15 ACID-BASE EQUILIBRIAgencheminkaist.pe.kr/Lecturenotes/CH101/Chap15_2021.pdf · 2021. 2. 15. ·...
Transcript of 15 ACID-BASE EQUILIBRIAgencheminkaist.pe.kr/Lecturenotes/CH101/Chap15_2021.pdf · 2021. 2. 15. ·...
General Chemistry II
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ACID-BASE EQUILIBRIA15CHAPTER
General Chemistry II
15.1 Classifications of Acids and Bases
15.2 Properties of Acids and Bases in Aqueous
Solutions: The Brønsted-Lowry Scheme
15.3 Acid and Base Strength
15.4 Equilibria Involving Weak Acids and Bases
15.5 Buffer Solutions
15.6 Acid-Base Titration Curves
15.7 Polyprotic Acids
15.8 Organic Acids and Bases: Structure and Reactivity
15.9 Exact Treatment of Acid-Base Equilibria
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Cyanidin is blue in the basic sap of the cornflower
and red in the acidic sap of the poppy.
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Acid Base
Arrhenius [H3O+ ] > KW
1/2 [OH- ] > KW1/2
Brønsted-Lowry donates H+ accepts H+
Lewis accepts donates
lone-pair electrons lone-pair electrons
15.1 CLASSIFICATIONS OF ACIDS AND BASES
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Arrhenius Acids and Bases
Acid: A substance that, when dissolved in water, increases the
concentration of hydronium ion (H3O+) above the value in
pure water.
HCl(aq) + H2O H3O+(aq) + Cl-(aq)
Base: A substance that increases the concentration of hydroxide
ion (OH–).
NaOH(aq) Na+(aq) + OH-(aq)
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Acid: A substance that can donate a proton
Base: A substance that can accept a proton
Brønsted-Lowry conjugated acid-base pairs:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
acid1 base2 base1 acid2
Brønsted-Lowry Acids and Bases
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Lewis Acids and Bases
Acid: Any species that accepts lone-pair electrons
Base: Any species that donates lone-pair electrons
Competition between two bases for a proton by offering
electron pairs:
2 3
2 11 2
HF( ) H O ( )
acid
H O( ) F ( )
bas acid
e ba
se
aq a aql q
Reactions without proton transfers
~ Octet-deficient compound (BF3) ← strong Lewis acid
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~ removing H2O from oxoacids or hydroxides
Acid anhydrides: Oxides of most of the nonmetals
N2O5(s) + H2O(l) HNO3(aq)
Base anhydrides: Oxides of Group I & II metals
CaO(s) + H2O(l) Ca(OH)2(s)
Amphoteric: Oxides of Group III & V metals
Al2O3(s) + 6 H3O+(aq) 2 Al3+(aq) + 9 H2O(l)
Al2O3(s) + 2 OH- (aq) + 3 H2O(l) 2 Al(OH)4- (aq)
Anhydrides of Acids and Bases
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Fig. 15.2 Acidity and basicity of oxides of main group elements.
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Autoionization of Water
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
acid1 base2 acid2 base1
2 H2O(l) H3O+(aq) + OH-(aq)
14
w 3 [H O ][OH ] 1.0 10 K
[H3O+] = [OH-] = 1.0 x10–7 M
for pure water at 25°C
15.2 PROPERTIES OF ACIDS AND BASES IN
AQUEOUS SOLUTIONS: THE BRØ NSTED-
LOWRY SCHEME
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Strong Acids and Bases
Strong Acids
~ ionizes fully in aqueous solution producing H3O+
H2O(l) + HCl(aq) H3O+(aq) + Cl-(aq)
base2 acid1 acid2 base1
Leveling Effect of water on HCl, HBr, HI, H2SO4, HNO3, HClO4
~ too strong to tell the difference in water
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Strong Bases
~ ionizes fully in aqueous solution producing OH-,
amide ion (NH2-), hydride ion (H-), NaOH, ...
H2O(l) + NH2-(aq) OH-(aq) + NH3(aq)
acid1 base2 base1 acid2
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The pH Function
10 3 pH log [H O ]
Fig. 15.4 pH’s of many everyday materials
aqueous solution at 25°C
pH + pOH = 14
pH < 7 acidic (can be negative)
pH = 7 neutral
pH > 7 basicFig. 15.3 A simple pH meter
with a digital readout.
