ACID-BASE EQUILIBRIUM
WHAT IS AN ACID AND A BASE?
Arrhenius Theory
 Acids – are solutes that
produce hydrogen ions
H + in aqueous solutions
ex. HCl (aq)  H + (aq) + Cl - (aq)
 Bases – are solutes that
produce hydroxide ions,
OH - in aqueous solutions
ex. NaOH (s)  Na + (aq) + OH (aq)
Bronsted-Lowry Theory
 A Bronsted-Lowry acid is
a proton donor
 A Bronsted-Lowry base is
a proton acceptor
Arrhenius Acids
Arrhenius Bases
HBr (aq)  H + (aq) + Br - (aq)
LiOH (aq)  Li + (aq) + OH - (aq)
H 2 SO 4(aq)  H + (aq) + HSO 4 -
KOH (aq)  K + (aq) + OH - (aq)
(aq)
HClO 4(aq) 
(aq)
H+
(aq)
+ ClO 4
-
Ba(OH) 2(aq)  Ba 2+ (aq) +
2OH - (aq)
PROBLEMS WITH ARRHENIUS THEORY
 The Salt problem: The theory failed to account for the
basic properties of compounds that do not contain the
hydroxide ion, such as ammonia (NH 3(aq) )
 NH 3(g) + H 2 O (l)  NH 4 + (aq) + OH - (aq)
The Solvent Problem The theory failed to account for a solvent ’s
key role in acid-base properties
- for example, the aqueous solution of hydrogen chloride
(hydrochloric acid) conducts electricity, while hydrogen
chloride in an organic solvent does not [therefore, latter
solution does not contain ions].
- The solvent plays a key role in acid -base properties.
BRONSTED-LOWRY THEORY
\
Acid (H+ donor)
HCl (g) + H 2 O (l)  H 3 O + (aq) + Cl - (aq)
B a s e ( H + a c c e p to r )
H+ donor (acid)
CH3COOH(aq) + H2O(l) ↔ CH3COO-(aq) + H3O+(aq)
H+ acceptor (base)
NH3(g) + H2O(l) ↔ NH4+ (aq) + OH-(aq)
H+ acceptor (base)
H+ donor (acid)
AMPHOTERIC
 ** A substance can be classified as a Bronsted-Lowry acid or
base only for a specific reaction
 Protons may be gained in a reaction with one substance, but lost in a
reaction with another
H 2O (l) + H 2O (l)  H 3O +(aq) + OH −(aq)
H+ acceptor (base)
H+ donor (acid)
Water is amphoteric, meaning it can act as both an acid and a
base.
 one molecule of water acts as an acid, donating a H+ ion and forming
the conjugate base, OH - , and a second molecule of water acts as a
base, accepting the H + ion and forming the conjugate acid, H3O + .
CONJUGATE ACID-BASE PAIRS
 According to the Bronsted-Lowry concept, acid -base reactions
involve the transfer of a proton
 Usually reversible and result in an acid-base equilibrium
 Every base has a corresponding acid (conjugate acid) and every
acid has a corresponding base (conjugate base)
 Two molecules or ions that are related by the transfer of a proton
are called a conjugate acid-base pair.
 Conjugate acid-base pairs differ in formula by one hydrogen ion
 H2SO4 – HSO4 H2O – OH-
COMMON CONJUGATE ACID-BASE PAIRS
COMMON CONJUGATE ACID-BASE PAIRS
HCl (g) + H 2 O (l)  H 3 O + (aq) + Cl - (aq)
CH 3 COOH (aq) + H 2 O (l) ↔ CH 3 COO - (aq) + H 3 O + (aq)
NH 3(g) + H 2 O (l) ↔ NH 4 + (aq) + OH - (aq)
H 2 SO 4(aq) + H 2 O (l)  HSO 4 - (aq) + H 3 O + (aq)
HCl (aq) + NH 3(g)  Cl - (aq) + NH 4 + (aq)
A COMPETITION FOR PROTONS
 The stronger the acid, the weaker its conjugate base, and
conversely, the weaker an acid, the stronger its conjugate
base.
