
Cross multiply to isolate [H+]
[H+] = Ka [HA] / [A-] = [H+] = Ka ([HA] / [A-])
Take –log of both sides
-log [H+] = -log Ka -log ([HA] / [A-])
-log anything = panything
pH = pKa – log ([HA] / [A-])
change sign & invert log
Henderson-Hassalbach Equation: pH = pKa + log ([A-] / [HA])
Henderson-Hassalbach Equation: pOH = pKb + log ([M+] / [MOH])
Buffer Equilibria problems:
 Calculate the salt to acid ratio to make a buffer solution with pH = 5.0 (Ka for HQ = 2.0 x 10-5)
(Ka for HA = 3.0 x 10-5)
Drill: Calculate the salt to acid ratio to make a buffer solution with pH = 4.7
Buffer Equilibria problems:
 Calculate the salt to base ratio to make a buffer solution with pH = 9.48 (Kb for MOH = 2.0 x 10-5)
Equivalence Point: Point at which the # of moles of the two titrants are equal
14
14
12
12
10
10
[OH-] = [A-2]
8
8 [HA-] = [A-2]
6
6
4
4
2
2
0
0.00
0
10.00
20.00
30.00
40.00
50.00
[H2A] = [OH-]
[H2A] = [HA-]
0
20
40
60
80
Problems:
 Calculate the HCO3- to H2CO3 ratio in blood with pH = 7.40 (Ka1 for H2CO3 = 4.4 x 10-7)
 150 ml of 0.10 M NaOH is added to 100.0 ml of 0.10 M H2CO3. Calculate pH.
(Ka1 for H2CO3 = 4.4 x 10-7) (Ka2 for H2CO3 = 4.8 x 10-11)
100