Polyprotic Acids & Bases
A polyprotic acid can donate more than one H+
Carbonic acid: H2CO3(aq); dissolved CO2 in water
Sulfuric acid: H2SO4(aq)
Phosphoric acid: H3PO4(aq)
A polyprotic base: can accept more than one proton
Carbonate ion: CO32-(aq)
Sulfate ion: SO42-(aq)
Phophate ion: PO43-(aq)
Treat each step of protonation or deprotonation sequentially
H2CO3 (aq) + H2O(l)  H3O+(aq) + HCO3-(aq) Ka1 = 4.3 x 10-7
HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq)
Ka2 = 4.8 x 10-11
Typically:
Ka1 >> Ka2 >> Ka3 >>…
Harder to loose a positively charged proton from a negatively
charged ion, because of attraction between opposite
charges.
Calculate the pH of 0.010 M H2SO4(aq) at 25oC.
Sulfuric acid is the only common polyprotic acid where the
first deprotonation step is complete. The second
deprotonation step is much weaker and adds slightly to the
H3O+(aq) concentration.
For the first step assume all H2SO4(aq) deprotonates
H2SO4 (aq) + H2O(l)  H3O+ (aq) + HSO4-(aq)
From the first step [H3O+(aq)] = 0.010 M
Second deprotonation
HSO4- (aq) + H2O(l)  H3O+ (aq) + SO42- (aq) Ka2 = 0.012
Initial
Change
Equilibrium
HSO4- (aq) SO42- (aq)
0.010
0
-x
+x
0.010-x
x
H3O+ (aq)
0.010
0.010 + x
0.010 + x
Ka2 = ([H3O+ (aq)])([SO42- (aq)]) / ([HSO4- (aq) ])
0.012 = (0.010+x)(x) / (0.010-x)
Solve the quadratic equation for x. Ka2 is large; cannot
assume that x << 0.010
[H3O+ (aq)] = 1.4 x 10-2 M
pH = 1.9
Determine the pH of 0.20 M H2S(aq) at 25oC
H2S (aq) + H2O(l)  H3O+ (aq) + HS- (aq)
HS- (aq) + H2O(l)  H3O+ (aq) + S2- (aq)
Ka1 = 1.3 x 10-7
Ka2 = 7.1 x 10-15
For the first deprotonation step determine [H3O+(aq)] using
equilibrium tables. [H3O+(aq)] = 1.6 X 10-4 M
Can assume that x << 0.20 since Ka1 is small
Second deprotonation constant is very small, so ignore
addition of H3O+(aq) due to second step.
pH determined by first step alone. pH = 3.8
Composition and pH
For a solution of H2CO3(aq): at low pH the fully protonated
species (H2CO3) dominates; at high pH the fully
deprotonated form (CO32-) dominates; and at intermediate
pH the intermediate species (HCO3-) dominates.
LeChatelier’s principle at work
H2CO3 (aq) + H2O(l)  H3O+(aq) + HCO3-(aq) Ka1 = 4.3 x 10-7
HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq)
Ka2 = 4.8 x 10-11
H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq)
Ka1 = [H3O+(aq)] [HCO3-(aq)] / [H2CO3(aq) ]
HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq)
Ka2 = [H3O+(aq)] [CO32-(aq)] / [HCO3-(aq) ]
Define a(X): fraction of species X
a(X) =
X
[H2CO3(aq) ] + [HCO3-(aq)] + [CO32-(aq)]
The fraction of deprotonated species increases as the pH
increases
Determine the concentration of H2CO3(aq), HCO3-(aq), CO32(aq), H3O+(aq) present and the pH at equilibrium in a solution
that is initially 0.010 M in H2CO3. (Ka1 = 4.3 x 10-7, Ka2 = 4.8 x
10-11)
Step 1
H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq)
Initial
Change
H2CO3(aq)
0.010
H3O+(aq)
0
-x
Equilibrium 0.010-x
x
x
x
x
4.3 x 10-7 = x2 / (0.010-x)
Assume x << 0.010; x = 6.6 x 10-5
[H2CO3(aq)] ≈ 0.010 M; [H3O+(aq)] = 6.6 x 10-5 M;
[HCO3-(aq)] = 6.6 x 10-5 M
HCO3-(aq)
0
HCO3-(aq) + H2O(l)  H3O+(aq) + CO32-(aq)
Ka2 = [H3O+(aq)] [CO32-(aq)] / [HCO3-(aq) ]
Initial
Change
HCO3-(aq)
6.6 x 10-5
-y
Equilibrium 6.6 x 10-5 - y
H3O+(aq)
6.6 x 10-5
CO32-(aq)
0
6.6 x 10-5 + y
y
6.6 x 10-5 + y
y
4.8 x 10-11 = (6.6 x 10-5 + y) y / (6.6 x 10-5 - y)
Assume y << 6.6 x 10-5
4.8 x 10-11 = (6.6 x 10-5 y / (6.6 x 10-5 ); y = 4.8 x 10-11
At equilibrium: [H2CO3(aq)] ≈ 0.010 M; [H3O+(aq)] = 6.6 x 10-5
M; [HCO3-(aq)] = 6.6 x 10-5 M; [CO32-(aq)] = 4.8 x 10-11 M
pH = 4.18
Buffers
Buffer solutions : resists change in pH even with addition of
small amounts of acid or base.
