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9 Aqueous Solutions and Chemical Equilibrium
9A The Chemical Composition of Aqueous Solutions
9A-1 Classifying Solutions of Electrolytes
Table 9-1 Classification of Electrolytes
Strong
Weak
1. Inorganic acids such as HNO3,
HClO4, H2SO4, HCl, Hl, HBr,
HClO3, HBrO3
2. Alkali and alkaline-earth
hydroxides
1. Many inorganic acids, including H2CO3,
H3PO4, H2S, H2SO3
2. Most organic acids
3. Ammonia and most organic bases
4. Halides, cyanides, and thiocyanates of Hg,
Zn, and Cd
3. Most salts
Salt: reaction of an acid with a base, NaCl, Na2SO4, and NaOOCCH3
9A-2 Acids and Bases
1923, J. N. Brønsted (Denmark) and J. M. Lowry (England),
Brønsted - Lowry theory:
Acid → proton donor
Base → proton acceptor.
What are Conjugate Acids and Bases?
Acid1 ↔ base1 + proton; Acid1 and base1: conjugate acid/base pair
Base2 ↔ proton + Acid2
Acid1 + Base2 ↔ base1 + Acid2
neutralization
NH3
base1
+
H 2O
Acid2
↔
NH4+
conjugate acid1
+
OHconjugate base2
H 2O
base1
+ HNO2
Acid2
↔
H3O+
conjugate acid1
+
NO2conjugate base2
NO2base1
+
↔
HNO2
conjugate acid1
+
OHconjugate base2
H 2O
Acid2
9A-3 Amphiprotic Species
Species that posses both acidic and basic properties are amphiprotic.
Amphiprotic solvent (that can act either as an acid or as a base):
Water, methanol, ethanol, and anhydrous acetic acid.
NH3
base1
+ CH3OH ↔
Acid2
NH4+
conjugate acid1
+
CH3Oconjugate base2
CH3OH + HNO2
Acid2
base1
↔
CH3OH2+
conjugate acid1
+
NO2conjugate base2
H2PO4- + H3O+
Acid2
base1
↔
H3PO4
acid1
+
H 2O
base2
H2PO4- +
acid1
↔
HPO42base1
+
H 2O
Acid2
OHbase2
Zwitterions: a species that bears both a positive and a negative charge.
NH2CH2COOH
glycine
↔
R’COOH + R”NH2
base2
acid1
↔
NH3+CH2COO zwitterion
R’COObase1
+
R”NH3+
Acid2
9A-4 Autoprotolysis
self-ionization (autoionization) to form a pair of ionic species
base1
Acid2
autoprotolysis
acid1
H 2O
←→
H 3O +
+
OH-
CH3OH +
CH3OH
←→
CH3OH2+
+
CH3O-
HCOOH +
HCOOH
←→
HCOOH2+
+
HCOO-
NH3
←→
NH4+
+
NH2-
H 2O
NH3
+
+
36
base2
9A-5 Stength of Acids and Bases
Strongest acid
Weakest acid
HClO4 + H2O →
HCl + H2O →
H3PO4 + H2O ↔
Al(H2O)63+ + H2O ↔
HC2H3O2 + H2O ↔
H2PO4- + H2O ↔
NH4+ + H2O ↔
H3O+ + ClO4H3O+ + ClH3O+ + H2PO4H3O+ + AlOH(H2O)52+
H3O+ + C2H3O2H3O+ + HPO42H3O+ + NH3
Weakest base
Strongest base
Fig. 9-1 Dissociation reactions and relative strengths of some common acids and their
conjugate bases. Note that HCl and HClO4, are completely dissociated in water.
The tendency of a solvent to accept or donate protons determines the strength of a solute
acid or base dissolved in it.
In H2O: HCl and HClO4 → strong acid
In anhydrous acetic acid: HClO4 stronger than HCl (dissociation >5000x)
CH3COOH +
base1
HClO4 ⇔
Acid2
CH3COOH2+
Acid1
+ ClO4Base2
differentiating solvent: Acetic acid
(various acids dissociate to different degree → different strength)
in acetic acid: HClO4 > HCl > HNO3 > H2SO4
leveling solvent: Water for HClO4, HCl, HNO3, and H2SO4.
(several acids are completely dissociated → the same strength)
9B Chemical Equilibrium
9B-1 The Equilibrium State
(4-5)
H3AsO4 + 3I- + 2H+ ⇔ H3AsO3 + I3- + H2O
orange-red color appear
H3AsO3 + I3- + H2O ⇔ H3AsO4 + 3I- + 2H+
orange-red color disappear
The Le Châtelier principle: the position of chemical equilibrium always shifts in a
direction as to relieve a stress that is applied to the system.