 Acid-base reactions – competition for protons between two
bases
 Acetic acid equilibrium – between acetate and water
CH 3 COOH (aq) + H 2 O (l) ↔ CH 3 COO - (aq) + H 3 O + (aq)
 Ability of acetate ion to hold on to its proton (H + ion) is much
higher than the ability of H 2 O to pull the proton away
 Percent ionization Is low
 Equilibrium lies far left – acetic acid is a weak acid (acetate ion is
a strong base!)
AUTOIONIZATION OF WATER
 Autoionization is when water molecules ionize one another.
 The transfer of a proton (H + ) from one molecule of water to the other,
producing a hydronimum ion (H 3 O + )
 H 2O (l) + H 2O (l)  H 3O +(aq) + OH −(aq)
 Simplified into…
 H 2O (l)  H +(aq) + OH −(aq)
H2O(l)  H+(aq) + OH−(aq)
The equillibrium constant for the ionization of
water:
Kw = [H+(aq)][OH-(aq)]
@ SATP; Kw= 1.0 x 10-14
*an endothermic process ∴ ↑ 𝐾𝑤 ↑ 𝑇
RECALL FROM GR. 11…
[H + (aq) ] = [OH - (aq) ] = pH
7
NEUTRAL
[H +
ACIDIC
(aq) ]
>
[OH -
(aq) ]
= low
pH
[H + (aq) ] < [OH - (aq) ] =
high pH
BASIC
STRONG ACIDS
 HClO 4(aq), HI (aq), HBr (aq), HCl (aq) , HNO 3(aq) , H 2 SO 4(aq), etc..
 % ionization for strong acids > 99% [We assume 100%
dissociation in calculations]
 1. A 0.15 mol/L solution of hydrochloric acid at
SATP. Calculate the concentration of the
hydroxide ions using K w = [H + (aq) ][OH - (aq) ].
 HCl (aq)  H + (aq) + Cl - (aq)
 Hydrochloric acid is a strong acid – assume 100%
ionization [*tiny contribution made by
autoionization of water can be ignored]
 1: 1: 1 mole ratio
 If [HCl (aq) ] = 0.15 mol/L then.. [H + ] = 0.15 mol/L
 @ SATP; K w = [H + (aq) ][OH - (aq) ]
 K w = 1.0 x 10 -14
 [OH - (aq) ] = K w / [H + (aq) ] = 1.0 x 10 -14 / 0.15 mol/L
 = 6.7 x 10 -14 mol/L
 The concentration of hydroxide ions is 6.7 x 10 -14
mol/L.
 2. Calculate the hydroxide ion concentration
in a 0.25 mol/L HBr (aq) solution.
 HBr (aq)  H + (aq) + Br - (aq)
 Hydrobromic acid is a strong acid – assume
100% ionization [*tiny contribution made by
autoionization of water can be ignored]
 1: 1: 1 mole ratio
 If [HBr (aq) ] = 0.25 mol/L then.. [H + ] = 0.25 mol/L
 @ SATP; Kw = [H + (aq) ][OH - (aq) ]
 Kw = 1.0 x 10 -14
 [OH - (aq) ] = K w / [H + (aq) ] = 1.0 x 10 -14 / 0.25 mol/L
= 4.0 x 10 -14 mol/L
 The concentration of hydroxide ions is 4.0 x 10 -14
mol/L.
STRONG BASES
 LiOH (aq), NaOH (aq), KOH (aq_, RbOH (aq) , CsOH (aq) , etc..
 Dissociate completely in water to release hydroxide
ions
 All group 1 hydroxides are strong bases
 When dissolved in water, these bases produce 1 mole of
hydroxide ion for every mole of metal hydroxide
 NaOH (aq)  Na + (aq) + OH - (aq)
 3. Determine the hydrogen ion and hydroxide ion
concentration in 500.0mL of an aqueous solution,
containing 2.6 g of dissolved sodium hydroxide.