Buffer solutions are mixed solutions: mixture of a weak acid
and its conjugate base or a weak base and its conjugate
acid.
Human blood has a pH maintained at pH = 7.4 due to a
combination of carbonate, phosphate and protein buffers.
The ocean is buffered to a pH of ~ 8.4 by buffering that
depends the presence of hydrogen carbonates and
silicates.
Buffer Action
An acid buffer is an aqueous solution of a weak acid and its
conjugate base.
It buffers solutions on the acid side of neutral (pH < 7).
Example: solution of CH3COOH(aq) + CH3COONa(aq)
CH3COOH / CH3COOA base buffer is an aqueous solution of a weak base and its
conjugate acid.
It buffers solutions on the basic side of neutral (pH > 7).
Example: NH3(aq) + NH4Cl(aq)
NH3/ NH4+
Buffer solution of CH3COOH(aq) / CH3COO- (aq)
If a small amount of strong acid is added:
H3O+(aq) + CH3COO-(aq)  CH3COOH(aq) + H2O(l)
K for this reaction = 1/Ka(CH3COOH(aq)) = 5.5 x 104
The CH3COO-(aq) acts as a “sink” for the added protons, and
the pH remains unchanged.
If a small amount of strong base is added
OH-(aq) + CH3COOH(aq)  CH3COO-(aq) + H2O(l)
K for this reaction = 1/Kb(CH3COO-(aq)) = 1.8 x 109
The CH3COOH(aq) acts as a “sink” for the added OH-, and the
pH remains unchanged
Designing a buffer
Make a solution with particular pH so that it buffers about this
pH.
Consider a solution of a weak acid and its conjugate base
HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
[H3
O+(aq)]
- log [H3
= Ka
O+(aq)]
pH = pKa + log
[HA(aq)]
[A-(aq)]
[HA(aq)]
= - log Ka - log
[A-(aq)]
[A- (aq)]
[HA(aq)]
Henderson-Hasselbach
equation
Note: for a solution of a weak base/conjugate acid, use Ka of
the conjugate acid
An optimal buffer is one in which the weak acid and its
conjugate base have equal concentrations
Select a weak acid that has its pKa as close as possible to the
desired pH
Having chosen the weak acid, use the Henderson Hasselbach
equation to determine the ratio of [A-(aq)] and [HA(aq)] that
will form solution that buffers around the desired pH
Calculate the pH of a buffer solution that is 0.040 M
CH3COONa (aq) and 0.080 M CH3COOH (aq).
pKa(CH3COOH(aq)) = 4.75
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
CH3COOH(aq)
Initial
0.080 M
Change
-x
Equilibrium
0.080 - x
Ka = (0.040 + x) x / (0.080 - x)
Assume x << 0.040
x = 3.6 x 10-5
pH = 4.44
H3O+(aq)
0
x
x
CH3COO-(aq)
0.040M
0.040 + x
0.040 + x
Suppose that a solution is made by dissolving 1.2 g NaOH(s)
(0.030 moles) in 500 mL of the buffer solution in the previous
problem. Calculate the pH of the resulting solution and the
change in pH. Assume the volume of the solution to be
constant.