The mass-action effect: shift in the position of an equilibrium caused by adding one of
the reactants or products to a system.
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Chemical reaction do not cease at equilibrium. Instead, the amounts of reactants and
products are constant because the rates of the forward and reverse processes are
identical.
Thermodynamics: a branch of chemical science, deals with the flow of heat and energy
in chemical reactions. The position of a chemical equilibrium is related to these energy
changes.
9B-2 Equilibrium-Constant expressions
wW + xX ⇔ yY + zZ
K=
[Y ] y [ Z ] z
[W ] w [ X ] x
K=
aYy × a Zz
aWw × a Xx
9B-3 Types of equilibrium constants encountered in analytical chemistry
Table 9-2 Equilibria and equilibrium constant of importance to analytical chemistry
Type of
Names and symbol
Typical
Equilibrium-constant
equilibrium
of equil. const.
Example
expression
Dissociation of Ion-product const.
Kw= [H3O+] [OH-]
2H2O ⇔ H3O+ + OHwater
Kw
Heterogenous
equilibrium between
Solubility
Ksp= [Ba2+] [SO42-]
a slightly soluble
BaSO4⇔ Ba2+ + SO42product, Ksp
substance and its ions
in a saturated soln
[H 3O + ][CH 3COO - ]
CH3COOH + H2O ⇔
Ka =
Dissociation Dissociation
H3O+ + CH3COO[CH 3COOH]
of a weak
constant, Ka or
[OH - ][CH 3COOH ]
Kb
acid or base
CH3COO- + H2O ⇔
Kb =
OH- + CH3COOH
[CH COO - ]
3
Formation
Formation of
a complex ion constant, βn
Oxidation/
reduction
equilibrium
Kredox
Distribution equilibrium
for a solute between
immiscible solvents
2+
-
Ni + 4CN ⇔ Ni(CN) 4
MnO4 + 5Fe2+ + 8H+ ⇔
Mn2+ + 5Fe3+ + H2O
Kd
2-
K redox =
I2(aq) ⇔ I2(org)
38
[Ni(CN) 24- ]
β4 =
[Ni 2+ ][CN - ] 4
[Mn 2 + ][Fe3+ ]5
[MnO 4− ][Fe2 + ]5 [H + ]8
Kd =
[I 2 ]org
[I 2 ]aq
9B-4 Applying the Ion-Product Constant for Water
2H2O ⇔ H3O+ + OH[H 3 O + ][OH - ]
K=
[H 2 O] 2
K[H2O]2 = Kw = [H3O+][OH-]
pKw = pH + pOH
At 25°C, pKw = 14.00
Table 4-3
Variation of Kw with Temperature
Temperature
Kw
0
0.114 × 10-14
25
1.01 × 10-14
50
5.44 × 10-14
100
49 × 10-14
pKw
(14)
Ex 9-1 Calculate the [H3O+] and [OH-] of pure water at 25°C and 100°C.
[H3O+] = [OH-], [H3O+]2 = [OH-]2 = Kw
At 25°C, [H3O+] = [OH-] =
1.00 × 10 −14 = 1.00 × 10 − 7
At 100°C, [H3O+] = [OH-] =
49 × 10 −14 = 7.0 × 10 − 7
Ex 9-2 Calculate the [H3O+] and [OH-] and the pH and pOH of 0.200 M aqueous NaOH
at 25°C.
[OH-]= 0.200 + [H3O+], [OH-] ≈ 0.200,
pOH = -log 0.200 =0.699
−14
Kw
1.00 × 10
[H 3 O + ] =
=
= 5.00 × 10 −14 M , pH =-log 5.00×10-14 =13.301
0.200
[OH ]
9B-4 Applying the Solubility-Product Constants
Ba2+(aq) + 2IO3-(aq)
Ba(IO3)2(s) ⇔
[Ba 2+ ][IO 3- ] 2
K=
; K [Ba(IO3)2(s)]= Ksp =[Ba2+][IO3-]2= 1.57×10-9
[Ba(IO3 ) 2 ( s )]
Ex 9-3 How many grams of Ba(IO3)2 (487 g/mol) can be dissolved in 500 mL of water
at 25°C?
[Ba(IO3)2] = [Ba2+],
[IO3-] = 2[Ba2+]
[Ba2+][IO3-]2 = 4[Ba2+]3 = 1.57×10-9
1.57 × 10 −9
= 7.32 × 10 − 4 M
solubility = [Ba ] =
4
2+
3
mass Ba(IO3)2 = 7.32 × 10-4 × 500 × 0.487 = 0.178 g
39
Ex 9-4 Calculate the [Ba(IO3)2] in a 0.0200 M Ba(NO3)2 solution.