 NaOH (aq)  Na + (aq) + OH - (aq)
 m of NaOH (a) solute = 2.6g
 v of NaOH (aq) solution = 500.0 mL
 n = m/M
 = 2.6g / (40 g / mol) = 0.065 mol
 c = n/v = 0.065 mol / 0.5L = 0.13 mol/L
 1 mol NaOH (aq) : 1 mol Na + (aq) : 1 mol OH - (aq)
 @ SATP; K w = [H + (aq) ][OH - (aq) ]
 K w = 1.0 x 10 -14
 [H + (aq) ] = K w / [OH - (aq) ] = 1.0 x 10 -14 / 0.13 mol/L
 = 7.7 x 10 -14 mol/L
 The concentration of hydrogen ions is 7.7 x 10 -14 mol/L
and the concentration of hydroxide ions is 0.13mol/L.
[H + ] & PH
 pH: A way to express the ACIDIT Y/BASICIT Y of an aqueous solution
 the amount of hydronium ion H 3 O + in solution
• pH > 7 basic
• pH = 7 neutral
• pH < 7 acidic
[ H + ( aq) ] < [ OH - (aq) ]
[H + ( aq) ] = [ OH - ( aq) ]
[H + ( aq) ] > [ OH - (aq) ]
 pH = -log[H 3 O + ]
 [H 3 O + ] = 10 -pH
 1 . ) If [H 3 O + ] = 1.0 x 10 -7 mol/L, then pH is…
 pH = -log[H 3 O + ]
 = -log(1.0 x 10 -7 mol/L)
 = 7.0
 2. Given the pH = 10.33, calculate the [H + (aq) ].
 [H 3 O + ] = 10 -pH
 = 10 -10.33
 = 4.7 x 10 -11 mol/L
[OH - ] & POH
 pOH = -log[OH - ]
 [OH - ] = 10 -pOH
 pH + pOH = 14
 Can simplify pH calculation – if either pH/pOH is known, the
other can be found! 




Calculate the pH, pOH and [OH - (aq) ] of a 0.042mol/L HNO 3(aq) solution.
HNO 3(aq) NO 3 - (aq) + H + (aq)
1:1:1 mole ratio
[HNO 3(aq) ]= [H + (aq) ] = 0.042 mol/L
 pH = -log [H + ] = -log(0.042 mol/L)
 = 1 .40
 pH + pOH =14
 pOH = 14 – pH = 14 – 1 .40 = 1 2.60
 [OH - ] = 10 -pOH
 = 10 -12.60
 = 2.5 x 10 -13 mol/L
 The pH of the solution is 1 .40; the pOH is 1 2.60; and the [OH - ] = 2.5 x
10 -13 mol/L.
• A solution was made by
dissolving 0.627 g Ba(OH) 2 in
100.0 mL final volume. If
Ba(OH) 2 is fully broken up into
its ions, what is the pOH and
the pH of this solution?
• G: 0.627 g of Ba(OH) 2 dissolved
in 100 .0 mL (final volume)
• M of Ba(OH) 2 = 171.63 g / mol
• R: pOH = ? pH = ?
• A: n = m / M , c = n/ v
• pOH = -log[OH - ], pH + pOH = 14
 m = 6 . 27 g/L x 0 .1 L = 0 . 6 27 g
 n = m / M = 0 . 6 27 g / ( 171 . 63 g /
m o l ) = 0 . 0 0 3 65 m o l
 B a ( OH ) 2  B a 2 + + 2 OH  M o l e ra t i o 1 B a ( O H ) 2 : 2 OH  n o f O H - = 2 x 0 . 0 0 3 6 5 = 0 . 0 07 31
mol
 [ O H - ] = 0 . 0 07 31 mol/0.1L =
0 . 07 31 M
 p O H = - l o g [ OH - ] = - l o g (0. 07 31)
 = 1 .14
 pH + pO H = 14
 pH = 1 2 . 9
 P : Th e s o l ut ion h a s a fi n a l pOH o f
1 .14 a n d a fi n a l pH o f 1 2 . 9 .
HOMEWORK
 Pg. 532 #1 , 2
 Pg. 537 #4-6
 Pg. 549 #17-19