Buffer Solutions
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
http://academic.cengage.com/chemistry/moore
Chapter 16
Additional Aqueous Equilibria
Stephen C. Foster • Mississippi State University
Buffer Action
Buffers are made with ~equal quantities of a
conjugate acid-base pair.
(e.g. CH3COOH + CH3COONa)
Buffer Action
1 L Solution
Initial
pH
after HCl
after NaOH
pure H2O
7.00
2.00
12.00
[CH3COOH] = 0.5 M
+ [CH3COONa] = 0.5 M
4.74
4.72
4.76
A buffered solution
Added OH- removed by the acid:
CH3COOH + OHCH3COO- + H2O
Added H3O+ removed by the conjugate base:
CH3COO- + H3O+
CH3COOH + H2O
The equilibrium maintains the acid/base ratio.
CH3COOH + H2O
CH3COO- + H3O+
pH remains stable.
The pH of Buffer Solutions
Blood pH
• Blood is buffered at pH = 7.40 ± 0.05.
 pH too low: acidosis.
 pH too high: alkalosis.
• CO2 generates the most important blood buffer.
H2CO3(aq) + H2O(ℓ)
Example
Add 0.010 mol of HCl or NaOH to:
Buffer Action
Buffers must contain:
• A weak acid to react with any added base.
• A weak base to react with any added acid.
• These components must not react with each other.
CO2(aq) + H2O(ℓ)
Buffer = chemical system that resists changes in pH.
H2CO3(aq)
H3O+(aq) + HCO3-(aq)
Depends on the [acid]/[base] – not absolute amounts.
HendersonHenderson-Hasselbalch equation
[H3O+] = Ka
[HA]
[A-]
log [H3O+] = log Ka + log
pH = pKa + log
Note: pH = pK
pKa
[A-]
[HA]
[HA]
[A-]
With pKa = -log Ka
when [HA] = [A-]
1
The pH of Buffer Solutions
Common Buffers
What is the pH of a 0.050 M (monoprotic) pyruvic acid
+ 0.060 M sodium pyruvate buffer? Ka = 3.2 x 10-3.
pH
weak acid
weak base
Ka(weak acid) pKa
4 lactic acid
lactate ion
1.4 x 10-4
3.85
5 acetic acid
acetate ion
1.8 x 10-5
4.74
6 carbonic acid
hydrogen carbonate ion
4.2 x 10-7
6.38
= 2.49 + 0.08
7 dihydrogen phosphate hydrogen phosphate ion
6.2 x 10-8
7.21
= 2.57
8 hypochlorous acid
hypochlorite ion
3.5 x 10-8
7.46
9 ammonium ion
ammonia
5.6 x 10-10
9.25
carbonate ion
4.8 x 10-11 10.32
pH = – log(3.2 x
10-3)
+ log(0.060/0.050)
What is the HPO42-/H2PO4- ratio in blood at pH =7.40.
Ka(H2PO4- ) = Ka,2(H3PO4) = 6.2 x 10-8.
7.40 = −log(6.2 x 10-8) + log([HPO42-]/[H2PO4-])
log([HPO42-]/[H2PO4-]) = 0.192
[HPO42-]/[H2PO4-] = 1.5
Addition of Acid or Base to a Buffer
1.0 L of buffer is prepared with [NaH2PO4] = 0.40 M and
[Na2HPO4] = 0.25 M. Calculate the pH of: (a) the buffer
(b) after 0.10 mol of NaOH is added.
Ka (H2PO4-) = 6.2 x 10-8
pH = pKa + log
[A-]
[HA]
pH = 7.21 + log 0.25 = 7.01
0.40
Buffer Capacity
The amount of acid (or base) that can be added
without large pH changes.
[A-]/[HA] determines the buffer pH.
Magnitude of [A-] and [HA] determine buffer capacity
The buffer capacity for:
• Base addition = nconjugate acid
• Acid addition = nconjugate base
Useful buffer range: pH = pKa ±1
(10:1 or 1:10 ratio of [A-]/[HA]).