[Ba(IO3)2] = ½[IO3-], [Ba2+] = 0.0200 + ½[IO3-] ≈ 0.200
[Ba2+][IO3-]2 = (0.0200 + ½[IO3-])[IO3-]2 = 1.57 × 10-9
−9
0.0200 [IO3-]2 = 1.57 × 10-9 ; [IO3-] = 1.57 × 10 = 2.80 × 10-4 M
0.0200
solubility of Ba(IO3)2 = ½ [IO3-] = ½ (2.80 × 10-4 M) = 1.40 × 10-4 M
Ex 9-5 Calculate the molar solubility of [Ba(IO3)2] in a ( 200 mL 0.0100 M Ba(NO3)2 +
100 mL 0.100 M NaIO3) soln.
No. mmol Ba2+ = 200 mL × 0.0100 M = 2.00 mmol
No. mmol IO3- = 100 mL × 0.100 M = 10.0 mmol
No. mmol excess IO3- = 10.0 – 2(2.00) = 6.0 mmol
[IO3-] = 6.0 mmol/300 mL = 0.0200 M
molar solubility of Ba(IO3)2 = [Ba2+]
[IO3-] = 0.0200 + 2[Ba2+] ≈ 0.0200
1.57 × 10 −9
2+
= 3.93 × 10 −6 M
solubility of Ba(IO3)2 = [Ba ] =
2
0.0200
9B-6 Applying Acid and base Dissociation Constants
HNO2 + H2O ⇔ H3O+ + NO2-
Ka =
NH3 + H2O ⇔ NH4+ + OH-
[H 3O + ][ NO −2 ]
[HNO 2 ]
Kb =
[ NH +4 ][OH - ]
[ NH 3 ]
Dissociation constants for conjugated acid/base pairs
NH3 + H2O ⇔ NH4+ + OH[ NH +4 ][OH - ]
Kb =
[ NH 3 ]
NH4+ + H2O ⇔ NH3 + H3O+
[ NH3 ][H 3O + ]
Ka =
[ NH +4 ]
Ex 9-6 What is Kb for the equilibrium CN- + H2O ⇔ HCN + OH-10
HCN: Ka = 6.2 × 10
;
Kb =
K w 1.00 × 10 −14
=
= 1.61 × 10 −5
−10
Ka
6.2 × 10
40
Ka Kb = Kw
[H3O+] in a weak acid solution
HA + H2O ⇔ H3O+ + A+
-
[H O ][A ]
Ka = 3
[HA]
2 H2O ⇔ H3O+ + OHKw = [H3O+][OH-]
[A-] ≈ [H3O+]
-
CHA = [A ] + [HA]
= [H3O+] + [HA]
[HA] = CHA - [H3O+]
Ka =
[H 3O + ]2
C HA − [H 3O + ]
[H3O+]2 + Ka[H3O+] - KaCHA = 0
- K a + K a 2 + 4 K a C HA
[H 3O ] ==
2
+
if [H3O+] << CHA, K a =
[H 3O + ]2
C HA
[H 3O + ] = K a C HA
Ex 9-7 Calculate the [H3O+] in 0.120 M nitrous acid
+
HNO2 + H2O ⇔ H3O+ + NO2- K a = [H 3O ][NO 2 ] = 7.1 × 10 − 4
[HNO 2 ]
+
[H3O ] = [NO2 ], [HNO2] = 0.120 - [H3O+]
[H3O + ]2
if [H3O ] << 0.120 →
= 7.1 × 10− 4
0.120
+
[H 3O + ] = 0.120 × 7.1 × 10 − 4 = 9.2 × 10 −3 M
Ex 9-8 Calculate the [H3O+] in 2.0 × 10-4 M aniline hydrochloride, C6H5NH3Cl
C6H5NH3+ + H2O ⇔ C6H5NH2 + H3O+
Ka =
[H3O + ][C6 H5 NH 2 ]
[C6 H5 NH3+ ]
= 2.51× 10−5
[H3O+] = [C6H5NH2], [C6H5NH3+] = 2.0 × 10-4 - [H3O+]
+ 2
if [H3O+] << 2.0 × 10-4 → [H3O ] = 2.51× 10−5
2.0 × 10− 4
[H 3O + ] = 5.02 × 10−9 = 7.09 × 10−5 M (not <<<< 2.0 × 10-4)
[H 3O + ]2
2.0 × 10 − 4
+ 2
-5
+
-9
= 2.51 × 10 − 5 → [H3O ] + 2.51×10 [H3O ] – 5.02×10 = 0
− 2.51×10 −5 + (2.51×10 −5 ) 2 + 4 × 5.02 ×10 −9
[H 3O ] =
= 5.94 ×10 −5 M
2
+
41
[H3O+] in a weak base solution
+
NH3 + H2O ⇔ NH4+ + OH- ; K b = [NH 4 ][OH ]
