Polyprotic Acids
And Acid and Base Salts
Polyprotic Acids

So far, we have only dealt with acids that
can give up one proton.

Most acids encountered in biological
systems have multiple protons, depending
on the pH of the solution.

We call these ‘polyprotic’ acids or bases.
Diprotic acids and bases
H 2 A (aq) 
 H
HA
(aq)

(aq)

 H

(aq)
 HA
A

(aq)
2
(aq)
; K a1
; K a2
How do we calculate the pH of a solution of:
H2A, HA-, or A2- ?
Diprotic Acids

Treat this like a weak monoprotic acid:
H 2 A (aq) 
 H
[H

(aq)

(aq)
 HA (aq) ; K a1  6.3x10
][ HA
-
-
(aq)
[H 2 A (aq) ]
I
C
E
]
 6.3x103
H2A
HA-
H+
0.0750
-x
0.0750-x
0
+x
x
0
+x
x
3
Diprotic Acids
( x)( x)
 6.3x10 3
0.0750 - x
x 2  6.3x10 3 x  4.725 x10  4
cannot ignore x in this case

x  [H ]  [HA ]  0.019M
-
[H 2 A]  0.0750M  0.019M  0.056M
What about [A2-]?
Diprotic Acids

H 2 A (aq) 
 H (aq)
 HA - (aq) ; K a1  6.3x10 3

2
HA - (aq) 
 H (aq)
 A (aq)
; K a2  4.9 x1010
[H

(aq)
][ A
2(aq)
[HA - (aq) ]
]
 4.9 x10 10
The amount of H+ from dissociation of HA- is insignificant
relative to H+ from the dissociation of H2A, so total [H+] =
0.019M, and HA- also equals 0.019 M
10
10
4.9
x
10
[HA
]
(4.9
x
10
)(0.019)
(aq)
210
[A (aq) ] 


4
.
9
x
10
M

[H (aq) ]
(0.019)
Polyprotic Acids

Here are three successive ionizations of phosphoric
acid:
H3PO4(s) + H2O(l)
H2PO4−(aq)+ H2O(l)
HPO42−(aq)+ H2O(l)


H3O+(aq) + H2PO4−(aq)
H3O+(aq) + HPO42−(aq)
H3O+(aq) + PO43−(aq)
Ka1= 7.25×10−3
Ka2= 6.31×10−8
Ka3= 3.98×10−13
The first dissociation constants for phosphoric acid
is much greater than the second, about 100,000
times greater
This means nearly all the H+ ions in the solution
comes from the first step of dissociation.
Example
Calculate the H+, H3PO4, H2PO4-, HPO42-,
and PO43- concentrations at equilibrium in a
0.10 M H3PO4 solution, for which Ka1 = 7.1 x
10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13.
H3PO4(s) + H2O(l)
H2PO4−(aq)+ H2O(l)
HPO42−(aq)+ H2O(l)
H3O+(aq) + H2PO4−(aq)
H3O+(aq) + HPO42−(aq)
H3O+(aq) + PO43−(aq)
Ka1= 7.25×10−3
Ka2= 6.31×10−8
Ka3= 3.98×10−13
Example
X = .023
x2
0.10 - x
[H3PO4] = 0.10 - .023 = 0.077 M
Ka =
[H3O+] = [H2PO4-] = 0.023 M
Example
Substituting what we know about the H3O+ and
H2PO4- ion concentrations into the second
equilibrium expression gives:
[HPO42-] = 6.3 x 10-8
Example

Substituting what we know about the
concentrations of the H3O+ and HPO42- ions
into this expression gives
PO43- = 1.2 x 10-18
Salts



In general, salts are ionic compounds
composed of metallic ions and nonmetallic
ions
Salts dissociate in water. Salt solutions are
generally electrolytes.
An electrolyte is a substance that ionizes or
dissociates into ions when it dissolves in
water (conducts electricity)
Salt + Water



The reaction of a salt and water to form an acid and
base is called hydrolysis.
NaCl + H2O  NaOH + HCl
This is the reverse of a neutralization reaction in
which acid and bases react to form a salt and water.
When acids and bases react, the relative strength of
the conjugated acid-base pair in the salt determines
the pH of its solution.
Adding a “salt” to water


If the salt is from a strong acid or base, then
nothing will happen (like adding table salt to
water – no change in pH).
If it is a conjugate of a weak acid or base,
then the “salt” is itself also a weak base or
acid. So it hydrolyzes and makes some H+ or
OH-, which changes the pH.
Adding a “salt” to water



If the salt is from a result of a strong acid and
base then the pH is 7, for example KNO3.
A salt formed between a strong acid and a
conjugate of a weak base is an acid salt, for
example NH4Cl, and the pH will be acidic.
A salt formed between a conjugate of a weak
acid and a strong base is a basic salt, for
example NaCH3COO, and the pH will be
basic.
Acid-base properties of salt
solutions: hydrolysis
NaCl (aq)
NH4Cl (aq)
NaClO (aq)
Hydrolysis example
Which of the following salts, when added to
water, would produce the most acidic
solution?
a) KBr
b) NH4NO3
c) AlCl3
d) Na2HPO4
The answer is B
Example #1
Will an aqueous solution that is 0.20 M NH4F
be acidic, basic or neutral?
NH4+ is the conjugate of a weak base and F- is
the conjugate of a weak acid, so how the two
ions compare in their ability to affect the pH
must be determined.
NH4+ } ka = 5.6 x 10-10
F- } kb = 1.5 x 10-11

The acid is stronger than the base so the
solution will be slightly acidic.
Example #2
What is the pH of a 0.10 M solution of NaOCl?
HOCl, ka = 3.0x10-8
1. What type of salt is this? Na+ (from a strong
base so does not effect the pH) and OCl(from the weak acid HOCl)
2. Write the equilibrium expression for the
dissolved salt.
OCl- + H2O < HOCl- + OH-
Example #2 cont.
3.
Kb = kw/ka = 3.3 x 10-7
From ICE chart kb = x2/0.10
x= 1.8 x 10-4 M,
pOH = 3.74, pH =10.26
Example #3
What is the pH of a 0.20 M solution of
hydrazinium chloride, N2H5Cl? Hydrazine,
N2H4, is a weak base with kb = 1.7 x 10-6.
Answer: conj. Acid in solution write the acid
dissociation and use the ICE chart to
determine the H+ and the pH.
x = 3.4 x 10-5, pH = 4.47