Addition of Acid or Base to a Buffer
1.00 L Buffer: [NaH2PO4]= 0.40 M ; [Na2HPO4] = 0.25 M. (b) calculate pH after
0.10 mol of NaOH is added. Ka (H2PO4-) = 6.2 x 10-8
(b) 0.10 mol of NaOH, converts conj. acid to base:
H2PO4- + OH0.40
0.10
(0.40 – 0.10) 0
ninitial
nadded
nafter
pKa = -log(6.2 x 10-8) = 7.21
(a) No base added:
10 hydrogen carbonate
pH = pKa + log
[A-]
[HA]
→ HPO42- + H2O
0.25
(0.25 + 0.10)
Use n directly. [ ] = n/ V and V is
the same for both (cancels)
pH = 7.21 + log 0.35 = 7.28
0.30
Acid-Base Titrations
• Standard solution (titrant
titrant) is added from a buret.
• The equivalence point occurs when a stoichiometric
amount of titrant has been added.
• Use a pH meter.
• An indicator is used to find an end point.
point
• Color change observed.
• End point ≠ equivalence point (should be close...).
ntitrant = nanalyte
ntitrant = Vtitrant [titrant]
nanalyte= Vanalyte[analyte]
2
Detection of the Equivalence Point
Acid-Base Indicator = weak acid that changes color
with changes in pH.
HIn(aq) + H2O(ℓ)
color 1
H3O+(aq) + In-(aq)
color 2
Observed color will vary (depends on [HIn] and [In-] in
solution).
Ka =
Detection of the Equivalence Point
The acidic form dominates when the [HIn] >> [In-]
If
pH = pKa -1
Basic color shows when [In-] >> [HIn]
[In-] = 10
[HIn]
Ka =
- pKa = 1 - pH
Detection of the Equivalence Point
[H3O+][In-]
[H3O+]
=
[HIn]
10
Ka =
- pKa = - pH - 1
If
[H3O+][In-]
[HIn]
[HIn] = 10
[In-]
[H3O+][In-]
= 10 [H3O+]
[HIn]
pH = pKa +1
Titration of Strong Acid with Strong Base
AcidAcid-base titration curve = plot of pH vs Vtitrant added.
Titrate 50.0 mL of 0.100 M HCl with 0.100 M NaOH
Red
pH ≤ 4
Yellow
pH ≥ 6
Initial pH = -log(0.100) = 1.000 (fully ionized acid)
Moles of acid = 0.0500 L
Methyl Red
Bromothymol blue
Phenolphthalein
0.100 mol
L
= 5.00 x 10-3 mol
Titration of Strong Acid with Strong Base
Titration of Strong Acid with Strong Base
After 40.0 mL of base added (before equivalence)
Base removes H3O+ :
H3O+
+
OH2 H2O
After Equivalence (50.2 mL added)
All acid consumed.
original nacid – total nbase added
[H3O+] =
Vacid (ℓ) + Vbase added (ℓ)
-3
-3
[H3O+] =(5.00 x 10 – 4.00 x 10 ) mol= 0.0111 M
(0.0500 + 0.0400) L
pH = -log(0.0111) = 1.95
At Equivalence (50.0 mL added)
All acid and base react. Neutral salt. pH = 7.00.
nOH- added = 0.0502 L
0.100 mol
= 5.02 x 10-3 mol
L
original nacid = 0.0500 L 0.100 mol
L
= 5.00 x 10-3 mol
nOH- remaining = 0.02 x 10-3 mol
Vtotal = (0.0500 + 0.0502) L = 0.1002 L
pOH = - log(0.02 x10-3 / 0.1002) = 3.70
pH = 14.00 – 3.70 = 10.30
3
Titration of Strong Acid with Strong Base
Vtitrant/mL Vexcess/mL Vtotal/mL
[OH-]/mol L-1
Titration of Strong Acid with Strong Base
pH
50.0
0
100.0
0
7.00
50.1
0.1
100.1
0.0001
10.00
50.2
0.2
100.2
0.0002
10.30
51
1
101.0
0.0010
11.00
55
5
105.0
0.0048
11.67
60
10
110.0
0.0091
11.96
70
20
120.0
0.0167
12.22
80
30
130.0
0.0200
12.30
0.1 mL of base
increased the
pH by 3 units!
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
More complicated.
• Weak acid is in equilibrium with its conjugate base.