[ NH 3 ]
+
+
2H2O ⇔ H3O + OH ; Kw = [H3O ][OH-]
[OH − ] = K bC B
Ex 9-9 Calculate the [OH-] in 0.0750 M NH3 solution
NH3 + H2O ⇔ NH4+ + OH-
[NH +4 ][OH - ] K w 1.0 × 10 −14
= 1.75 × 10 −5 M
=
=
Kb =
−
10
[ NH3 ]
K a 5.70 × 10
[NH4+]= [OH-], [NH4+]+ [NH3] = C NH 3 = 0.0750M
-
[NH3] =0.0750 - [OH ] →
[OH − ]2
-
0.0750 − [OH ]
= 1.75 × 10 −5
if [OH-] << 0.0750 → [OH-]2 ≈ 0.0750 × 1.75 × 10-5
[OH-] = 1.15 × 10-3 M
Ex 9-10 Calculate the [OH-] in 0.0100 M sodium hypochlorite solution
HOCl: Ka = 3.0 × 10-8
OCl- + H2O ⇔ HOCl + OH-
Kb =
[HOCl][OH- ]
[OCl− ]
=
K w 1.0 × 10 −14
=
= 3.33 × 10− 7 M
−
8
K a 3.0 × 10
[HOCl]= [OH-], [OCl-]+ [HOCl] = 0.0100M
[OCl-] = 0.0100 - [OH-] ≈ 0.0100 (assume [OH-] << 0.0100)
[OH − ]2
= 3.33 × 10− 7 →
0.0100
[OH-] = 5.8 × 10-5 M
9C Buffer Solutions
Buffer: a mixture of a weak acid and its conjugate base or a weak base and its
conjugate acid that resists changes in pH of a solution
9C-1 Calculation of the pH of Buffer Solutions
*Weak Acid/Conjugate Base Buffers
42
A- + H2O ⇔ OH- + HA
HA + H2O ⇔ H3O+ + AK w [OH − ][HA]
[H 3O + ][A - ]
=
Kb =
Ka =
[HA]
Ka
[A - ]
[HA] = CHA - [H3O+] + [OH-], [HA] ≈ CHA
C
[H 3O + ] = K a HA
C NaA
[A-] = CNaA + [H3O+] - [OH-], [A-] ≈ CNaA
Ex. 9-11 What is the pH of a solution that is 0.400 M in HCOOH and 1.00 M in
HCOONa?
HCOOH + H2O ⇔ H3O+ + HCOO- Ka = 1.80 × 10-4
HCOO- + H2O ⇔ HCOOH + OH- Kb = K w = 5.56 × 10-11
Ka
Ka =
[H 3O + ][HCOO - ]
= 1.80 × 10 − 4 [HCOO ] ≈ CHCOO- = 1.00;
[HCOOH]
[H 3O + ] = 1.80 × 10 − 4 ×
[HCOOH] ≈ CHCOOH = 0.400
[H3O+] << CHCOOH and [H3O+] << CHCOO- are valid.
0.400
= 7.20 × 10 − 5
1.00
pH = -log (7.20 × 10-5) = 4.14
*Weak Base/Conjugate Acid Buffers
Ex. 9-12 Calculate the pH of a solution that is 0.200 M in NH3 and 0.300 M in
NH4Cl.
NH4+ + H2O ⇔ NH3 + H3O+
Ka = 5.70 × 10-10
NH3 + H2O ⇔ NH4+ + OH- Kb =
K w 1.00 × 10 −14
=
= 1.75 × 10-5
Ka
5.70 × 10-10
[NH4+] = CNH Cl + [OH-] - [H3O+] ≈ CNH Cl + [OH-]
4
4
+
[NH3] = CNH + [H3O ] - [OH ] ≅ CNH - [OH-]
3
3
[NH3] ≈ CNH = 0.200
[NH4+] ≈ CNH Cl = 0.300;
4
[H 3O + ] =
−
[OH ] =
3
K a × [ NH 4+ ] 5.70 × 10−10 × 0.300
=
= 8.55 × 10−10
[ NH3 ]
0.200
1.00 × 10−14
8.55 × 10
−10
= 1.17 × 10− 5 << CNH4Cl or CNH3
pH = - log (8.55 × 10-10) = 9.07
43
9C-2 Properties of Buffer solutions
*The Effect of Dilution:
Fig. 9-4 The effect of dilution of the pH of
buffered and unbuffered solutions. The
dissociation constant for HA is 1.00 × 10-4.
