Transcript General Chemistry

Chapter 16
Acids and Bases
Johannes N. Bronsted
Thomas M. Lowry
1879-1947.
1874-1936.
Both independently developed Bronsted-Lowry
theory of acids and bases.
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Acids and Bases: A Brief Review
Classical Acids:
Taste sour
Donate H+ (called “H-plus” or “proton”)
Turn litmus red
Generally formed from H-Z, where Z = nonmetal
Classical Bases:
Taste bitter and feel soapy.
Donate OH- (called “O-H-minus” or “hydroxide”)
Turn litmus blue
Generally formed from MOH, where M = metal
Neutralization:
Acid + Base  Salt + water
H-Z + MOH  MZ + HOH
H+ in water is actually
in the form of H3O+,
“hydronium”
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Brønsted-Lowry Acids and Bases
Proton Transfer Reactions
• Brønsted-Lowry acid/base definition:
acid donates H+
base accepts H+.
• Brønsted-Lowry base does not need to contain OH-.
• Consider HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq):
– HCl donates a proton to H2O. Therefore, HCl is an acid.
– H2O accepts a proton from HCl. Therefore, H2O is a base.
• Water can behave as either an acid or a base.
• Amphoteric substances can behave as acids and bases.
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Brønsted-Lowry Acids and Bases
Conjugate Acid-Base Pairs
• Whatever is left of the acid after the proton is donated
is called its conjugate base.
• Similarly, whatever remains of the base after it
accepts a proton is called a conjugate acid.
• Consider HCl(aq)  H2O(l )  H3O (aq)  Cl  (aq)
–After H2O (base) gains a proton it is converted into H3O+
(acid). Therefore, H2O and H3O+ are conjugate acid-base
pairs.
–After HCl (acid) loses its proton it is converted into Cl(base). Therefore HCl and Cl- are conjugate acid-base pairs.
•Conjugate acid-base pairs differ by only one proton.4
Brønsted-Lowry Acids and Bases
Conjugate Acid-Base Pairs
In each of following Bronsted-Lowry Acid/Base reactions,
Which is acid? base? conjugate acid? conjugate base?
HCl + H2O  H3O+ + Clacid
base
conj. acid
conj. base
NH3 + H2O  NH4+ + OHbase
acid
conj. acid
conj. base
CH3NH2 + H2SO4  CH3NH3+ +
base
acid
conj. acid
HSO4conj. base
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Brønsted-Lowry Acids and Bases
Relative Strengths of Acids and
Bases
• The stronger the acid, the
weaker the conjugate base.
• The stronger the base, the
weaker the conjugate acid.
• H+ is the strongest acid that
can exist in equilibrium in
aqueous solution.
• OH- is the strongest base
that can exist in equilibrium
in aqueous solution.
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Strong and Weak Acids and Bases
Strong acids: completely ionized in water:
HCl, HNO3, H2SO4 (also HBr, HI)
Strong bases: completely ionized in water:
MOH, where M = alkali
M(OH)2, where M = alkaline earth
Weak acids: incompletely ionized in water:
any acid that is not strong - acetic acid, etc.
Ka is finite.
Weak bases: incompletely ionized in water:
any base that is not strong – NH3, etc.
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Kb is finite.
The Autoionization of Water
The Ion Product of Water
• In pure water the following equilibrium is established
H3O+(aq) + OH-(aq)
H2O(l) + H2O(l)

-
[H3O ][OH ]
Keq 
2
[H2O]
but [H2O]2 = constant

K eq [H 2O]  K w  [H 3O ][OH ]
2

-
14
K w  [H 3O ][OH ]  1.0 10
-
(at 25oC)
This is called the autoionization of water
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The Autoionization of Water
In pure water at 25oC, [H3O+][OH-] = 1 x 10-14
and also, [H3O+] = [OH-] = 1 x 10-7
(From now on, for simplification, let’s use the abbreviation:
[H+] = [H3O+] which means [H]+ = [OH-] = 1 x 10-7
We define
pH = -log [H+]
and
pOH = -log [OH-]
In pure water at 25oC, pH = pOH = 7.00
pH + pOH = 14
pKw = 14
Acidic solutions have pH < 7.00
Basic solutions have pH > 7.00
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The pH Scale
Conc. Drano
Battery acid
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The pH Scale
(a) For [H+] = 3.4 x 10-5M, calculate pH, pOH and [OH-]
pH = 4.47, pOH = 9.53, [OH-] = 2.95 x 10-10
(b) For [OH-] = 4.4 x 10-3M, calculate pH, pOH, and [H+]
pOH = 2.36, pH = 11.64, [H+] = 2.27 x 10-12
(c) For pH= 8.9, calculate, pOH, [H+], [OH-]
pOH = 5.1, [H+] = 1.26 x 10-9, [OH-] = 7.94 x 10-6
(d) For pOH= 3.2, calculate pH, [H+], [OH-]
pH = 10.8, [H+] = 1.58 x 10-11, [OH-] = 6.3 x 10-4
The pH meter is the most accurate way to measure pH
values of solutions.
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Use this “decision tree” to calculate pH
values of solutions of specific solutions.
1. Is it pure water? If yes, pH = 7.00.
2. Is it a strong acid? If yes, pH = -log[HZ]
3. Is it a strong base? If yes, pOH = -log[MOH]
or pOH = -log (2 x [M(OH)2])
4. Is it a weak acid? If yes, use the relationship
Ka = x2/(HZ – x), where x = [H+]
5. Is it a weak base? If yes, use the relationship
Kb = x2/(base – x), where x = [OH-]
6. Is it a salt (MZ)? If yes, then decide if it is neutral, acid,
or base; calculate its K value by the relationship
KaKb = Kw, where Ka and Kb are for a conjugate system;
then treat it as a weak acid or base.
7. Is it a mixture of a weak acid and its weak conjugate base?
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It is a buffer; see next chapter.
2. Strong Acids
Calculate the pH of 0.2 M HCl .
pH = -log[H+] = -log[0.2] = 0.70
3. Strong Bases
Calculate the pH of 0.2 M NaOH .
pOH = -log[OH-] = -log[0.2] = 0.70
pH = 14 – 0.70 = 13.30
Calculate the pH of 0.2 M Ba(OH)2.
pOH = -log[OH-] = -log[0.4] = 0.40
pH = 13.60
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4. Weak Acids
Weak acids are only partially ionized in solution, and
are in equilibrium:
H3O+(aq) + A-(aq)
HA(aq) + H2O(l)
(shorthand):
HA(aq)
[H 3O  ][A- ]
Ka 
[HA]
H+(aq) + A-(aq)
OR:
[H  ][A- ]
Ka 
[HA]
(shorthand)
•Ka is the acid dissociation constant.
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4. Weak Acids
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4. Weak Acids
Calculate pH of 0.2 M solution of acetic acid, HC2H3O2
From previous slide, Ka= 1.8 x 10-5
From now on, we’ll use the shorthand notation for HC2H3O2
= HAc
Denote the acetate ion, C2H3O2- as “Ac-”
Equilibrium is then:
Initial conc (M):
change:
at equilibrium:
HAc
0.2
-x
0.2-x
0
+x
x
2
x
[H ][Ac ]
Ka =