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Hydrolysis (ionization) of a weak acid
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Acid ionization (hydrolysis) constant, Ka
The stronger an acid, the larger Ka (smaller pKa).
pKa of strong acids < 0
pKa of H3O+ = 0
pKa of weak acids > 0
pKa of H2O = 14
15.3 ACID AND BASE STRENGTH
𝐾𝑎 =𝐻3𝑂
+ 𝐴−
𝐻𝐴, 𝑝𝐾𝑎 = − log10 𝐾𝑎
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Hydrolysis of a weak base
A-(aq) + H2O(l) HA(aq) + OH-(aq)
or B(aq) + H2O(l) BH+(aq) + OH-(aq)
[HA] [OH ][H O ]
[H
[HA][OH ]=
[A ] [ O
[HA]
[H O ][A] ]A ]K
+
3
+
w
a3
b +
3
w
-
-
-
- -
K K
K
w a b w a b , p p p K K K K K K
Base hydrolysis constant, Kb
𝐾𝑏 =𝐻𝐴 𝑂𝐻−
𝐴−, 𝑝𝐾𝑏 = − log10 𝐾𝑏
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Fig. 15.5 The relative strength
of some acids and their
conjugate bases.
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~ Prediction of the direction of net hydrogen ion transfer
HF(aq) + H2O(l) H3O+(aq) + F-(aq), Ka = 6.6×10-4
HCN(aq) + H2O(l) H3O+(aq) + CN-(aq), Ka' = 6.17×10-10
HF a stronger acid than HCN
→ Equilibrium is strongly to the right.
HF(aq) + CN-(aq) HCN(aq) + F-(aq), K = Ka/Ka' =1.1×106
acid1 base2 acid2 base1
(strong) (strong) (weak) (weak)
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Molecular Structure and Acid Strength
Fig. 15.6 (a) Basic compound, electropositive X, breaking X – O bond.
(b) Acidic compound, electronegative X, breaking O – H bond.
–X–O–H group ~ Oxoacid (electronegativity of X, pKa)
NaOH (0.93, basic)
HClO3 (3.16, –3) > HNO3 (3.04, –1.3) > HIO3 (2.66, 0.80)
H3PO4 (2.19, 2.12) > H3AsO4 (2.18, 2.30)
H2SO3 (2.58, 1.81) > H2CO3 (2.55, 6.37)
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Fig. 15.7 Lewis diagram for H3PO3. (a) Wrong triprotic structure.
(b) Correct diprotic structure. Assignment of the formal charge to P
and the lone O. P – H bond is not breaking.
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Indicators
Organic weak acid that has a different color from its
conjugate base
HIn(aq) + H2O(l) H3O+(aq) + In-(aq)
+[H O ][In ]
[HIn]3
a
K K
3
a
+[H O ][HIn]=
[In ]
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Indicators
Organic weak acid that has a different color from its
conjugate base
HIn(aq) + H2O(l) H3O+(aq) + In-(aq)
+[H O ][In ]
[HIn]3
a
K K
3
a
+[H O ][HIn]=
[In ]
Range of color change: pH ~ pKa ± 1
Fig. 15.8 bromophenol red, thymolphthalein, phenolphthalein, bromocresol green
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Fig. 15.9 Indicators change
their colors at very different
pH values.
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Fig. 15.10 Natural indicator: Red cabbage extract in a natural pH indicator.
The color changes from red to violet to yellow as the solution becomes
less acidic.
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Weak Acids
HOAc H3O+ AcO–
--------------------------------------------------------------------------Initial 1.000 ~ 0 0Change –y + y + y
---------------- ------ -----Equilibrium 1.000 – y y y--------------------------------------------------------------------------
+[H O ][Ac ]1.76 10
[HAc] 1.000
253
a
Ky
y
→ y = 4.20×10–3
[H3O+] = y = 4.20×10–3 M → pH = 2.38
𝑦2
1000 − 𝑦≈
𝑦2
1000= 1.76 × 10−5
Fraction ionized = [Ac–] / [HAc]0 = y / 1.000 = 4.20×10–3 → 0.42%
EXAMPLE 15.6
15.4 EQUILIBRIA INVOLVING WEAK ACIDS
AND BASES
Calculate the pH and the fraction of HOAc ionized at equilibrium.