Initial solute concentrations are 1.00 M
Concentration of reagents, M
*The Effect of Added Acids and bases
Ex. 9-13 calculate the pH change that takes place when a 100-mL portion of (a)
0.0500 M NaOH and (b) 0.0500 M HCl is added to 400 mL of the buffer
solution that was described in Ex. 9-12..
(a) Addition of NaOH converts part of the NH4+ in the buffer to NH3:
NH4+ + OH- ⇔
C NH 3 =
NH3 + H2O
Ka = 5.70 × 10-10
400 × 0.200 + 100 × 0.0500 85.0
=
= 0.170M
500
500
C NH 4 Cl =
400 × 0.300 − 100 × 0.0500 115
=
= 0.230M
500
500
[ H 3O + ] = 5.70 × 10 −10 ×
0.230
= 7.71 × 10 −10
0.170
pH = - log (7.71 × 10-10) = 9.11,
ΔpH = 9.11 - 9.07 = 0.04
(b) Addition of HCl converts part of the NH3 in the buffer to NH4+:
NH3 + H3O+ ⇔ NH4+ + H2O
400 × 0.200 − 100 × 0.0500 75.0
=
= 0.150M
500
500
400 × 0.300 + 100 × 0.0500 125
C NH 4 Cl =
=
= 0.250M
500
500
0.250
[H 3O + ] = 5.70 × 10 −10 ×
= 9.50 × 10 −10
0.150
C NH 3 =
pH = - log (9.50 × 10-10) = 9.02,
ΔpH = 9.02 - 9.07 = - 0.05
44
*The Composition of Buffer Solutions as a function of pH; Alpha Values
In a HAc/NaAc buffer solution
cT = cHOAc + cNaOAc,
[OAc- ]
[HOAc]
α0 =
, α1 =
,
CT
CT
α0 + α1 = 1
HOAc + H2O ⇔ H3O+ + OAc[H 3 O + ][OAC - ]
K a [ HOAc]
Ka =
, [OAC ] =
,
[H 3O + ]
[HOAc]
⎛ [H 3O + ] + K a
K a [ HOAc]
cT = [ HOAc] +
= [ HOAc]⎜⎜
+
[H 3O + ]
⎝ [H 3O ]
[ H 3 O + ][OAC - ]
[ HOAc] =
Ka
⎞
[H 3O + ]
[HOAc]
⎟⎟ ⇒ α 0 =
=
CT
[H 3 O + ] + K a
⎠
+
[ H 3 O + ][OAc - ]
- ⎛ [H 3 O ] + K a
cT =
+ [OAC ] = [OAC ]⎜⎜
Ka
Ka
⎝
⎞
Ka
[OAc - ]
⎟⎟ ⇒ α 1 =
=
CT
[H 3O + ] + K a
⎠
Alpha values depend only on [H3O+] and Ka and are independent on cT.
*Buffer Capacity, β:
the number of moles of strong acid or strong base that causes 1.00 L of the buffer to
undergo a 1.00-unit change in pH.
β=
dC b
dC a
=−
dpH
dpH
Fig. 9-5. Variation in α with pH. Note that most of the
transition between α0 and α1 occurs within ±1 pH unit of
the crossover point of the two curves. The crossover point
where α0 = α1 = 0.5 occurs when pH = pKHOAc = 4.74
45
Fig. 9-6. Buffer capacity as
a function of the ratio
CNaA/CHA.
Preparation of Buffers
Ex. 9-14 Describe how you might prepare approximately 500.0 mL or a pH 4.5 buffer
solution from 1.0 M acetic acid (HOAc) and sodium acetate (NaOAc).
pH 4.5 ⇒ [H3O+] = 10-4.5 = 3.16 × 10-5 M
[ H 3 O + ][OAC - ]
[OAC - ] 1.75 × 10 −5 1.75 × 10 −5
−5
=
=
= 0.5534
= 1.75 × 10 ⇒
Ka =
[HOAc]
[H 3 O + ]
3.16 × 10 −5
[ HOAc]
[OAc-] = 0.5534 × 1.0 M = 0.5534 M
mass NaOAc = 0.5534 M × 0.5 L × 82.034 g/mol NaOAc = 22.7 g NaOAc
Dissolve 22.7 g NaOAc in the HOAc solution (in 500 mL volumetric flask). Check
the pH with a pH meter, adjust the pH by adding a small amount of acid or base.
46