Learning goals and key skills:
Understand the nature of the hydrated proton, represented as either H+(aq) or H3O+(aq) Define and identify Arrhenuis acids and bases.
Define and identify Bronsted-Lowry acids and bases, and identify conjugate acid-base pairs.
Relate the strength of an acid to the strength of its conjugate base.
Understand how the equilibrium position of a proton transfer reaction relates the strengths of acids and bases involved.
Describe the autoionization of water and understand how [H3O+] and [OH-] are related Calculate the pH of a solution given [H3O+] or [OH-]
Calculate the pH of a strong acid or strong base given its concentration
Calculate Kaor Kbfor a weak acid or weak base given its concentration and the pH of the solution
Calculate pH of a weak acid or weak base or its percent ionization given its concentration and Kaor Kb.
Calculate Kbfor a weak base given Kaof its conjugate acid, and similarly calculate Kafrom Kb.
Predict whether and aqueous solution of a salt will be acidic, basic, or neutral Predict the relative strength of a series of acids from their molecular structures Define and identify Lewis acids and bases.
Chapter 16 Acid-Base Equilibria
Acids and Bases
Arrhenius
-An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions.
-A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions.
Brønsted-Lowry -An acid is a proton donor.
-A base is a proton acceptor.
Acids and bases may be inorganic (7 strong acids, 8 strong bases) or organic (acids have –COOH group):
Amphiprotic – a substance that is capable of acting as
an acid or a base e.g. HCO
-, HSO
-,H O
What happens when an acid dissolves in water?
HCl
(aq)+ H
2O
(l)→ Cl
–(aq)+ H
3O
+(aq)• Water acts as a Brønsted-Lowry base and abstracts a proton (H
+) from the acid.
• As a result, the conjugate base of the acid and a hydronium ion are formed.
Conjugate acids and bases
HCN
(aq)+ H
2O
(l)CN
-(aq)+ H
3O
+(aq)HClO
(aq)+ H
2O
(l)ClO
-(aq)+ H
3O
+(aq)NH
3 (aq)+ H
2O
(l)NH
4+(aq)
+ OH
-(aq)⇌
⇌
⇌
Acid and base strength
•Strong acids are completely dissociated in water.
–Their conjugate bases are weak.
•Weak acids only dissociate partially in water.
–Their conjugate bases
are strong.
Acid and Base Strength
• In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.
HCl
(aq)+ H
2O
(l)→ H
3O
+(aq)+ Cl
-(aq)• H
2O is a much stronger base than Cl
-, so the equilibrium lies so far to the right that K is not measured (K>>1).
• Acetate is a stronger base than H
2O, so the equilibrium favors the left side (K<1).
CH
3CO
2H
(aq)+ H
2O
(l)⇌ H
3O
+(aq)+ CH
3CO
2-(aq)Auto-ionization of water
K
w= [H
3O
+] [OH
−]
• This special equilibrium constant is referred to as the ion-product constant for water, K
w.
• At 25°C, K
w= 1.0 × 10
−14• As we have seen, water is amphoteric.
• In pure water, a few molecules act as bases and a few act as acids.
• This is referred to as autoionization.
H
2O
(l)+ H
2O
(l)⇌ H
3O
+(aq)+ OH
-(aq)The ion-product constant for water Temperature K
w= [H
3O
+] [OH
−]
0 °C 0.114 × 10
-1425 °C 1.008 × 10
−1450 °C 5.476 × 10
-1475 °C 15.85 × 10
-14100 °C 51.3 × 10
-14pH and pOH scale pH = -log [H 3 O + ] pOH = -log [OH - ]
• At 25 °C in pure water,
K
w= [H
3O
+] [OH
−] = 1.0 × 10
−14[H
3O
+] = 1.0 × 10
-14= 1.0 × 10
-7Since in pure water [H
3O
+] = [OH
-],
pH = -log [H 3 O + ]
These are the pH values for several common substances.