HOAc(aq) + H2O(l) H3O+(aq) + AcO-(aq)
1 M
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HOAc H3O+ AcO–
--------------------------------------------------------------------------Initial 0.00100 ~ 0 0Change –y + y + y
---------------- ------ -----Equilibrium 0.00100 – y y y--------------------------------------------------------------------------
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Weak Bases
H2O(l) + NH3(aq) NH4+(aq) + OH–(aq)
--------------------------------------------------------------------------
Initial 0.0100 0 ~ 0
Change –y + y + y
---------------- ------ -----
Equilibrium 0.0100 – y y y
--------------------------------------------------------------------------
+[NH ][OH ]
1.8 10[NH ] 0.0100
254
b
3
Ky
y
y = 4.15 ×10–4 M = [OH–]
[H3O+] = Kw / [OH–] = 2.4 ×10–11 M → pH = 10.62
EXAMPLE 15.8 Calculate the pH of an aqueous solution of ammonia.0.01 M
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Hydrolysis
AcO–(aq) + H2O(l) HOAc(aq) + OH–(aq) --------------------------------------------------------------------------
Initial 0.100 0 ~ 0
Change –y + y + y
---------------- ------ -----
Equilibrium 0.100 – y y y
--------------------------------------------------------------------------
EXAMPLE 15.9 Hydrolysis of NaOAc
NaOAc(s) + H2O(l) Na+(aq) + AcO–(aq)
0.1 M
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Hydrolysis of a weak base
A-(aq) + H2O(l) HA(aq) + OH-(aq)
or B(aq) + H2O(l) BH+(aq) + OH-(aq)
[HA] [OH ][H O ]
[H
[HA][OH ]=
[A ] [ O
[HA]
[H O ][A] ]A ]K
+
3
+
w
a3
b +
3
w
-
-
-
- -
K K
K
w a b w a b , p p p K K K K K K
Base hydrolysis constant, Kb
𝐾𝑏 =𝐻𝐴 𝑂𝐻−
𝐴−, 𝑝𝐾𝑏 = − log10 𝐾𝑏
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Hydrolysis
AcO–(aq) + H2O(l) HOAc(aq) + OH–(aq) --------------------------------------------------------------------------
Initial 0.100 0 ~ 0
Change –y + y + y
---------------- ------ -----
Equilibrium 0.100 – y y y
--------------------------------------------------------------------------
y = 7.5 × 10–6 M = [OH–]
[H3O+] = Kw / [OH–] = 1.3 × 10–9 M → pH = 8.89
[HAc][OH ]5.7 10
[Ac ] 0.100
2
w
a
10
b
KK
K
y
y
EXAMPLE 15.9 Hydrolysis of NaOAc
NaOAc(s) + H2O(l) Na+(aq) + AcO–(aq)
0.01 M
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Buffer solution ~ maintains an approximately constant pH
Weak acid + Salt containing its conjugate base
(eg. HOAc/NaOAc)
15.5 BUFFER SOLUTIONS
- Controlling the solubility of ions
- Maintaining pH of biochemical and physiological
processes
blood pH 7.4 (7.0 – 7.8)
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Calculations of Buffer Action
HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq)
---------------------------------------------------------------------------------------
Initial 1.00 ~ 0 0.500
Change –y + y + y
--------------- ------ -------------
Equilibrium 1.00 – y y 0.500 + y
---------------------------------------------------------------------------------------
EXAMPLE 15.10 Calculate the pH of a solution of HCOOH and NaHCOO.
Buffer solution of HCOOH (1.00 mol) / NaHCOO (0.500 mol) in 1L
[H O ][HCOO ]1.77 10
[HCOOH] 1.00
+43
a
0.500
Ky +y
y
1.77 101.00 1.00
40.500 0.500y +y y
y
y = [H3O+] = 3.54×10–4 M → pH = 3.45
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Testing the buffer strength.
Buffer solution of HCOOH (1.00 mol) / NaHCOO (0.500 mol) in 1L
+ 0.10 mol of HCl
EXAMPLE 15.11
1. Before considering ionization of HCOOH….
HCl ionizes completely → reacts with HCOO– to give HCOOH
[HCOO–]0 = 0.500 – 0.10 = 0.40 M
[HCOOH]0 = 1.00 + 0.10 = 1.10 M
2. Now consider ionization of HCOOH
HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq)
-----------------------------------------------------------------------------------------Initial 1.10 ~ 0 0.40Change –y + y + y
--------------- ------ -------------Equilibrium 1.10 – y y 0.40 + y-----------------------------------------------------------------------------------------
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K+
43a
0.40[H O ][HCOO ]1.77 10
[HCOOH]
y +y
y1.10
40.40 0.401.77 10
y +y y
y1.10 1.10
y = [H3O+] = 4.9 ×10–4 M → pH = 3.31
Addition of 0.100 mol HCl to
Buffer solution of Ex. 15.10: pH = 3.45 → 3.31
Pure water: pH = 7 → 1
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Designing Buffers
2 3HA( ) H O( ) H O ( ) A ( )aq l aq aq
3a
03
0
[H O ][A ] [H O ][A ]
[HA] [HA]K
a 100
0
[HA] pH p log
[A ]K
Determining pH of the buffer solution
1. Choose a weak acid whose pKa ≈ pH
2. Fine-tuning of pH by adjusting the ratio of [HA]0 / [A–]0
Henderson-Hasselbalch Equation
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a 100
0
[HA] pH p log
[A ]K
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Capacity of the Buffer Solution
Fig. 15.12 pH change of buffer solutions as a strong base (NaOH) is added.