[HAc]
0.2 x

 H+ + Ac0
+x
x

But, this is a quadratic equation!!
We can assume x is small if Ka < 10-4.
Then,
x2/0.2= 1.8 x 10-5
x = 1.90 x 10-3 = [H+] = [Ac-]
pH = -log (1.90 x 10-3) = 2.72
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4. Weak Acids
Polyprotic Acids
• Polyprotic acids have more than one ionizable proton.
• The protons are removed in steps not all at once:
H2SO3(aq)
H+(aq) + HSO3-(aq) Ka1 = 1.7 x 10-2
HSO3-(aq)
H+(aq) + SO32-(aq)
Ka2 = 6.4 x 10-8
•It is always easier to remove the first proton in a
polyprotic acid than the second.
•Therefore, Ka1 > Ka2 > Ka3 etc.
•Most H+(aq) at equilibrium usually comes from the
first ionization (i.e. the Ka1 equilibrium).
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4. Weak Acids
Polyprotic Acids
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Carbonic acid, H2CO3, Ka1=4.3 x 10-7
H2CO3  H+ + HCO3HCO3- 
H+ + CO32-
Ka1 = 4.3 x 10-7
Ka2 = 5.6 x 10-11
Since Ka1>>Ka2, nearly all H+ ions come from 1st equilibrium.
Therefore, the 1st equilibrium determines the pH.
Calculate pH of 0.4 M H2CO3 solution.
Ka = x2/HZ-x
4.3 x 10-7 = x2/0.4
x = [H+] = 4.15 x 10-4
pH = -log (4.1 x 10-4) = 3.38
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5. Weak Bases
• Weak bases remove protons from substances.
• There is an equilibrium between the base and the
resulting ions:
Weak base + H2O
conjugate acid + OH-
Example:
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
The base dissociation constant, Kb is defined as:

4
-
[ NH ][OH ]
Kb 
[ NH3 ]
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5. Weak Bases
• The larger Kb the stronger the base.
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5. Weak Bases
What is pH of 0.15 M solution of methylamine, NH2CH3, a weak base?
For convenience, denote it as “B”
Then:
B + H 2O
Initially: 0.15
change:
-x
At equil: 0.15-x
Then:

BH+ +
OH- Kb =4.4 x 10-4
0
+x
x
0
+x
x
x2
x2

 4.4x10 4
.15  x .15
3

x  8.12 x 10  [OH ]
pOH  log[OH ]  log(8.12x103 )  2.09
pH  14  pOH  14  2.09  11.9
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Relationship Between Ka and Kb
• For a conjugate acid-base pair
Ka  Kb = Kw (constant)
• Therefore, the larger the Ka, the smaller the Kb. That
is, the stronger the acid, the weaker the conjugate
base.
• Taking negative logarithms:
-log Ka- log Kb= -log Kw
pKa + pKb = pKw
For HAc (acetic acid), the pKa = -log (1.8 x 10-5) = 4.74.
Thus, the pKb for Ac- (acetate 0 is 14 – 4.74 = 9.26
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6. Salts
Salts may be acidic, basic or neutral.
Salts made by a strong acid and a strong base are neutral,
e.g., NaCl, KNO3.
Salts made by a weak acid and a strong base are weakly basic,
e.g., sodium acetate, NaAc, NaHCO3.
Salts made by a strong acid and a weak base are weakly acidic,
e.g., NH4Cl.
Calculate the pH of a 0.35 M solution of sodium acetate.
Since Ka Kb = Kw, where Ka = 1.8 x 10-5 for acetic acid,
Then Kb for NaAc (sodium acetate) is
Kb = Kw/Ka = 1.00 x 10-14/1.8 x 10-5 = 5.56 x 10-10.
Now treat this as a weak base problem,
Kb = x2/base = 5.56 x 10-10 = x2/0.35
x = [OH-] = 1.39 x 10-5
pOH = 4.85 and pH = 9.14
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