Three ways to measure pH
• Litmus paper red-to-blue:
basic, pH > 8
blue-to-red:
acidic, pH < 5
• An indicator
• A pH meter
Seven strong acids
Strong acids completely ionize.
hydrochloric acid HCl
(aq)→ H
+(aq)+ Cl
-(aq)hydrobromic acid hydroiodic acid
chloric acid
perchloric acid HClO
4 (aq)→ H
+(aq)+ ClO
4-(aq)nitric acid
sulfuric acid H
2SO
4 (aq)→
“100%”
“100%”
“100%”
Anion name Acid name
S
2-, sulfide ion H
2S, hydrosulfuric acid CN
-, cyanide ion HCN, hydrocyanic acid
CO
32-, carbonate ion H
2CO
3, carbonic acid CrO
42-, chromate ion H
2CrO
4, chromic acid
BrO
-, hypobromite ion HBrO, hypobromous acid NO
2-, nitrite ion HNO
2, nitrous acid
Acids: nomenclature
Anion name Acid name
_______ ide ion
hydro_____ ic acid Cl
-, chloride ion HCl, hydrochloric acid
_______ ate ion _______ ic acid ClO
3-, chlorate ion HClO
3, chloric acid
_______ ite ion _______ ous acid ClO
2-, chlorite ion HClO
2, chlorous acid
Acids: nomenclature
Strong bases
Strong bases completely dissociate.
Hydroxides of the
• alkali metals (Li, Na, K, Rb, Cs)
NaOH
(aq)→ Na
+(aq)+ OH
-(aq)• heavy alkaline earth metals (Ca, Sr, Ba) (although these have limited solubility)
“100%”
K a and K b
HA
(aq)+ H
2O
(l)H
3O
+(aq)+ A
-(aq)[H
3O
+] [A
-] K
c= [HA]
This equilibrium constant is called the acid-dissociation constant, K
a.
[H
3O
+] [A
-] K
a= [HA]
⇌
Dissociation Constants
The greater the value of K
a, the stronger is the acid.
Calculating K a from the pH
The pH of a 0.100 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K
afor formic acid at this temperature.
We know that [H
3O
+] [HCOO
-] [HCOOH]
K
a=
To calculate K
a, we need the equilibrium concentrations of all three things.
We can find [H
3O
+], which is the same as [HCOO
-], from the pH.
HCOOH
(aq)+ H
2O
(l)⇌ H
3O
+(aq)+ HCOO
-(aq)Calculating K a from pH
Now we can set up an ICE table…
[HCOOH], M [H3O+], M [HCOO-], M
Initially 0.100 0 0
Change - 4.17 ×10-3 + 4.17 ×10-3 + 4.17 ×10-3 Equilibrium 0.10 - 4.17 ×10-3
= 0.0958
4.17 ×10-3 4.17 ×10-3
Calculating Percent Ionization
concentration ionized original concentration
In this example,
Percent Ionization = × 100%
[H
3O
+]
eq= 4.2 × 10
-3M [HCOOH]
initial= 0.10 M
[H
3O
+]
eq[HA]
initialPercent Ionization = × 4.2 × 10
-3100%
0.10
= 4.2%
× 100%
Percent Ionization =
Calculating pH from K a
Calculate the pH of a 0.30 M solution of acetic acid, HC
2H
3O
2, at 25°C.
HC
2H
3O
2(aq)+ H
2O
(l)⇌ H
3O
+(aq)+ C
2H
3O
2-(aq)K
afor acetic acid at 25°C is 1.8 × 10
-5.
[H
3O
+] [C
2H
3O
2-]
[HC
2H
3O
2]
K
a=
Calculating pH from K a
We next set up an ICE table…
[HC2H3O2], M [H3O+], M [C2H3O2-], M
Initially 0.30 ≈0 0
Change -x +x +x
Equilibrium 0.30 - x ≈0.30 x x
We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.
• 100*K
a= 100*(1.8 × 10
-5) = 0.0018
• 0.30-x = 0.30-0.0018 ≈ 0.30 (using sign. figures) In general, the approximation that [HA]
eqis effectively equal to [HA]
0is valid whenever
[HA]
0is greater than 100*K
a.
Polyprotic Acids…
…have more than one acidic proton
If the difference between the K
afor the first dissociation and subsequent K
avalues is 10
4or more, the pH generally depends only on the first dissociation.
Example (polyprotic acids)
H
3PO
4 (aq)+ H
2O
(l)⇌ H
3O
+(aq)+ H
2PO
4- (aq)K
a1= 7.5×10
-3H
2PO
4-(aq)
+ H
2O
(l)⇌ H
3O
+(aq)+ HPO
42- (aq)K
a2= 6.2×10
-8HPO
42-(aq)