Example
Titrate 50.0 mL of 0.100 M acetic acid with 0.100 M
NaOH. What is the pH after the following titrant
additions: 0 mL, 40.0 mL, 50.0 mL, and 50.2 mL?
40.0 mL added
0 mL added
HA + H2O
Ka = 1.8 x 10-5 =
2
[H3O+][A-]
≈ x
[HA]
(0.100)
50 mL of
0.100 M
ninitial
nadded
nleft
HA(aq) + OH-(aq)
+
H2O(ℓ)
0.00500
0.00400
0.00100
0.00400
Each OH- removes 1 HA…
nleft = 0.00500 - 0.00400
H3O+ + A-
A-(aq)
…and makes 1 A40 mL of 0.100 M
x= 0.0013
Vtotal = (50.0 + 40.0) mL = 90.0 mL = 0.0900 L
pH = -log(0.0013) = 2.88
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
40.0 mL added
(c) 50.0 mL added
Equivalence. All HA is converted to A-.
Henderson-Hasselbalch:
pH = -log(1.8 x 10-5) + log
[A-] = nA-/ V
(0.00400/0.0900)
(0.00100/0.0900)
[HA] = nHA/ V
HA(aq) + OH-(aq)
ninitial
nadded
nleft
A-(aq)
+
H2O(ℓ)
0.00500
0.00500
0
0.00500
Note: V cancels (could be omitted)
pH = 4.74 + log
0.0400
= 5.34
0.0100
Vtotal= 100.0 mL, so
[A-] = 0.00500 mol/0.100 L = 0.0500 M
A- is basic!
4
Titration of Weak Acid with Strong Base
Titration of Weak Acid with Strong Base
Use Kb to solve for the [OH-] generated by [A-]:
(c) 50.2 mL added
0.2 mL of “extra” OH- dominates pH.
Ignore any contribution from A-.
Kb =
[OH-][HA]
[A-]
A+
(0.0500 - x)
Kb = Kw / Ka = 5.6 x 10-10
H2 O
Kb = 5.6 x 10-10 ≈
pOH = 5.28
HA
x
x2
0.0500
+
OHx
x = 5.3 x 10-6 M
[OH-] = (0.0002 L x 0.100 mol/L) / 0.1002 L
= 2.0 x 10-4 M
pOH = 3.7
pH = 14.0 – 3.7
= 10.3
pH = 14 - pOH = 8.72
Titration of Weak Acid with Strong Base
Titration of Weak Base with Strong Acid
Titrate of 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH.
Volume of 0.100 M HCl added (mL)
pH = pKa at 50% titration (25 mL) pKa (acetic acid) = 4.74.
Short vertical section compared to strong acid/strong base.
Solubility Equilibria and KSP
Some ionic compounds are slightly water soluble.
Saturation occurs at low concentration.
AgCl(s)
50.0 mL of 0.100 M ammonia titrated with 0.100M HCl.
• pH = pKa at 50% to equivalence (pKa = 9.25).
Solubility Equilibria and KSP
Example
Ksp(AgCl)=1.8 x 10-10. Calculate the solubility (mol/L).
Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-] = 1.8 x 10-10
The solubility product constant is:
Ksp = [ Ag+] [ Cl-]
At equilibrium,
[Ag+] = [Cl-] = S
Then:
1.8 x 10-10 = (S)(S) = S2
S = 1.3 x 10-5 M
As always, [ ]
of solid is
omitted
Solubility = 1.3 x 10-5 mol/L.
5
Factors Affecting Solubility
Common Ion Effect
pH and Dissolving Slightly Soluble Salts Using Acids
Insoluble salts containing anions of Bronsted-Lowry
bases dissolve in acidic solutions.
AgCl is slightly soluble in water:
Ca2+(aq) + CO32- (aq)
Ksp = 8.7 x 10-9
Pure water: CaCO3(s)
Acid:
Add more Ag+ or Cl-. Equilibrium will move left.
Called the common ion effect.
CO3
HCO3
Ag+(aq) + Cl-(aq)
AgCl(s)
-(aq)
2-(aq)
+ H3
+ H3
O+(aq)
O+(aq)
H2CO3(aq)
HCO3
-(aq)
+ H2O(ℓ)
H2CO3(aq) + H2O(ℓ)
CO2(g) + H2O(ℓ)
K ≈ 105
e.g. AgCl is less soluble in a NaCl solution than in
water. The “common ion” is Cl-.