Red line: 100 mL of a buffer that is 0.1 M in both HAc and Ac–.
Blue line: 100 mL of a buffer that is 1.0 M in both HAc and Ac–.
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Titration of a Strong Acid
with a Strong Base
Titration of 100.0 mL of
0.1000 M HCl with
0.1000 M NaOH at 25°C
H3O+(aq) + OH–(aq)
2 H2O(l)
15.6 ACID-BASE TITRATION CURVES
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Region I: Before the equivalence point
The pH determined by the excess H3O+.
2. V = 30.00 mL NaOH added
n(OH–) = (0.1000 mol/L)(0.0300 L) = 3.000×10–3 mol
n(H3O+) = (1.000×10–2 – 3.000×10–3 ) mol = 7.00×10–3 mol
Volume increased: Vtot = 100.0 mL + 30.00 mL = 0.1300 L
[H3O+] = n(H3O
+) / Vtot = (7.00×10–3 mol) / (0.1300 L)
= 0.0538 M → pH = 1.27
1. V = 0 mL NaOH added
[H3O+] = 0.1000 M → pH = 1.00
n(H3O+) = (0.1000 mol/L)(0.1000 L) = 1.000×10–2 mol
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Region II : At the equivalence point
The pH determined by the dissociation of water.
3. V = 100.0 mL NaOH added → pH = 7.00
Region III: Beyond the equivalence point
The pH determined by the excess OH–.
4. V = 100.05 mL NaOH added
n(OH–) = (0.1000 mol/L)(5×10–5 L) = 5×10–6 mol
Vtot = 0.1000 L (HCl) + 0.10005 L (NaOH)
= 0.20005 L
[OH–] = n(OH–) / Vtot = (5×10–5 mol) / (0.20005 L) = 2.5×10–5 M
→ [H3O+] = 4×10–10 M → pH = 9.4
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54 Titration of a Weak Acid with a Strong Base
At the equivalence point,
c0V0 = ctVe (monoprotic acid)
c0 : concentration of weak acid
V0 : volume of acid originally present
ct : concentration of OH– in the base titrant
Ve : volume of the base at the equivalence point
Titration of 100.0 mL of 0.1000 M HOAc
with 0.1000 M NaOH at 25°C
H3O+(aq) + OH–(aq) 2 H2O(l)
Region I: Initial solution (Weak acid solution)
1. V = 0 mL NaOH added → pH of 0.100 M HOAc solution
[H3O+] = 1.32×10–3 → pH = 2.88
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Weak Acids
HOAc H3O+ AcO–
--------------------------------------------------------------------------Initial 1.000 ~ 0 0Change –y + y + y
---------------- ------ -----Equilibrium 1.000 – y y y--------------------------------------------------------------------------
+[H O ][Ac ]1.76 10
[HAc] 1.000
253
a
Ky
y
→ y = 4.20×10–3
[H3O+] = y = 4.20×10–3 M → pH = 2.38
𝑦2
1000 − 𝑦≈
𝑦2
1000= 1.76 × 10−5
Fraction ionized = [Ac–] / [HAc]0 = y / 1.000 = 4.20×10–3 → 0.42%
EXAMPLE 15.6
15.4 EQUILIBRIA INVOLVING WEAK ACIDS
AND BASES
Calculate the pH and the fraction of HOAc ionized at equilibrium.
HOAc(aq) + H2O(l) H3O+(aq) + AcO-(aq)
1 M
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HOAc H3O+ AcO–
--------------------------------------------------------------------------Initial 0.00100 ~ 0 0Change –y + y + y
---------------- ------ -----Equilibrium 0.00100 – y y y--------------------------------------------------------------------------
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57 Titration of a Weak Acid with a Strong Base
At the equivalence point,
c0V0 = ctVe (monoprotic acid)
c0 : concentration of weak acid
V0 : volume of acid originally present
ct : concentration of OH– in the base titrant
Ve : volume of the base at the equivalence point
Titration of 100.0 mL of 0.1000 M HOAc
with 0.1000 M NaOH at 25°C
H3O+(aq) + OH–(aq) 2 H2O(l)
Region I: Initial solution (Weak acid solution)
1. V = 0 mL NaOH added → pH of 0.100 M HOAc solution
[H3O+] = 1.32×10–3 → pH = 2.88
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Fig. 15.14 A titration curve for the titration of a weak acid by a strong base.