The OH-(aq) will react with CH3COOH(aq)
CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(l)
Moles of OH- added = 0.030 moles
Moles of CH3COOH(aq) = (0.500 L) (0.080 M) = 0.040 mol
Amount of unreacted CH3COOH(aq) = 0.010 moles
Molarity of CH3COOH(aq) = 0.020 M
Moles of CH3COO-(aq) = initial amount + amount formed by
reaction of OH- (aq) and CH3COOH(aq)
= (0.040M x 0.500 L) + (0.030 moles) = 0.050 moles
Molarity of CH3COO-(aq) = 0.10 M
CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
pH = pKa + log ([CH3COO-(aq) ] / [CH3COOH(aq) ] )
= 5.45
If the solution had contained HCl at pH = 4.4, addition of the
NaOH would have raised the pH to 12.8
Titrations
Strong Acid - Strong Base
H3O+(aq) + OH-(aq)  2H2O(l)
pH changes slowly initially,
changes rapidly through pH = 7
(equivalence point) and then
changes slowly again
If the analyte is a strong acid, pH
increases as base is added
If the analyte is a strong base,
pH decreases as acid is added
Analyte: 25.00 mL of 0.250 M NaOH(aq)
Titrant: 0.340 M HCl(aq)
Determine the pH of the solution when 5.00mL of titrant
added
Answer: pH = 13.18
Determine the amount of titrant that must be added to reach
the equivalence point? Answer: 18.4 mL
Determine the pH of the solution after the addition of 20.4 mL
of titrant. Answer: pH = 1.82
Strong Acid-Weak Base and
Weak Acid - Strong Base
CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)
Slow change in pH before
equivalence point; solution
is a buffer
CH3COOH(aq)/CH3COO-(aq)
At halfway point
[HA] = [A-]
pH = pKa
At equivalence, pH
determined by CH3COO-(aq)
Changes in pH during a titration of a weak acid/base with a
strong base/acid:
Halfway to the stoichiometric point, the pH = pKa of the acid
The pH is greater than 7 at the equivalence point of the
titration of a weak acid and strong base
The pH is less that 7 at the equivalence point of the titration
of a weak base and strong acid
Beyond the equivalence point, the excess strong acid or base
will determine the pH of the solution
Titration of 100.0 mL of 0.1000 M CH3COOH(aq) with 0.1000 M
NaOH
Before addition of NaOH: pH determined by CH3COOH(aq)
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO-(aq)
Answer: pH = 2.88
Before the equivalence point: determine pH for a buffer
Addition of 30.00 mL of NaOH(aq)
The OH-(aq) reacts with the CH3COOH(aq). Determine
concentration of CH3COOH(aq) and CH3COO- (aq) in solution
after addition of the base. Answer: pH = 4.38
At half equivalence: [CH3COOH(aq)] = [CH3COO-(aq)]
pH = pKa
At equivalence: enough OH-(aq) added to react with all
CH3COOH(aq).
For this problem, equivalence is reached when 100.0mL of
OH- is added; i.e. 0.01000 moles of OH-(aq) added
Solution contains 0.01000 moles CH3COO-(aq) in 200.0 mL
solution; [CH3COO-(aq)] = 0.05000 M
pH determined by
CH3COO-(aq) + H2O(l)  CH3COOH(aq) + OH- (aq)
pH = 8.72
(note greater than 7.0)
Beyond equivalence: pH determined by excess OH-(aq)
Estimate the pH at the equivalence point of the titration of
25.00 mL of 0.100 M HCOOH(aq) with 0.150 M NaOH(aq)
(Ka(HCOOH) = 1.8 x 10-4)
At the equivalence point, enough NaOH(aq) has been added
to react with all the HCOOH(aq) forming HCOO-(aq)
The reaction:
HCOO-(aq) + H2O(l)  HCOOH(aq) + OHdetermines the pH at equivalence
Answer: 8.26