Solubility and the Common Ion Effect
Solubility and the Common Ion Effect
Example
Calculate the solubility of PbI2 in (a) water (b) 0.010 M
NaI. Ksp (PbI2) = 8.7 x 10-9.
(b) NaI supplies I - lowering the PbI2 solubility.
(a) Pure water:
Pb2+
S
PbI2
+
2 I2S
PbI2
[ ]initial (from NaBr)
[ ]change (from PbBr2)
[ ]equil
Pb2+
0
S
S
+
2 I0.010
2S
0.010 + 2S
Ksp = 8.7 x 10-9 = [Pb2+][I -]2 = S(2S + 0.010)2
Ksp = [Pb2+][I -]2 = (S)(2S)2 = 4S3 = 8.7 x 10-9
S = 1.3 x
10-3
M
Assume S << 0.010
8.7 x 10-9 ≈ S(0.010)2
S = 8.7 x 10-5 M
≈ 15 times lower than in (a)
Complex Ion Formation
Complex Ion Formation
Metal ions (Lewis acids) can react with Lewis bases.
Complex ion solubility can be evaluated using the
formation constant,
constant Kf.
Example
AgBr(s) + 2S2O32-(aq)
Ag(S2O3)23-(aq) + Br -(aq)
AgBr(s)
Ag+(aq) + 2 S2O32-(aq)
The complex is much more soluble than AgBr.
Black & white photography
net AgBr(s) + 2 S2O32-(aq)
Knet = KspKf = [Ag+][Br -]
Ag+(aq) + Br -(aq)
Ag(S2O3)23-(aq)
Ag(S2O3)23-(aq) + Br -(aq)
[Ag(S2O3)23- ]
[Ag(S2O3)23- ][Br -]
=
[Ag+][S2O32-]2
[S2O32-]2
light
AgBr in film
 Ag(s)
“Fix” a negative: complex and
dissolve unexposed Ag salts.
Ksp = 3.3 x10-13, Kf = 2 x 1013, so Knet = 7
The reaction is favored.
6
Amphoterism
Precipitation: Will It Occur?
Calculate Q, compare it to Ksp.
If Q > Ksp
• Q must decrease.
• remove ions, precipitate solid.
If Q = Ksp
• at equilibrium (saturated solution).
If Q < Ksp
• Q must increase.
• dissolve more solid (if present). Form more ions.
Precipitation: Will It Occur?
Precipitation: Will It Occur?
Mix 25.0 mL of 0.0025 M HCl and 10.0 mL of 0.010 M
AgNO3. Will AgCl precipitate? Ksp (AgCl) = 1.8 x 10-10.
Slowly add HCl(aq) to a solution that is 0.010 M in Cu+
and 0.500 M in Pb2+. Which salt will precipitate first?
Ksp for PbCl2 and CuCl are 1.7 x 10-5 and 1.9 x 10-7.
nCl-
= 0.0250 L (0.0025 mol/L)
nAg+ = 0.0100 L (0.010 mol/L)
= 6.25 x 10-5 mol
= 1.0 x 10-4 mol
CuCl: KSP= 1.9 x 10-7 = [Cu+][Cl-] = 0.010[Cl-]
Vtotal = (0.0250 + 0.0100) = 0.0350 L.
[Cl-] = 6.25 x 10-5 mol/ 0.0350 L = 1.79 x 10-3 mol/L
[Ag+]
= 1.0 x
10-4
mol / 0.0350 L = 2.86 x
10-3
mol/L
Q = [Ag+][Cl-] = (1.79 x 10-3)(2.86 x 10-3) = 5.1 x 10-6
Q > KSP
Find the minimum [Cl-] that will cause precipitation:
[Cl-] = 1.9 x 10-5 M
Least soluble, precipitates 1st
PbCl2: KSP = 1.7 x 10-5 = [Pb2+][Cl-]2 = 0.500[Cl-]2
[Cl-] = 5.8 x 10-3 M
Precipitation will occur
7