100. mL of 0.1000 M HOAc is titrated with 0.1000 M NaOH.
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Originally,
n(HOAc) = (0.1000 mol/L)(0.1000 L) = 1.000×10–2 mol
n(AcO–) generated by adding OH– :
n(AcO–) = n(OH–) = (0.1000 mol/L)(0.03000 L) = 3.000×10–3 mol
Amount of HOAc unreacted :
n(HOAc) = 1.000×10–2 mol – 3.000×10–3 mol = 7.000×10–3 mol
Region II: Before the equivalence point (Buffer solution)
2. V = 30.00 mL NaOH added ( 0 < V < Ve )
HOAc(aq) + OH–(aq) AcO–(aq) + H2O(l)
K = 1/ Kb = Ka / Kw = 2×109 >> 1
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Volume increased to 0.1300 L
Concentrations after adding 30.00 mL of NaOH:
[HOAc] = (7.000×10–3 mol) / (0.1300 L)
= 5.38×10–2 M
[AcO–] = (3.000×10–3 mol) / (0.1300 L)
= 2.31×10–2 M
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Calculations of Buffer Action
HCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq)
---------------------------------------------------------------------------------------
Initial 1.00 ~ 0 0.500
Change –y + y + y
--------------- ------ -------------
Equilibrium 1.00 – y y 0.500 + y
---------------------------------------------------------------------------------------
EXAMPLE 15.10 Calculate the pH of a solution of HCOOH and NaHCOO.
Buffer solution of HCOOH (1.00 mol) / NaHCOO (0.500 mol) in 1L
[H O ][HCOO ]1.77 10
[HCOOH] 1.00
+43
a
0.500
Ky +y
y
1.77 101.00 1.00
40.500 0.500y +y y
y
y = [H3O+] = 3.54×10–4 M → pH = 3.45
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Designing Buffers
2 3HA( ) H O( ) H O ( ) A ( )aq l aq aq
3a
03
0
[H O ][A ] [H O ][A ]
[HA] [HA]K
a 100
0
[HA] pH p log
[A ]K
Determining pH of the buffer solution
1. Choose a weak acid whose pKa ≈ pH
2. Fine-tuning of pH by adjusting the ratio of [HA]0 / [A–]0
Henderson-Hasselbalch Equation
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Volume increased to 0.1300 L
Concentrations after adding 30.00 mL of NaOH:
[HOAc] = (7.000×10–3 mol) / (0.1300 L)
= 5.38×10–2 M
[AcO–] = (3.000×10–3 mol) / (0.1300 L)
= 2.31×10–2 M
K
2
0a 10 10 2
0
[HAc] 5.38 10pH p log 4.75 log
[Ac ] 2.31 104.38
→ A buffer solution of
[HOAc]0 = 5.38×10–2 M and [NaOAc]0 = 2.31×10–2 M
At V = Ve/2, [HOAc]0 = [AcO–]0 → pH = pKa
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Region III: At the equivalence point
(equivalent to Hydrolysis of salts)
AcO– + H2O HOAc + OH–
3. V = Ve pH = 8.73
Region IV: Beyond the equivalence point
The pH determined by the excess OH–.
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Hydrolysis
AcO–(aq) + H2O(l) HOAc(aq) + OH–(aq) --------------------------------------------------------------------------
Initial 0.100 0 ~ 0
Change –y + y + y
---------------- ------ -----
Equilibrium 0.100 – y y y
--------------------------------------------------------------------------
y = 7.5 × 10–6 M = [OH–]
[H3O+] = Kw / [OH–] = 1.3 × 10–9 M → pH = 8.89
[HAc][OH ]5.7 10
[Ac ] 0.100
2
w
a
10
b
KK
K
y
y
EXAMPLE 15.9 Hydrolysis of NaOAc
NaOAc(s) + H2O(l) Na+(aq) + AcO–(aq)
Now 0.05 M instead of 0.1 M due to dilution
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Problem Sets
For Chapter 15,
10, 18, 24, 36, 50, 54, 94