General Chemistry II ACID-BASE EQUILIBRIA 15 CHAPTER 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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General Chemistry II
ACID-BASE EQUILIBRIA15CHAPTER
General Chemistry II
15.1 Classifications of Acids and Bases15.2 Properties of Acids and Bases in Aqueous
Solutions: The Brønsted-Lowry Scheme15.3 Acid and Base Strength15.4 Equilibria Involving Weak Acids and Bases15.5 Buffer Solutions15.6 Acid-Base Titration Curves15.7 Polyprotic Acids15.8 Organic Acids and Bases: Structure and Reactivity15.9 Exact Treatment of Acid-Base Equilibria
General Chemistry II
Cyanidin is blue in the basic sap of the cornflower and red in the acidic sap of the poppy.
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General Chemistry II
Acid Base
Arrhenius [H3O+ ] > KW1/2 [OH- ] > KW
1/2
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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General Chemistry II
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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General Chemistry II
Acid: A substance that can donate a protonBase: A substance that can accept a proton
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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General Chemistry II
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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General Chemistry II
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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General Chemistry II
Fig. 15.9 Indicators changetheir colors at very differentpH values.
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General Chemistry II
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 becomesless acidic.
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General Chemistry II
Weak Acids
HOAc H3O+ AcO–
--------------------------------------------------------------------------Initial 1.000 ~ 0 0Change –y + y + y
---------------- ------ -----Equilibrium 1.000 – y y 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
a0.500−
Ky + y
y
( ) ( ) −≈ ≈ ×−
1.77 101.00 1.00
40.500 0.500y + y yy
y = [H3O+] = 3.54×10–4 M → pH = 3.45
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General Chemistry II
General Chemistry II
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 HCOOHHCOOH(aq) + H2O(l) H3O+(aq) + HCOO–(aq)
-----------------------------------------------------------------------------------------Initial 1.10 ~ 0 0.40Change –y + y + y
--------------- ------ -------------Equilibrium 1.10 – y y 0.40 + y-----------------------------------------------------------------------------------------
Determining pH of the buffer solution1. Choose a weak acid whose pKa ≈ pH2. Fine-tuning of pH by adjusting the ratio of [HA]0 / [A–]0
Henderson-Hasselbalch Equation
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General Chemistry II
General Chemistry II
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General Chemistry II
a 100
0
[HA] pH p log [A ]
K −≈ −
General Chemistry II
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–.
c0V0 = ctVe (monoprotic acid)c0 : concentration of weak acidV0 : volume of acid originally presentct : concentration of OH– in the base titrantVe : volume of the base at the equivalence point
Titration of 100.0 mL of 0.1000 M HOAcwith 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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General Chemistry II
Weak Acids
HOAc H3O+ AcO–
--------------------------------------------------------------------------Initial 1.000 ~ 0 0Change –y + y + y
---------------- ------ -----Equilibrium 1.000 – y y y--------------------------------------------------------------------------
Calculate the pH and the fraction of HOAc ionized at equilibrium.HOAc(aq) + H2O(l) H3O+(aq) + AcO-(aq)
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1 M
General Chemistry II
HOAc H3O+ AcO–
--------------------------------------------------------------------------Initial 0.00100 ~ 0 0Change –y + y + y
---------------- ------ -----Equilibrium 0.00100 – y y y--------------------------------------------------------------------------
General Chemistry II
Titration of a Weak Acid with a Strong Base
At the equivalence point,
c0V0 = ctVe (monoprotic acid)c0 : concentration of weak acidV0 : volume of acid originally presentct : concentration of OH– in the base titrantVe : volume of the base at the equivalence point
Titration of 100.0 mL of 0.1000 M HOAcwith 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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General Chemistry II
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.