Polyprotic Acids / pH Calculation
A large number of acids can give two or more protons on ionization (dissociation) and these are referred to
as polyprotic acids. For example, with sulfurous
acid (H2SO3) we have the successive ionizations:
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Fig. I.1: Stepwise
dissociation of sulphurous acid (H2SO3)
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A polyprotic acid always dissociates
in a stepwise manner, one proton at a time. Note that the acid dissociation constants are labelled ka1 and ka2. The numbers on the constants refer to the particular
proton of the acid that is ionizing. Thus, ka1
always refers to the equilibrium involving removal of the first proton of a polyprotic acid. Note also that ka2 for sulfurous acid is
much smaller than ka1.
This can be explained by the fact that the second H+ has to leave
from a negatively charged species, HSO3- - electrostatic
attraction has to be overcomed – while the first H+ from a neutral
compound H2SO3.
The above observation is general: It is always easier to remove
the first proton from a polyprotic acid
than to remove the second and so on. The ka
values become successively smaller as successive protons are removed.
The acid dissociation constants for common polyprotic acids are given in Table I.1
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Table I.1: Stepwise
dissociation constants for several common polyprotic acids
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Polyprotic Acids and Ionic Equilibria
Depending on the pH of the solution, a polyprotic acid may exist predominantly as the undissociated acid or any one of its anionic forms. It is easy to calculate the fraction present of the species involved in the equilibrium as a function of [H+].
As an example let us calculate
the fraction of phosphoric acid
present as a function of the pH. Phosphoric
acid is typical of most weak polyprotic acids in that its successive ka values are very
different.
From the ionic equilibria shown above it is obvious that the relative
acid strengths are:
H3PO4
>> H2PO4- >> HPO4-2
This means that in a solution prepared by dissolving H3PO4
in water, only the first dissociation step makes an important contribution to
[H+]. This greatly simplifies the pH calculations for phosphoric
acid solutions as is illustrated in the Example I.1 below.
Let us though first derive
general expressions for [H3PO4], [H2PO4-],
[HPO4-2] as a function of [H+] in phosphoric acid
solutions.
Polyprotic Acids – Phosphoric Acid - General Expression of the Equilibrium Concentrations (of the species involved as a function of [H+])
Since there are six unknowns (the concentration of the acid, of the
three conjugate bases and of H+
and OH-] we need six equations to define the relations between these
quantities. These are
The equilibria ( equilibrium constant expressions):
[H+] * [H2PO4-] = ka1
* [H3PO4] (1)
[H+] * [HPO4-2 ] = ka2 * [H2PO4-] (2)
[H+] * [PO4-3] = ka3 * [HPO4-2] (3)
[H+] * [OH-] = kw (4)
The mass balance on phosphate:
[H3PO4] + [H2PO4-]
+[HPO4-2] + [PO4-3] = C (4a)
where C is the
initial concentration of phosphoric acid.
The fraction of the acid present as each species is the ratio of the
concentration of that species to the initial concentration C of the acid.
a3 = [H3PO4] / C
Note: The index on a shows
the number of protons attached to the molecule. It is more convenient to
calculate the reciprocal of this fraction, since this is directly expressible
in terms of the mass balance (4a).
1/a3 = C/[H3PO4] = 1 + [H2PO4-]
/ [H3PO4] + [HPO4-2] / [H3PO4]
+ [PO4-3] / [H3PO4] (5)
From equation (1): [H2PO4-]
/ [H3PO4] = ka1 / [H+] (6)
Multiplying (1) by (2) gives the third term of (5):
[HPO4-2] / [H3PO4] = (ka1*
ka2)/ [H+]2
(7)
Multiplying (1) by (2) by (3) gives the fourth term of (5):
[PO4-3] / [H3PO4] = (ka1* ka2* ka3)/
[H+]3 (8)
Substituting (6), (7) and (8) in (5) gives:
a3 = [H3PO4]
/ C = [ 1 + ka1 / [H+] + (ka1* ka2)/
[H+]2 + (ka1* ka2* ka3)/
[H+]3]-1 (9)
Combining (9) and (6):
a2 = [H2PO4-]
/ C = a3 * ka1 / [H+] (10)
Combining (9) and (7):
a1 = [HPO4-2]
/ C = a3 * (ka1* ka2) / [H+]2 (11)
Combining (9) and (8):
ao = [PO4-3]
/ C = a3 * (ka1* ka2* ka3) / [H+]3 (12)
There are very few situations where the above exact relations are used.
The above formulas should not be memorized since equilibrium concentrations for the species involved can be derived
from equilibrium constant expressions
and the assumption that all [H+] in solution is generated by the
dissociation of H3PO4 (Example I.1).
Example I.1
Calculate the pH of a 5.0 M H3PO4 solution and determine equilibrium concentrations of the
species involved
Solution
DATA
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[H3PO4]initial
= C = 5.0 M
ka1 =
7.5 * 10-3
ka2 =
6.2 * 10-8
ka3 =
4.8 * 10-13
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UNKNOWNS
|
[H3PO4] = ? [H2PO4-] =
?
[HPO4-2]
=
? [PO4-3] = ?
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Mathematical Approach
(Exact Solution)
STEP 1: Write down
the equilibrium reactions. Phosphoric acid H3PO4 is a
weak triprotic acid – see the ka ‘s in Fig. I.1. It dissociates partially in
water and most of the H+ is generated at the first dissociation
reaction (notice that ka1
>> ka2 >> ka3)
STEP 2: Write down the equilibrium constant expression and the equilibrium constant.
The dominant equilibrium will be the dissociation of H3PO4
and the equilibrium constant expression is given by:
ka1 = [H+]
* [H2PO4-] / [H3PO4] =
7.5 * 10-3 (13)
STEP 3: Write down
the equilibrium concentration of the
species involved in the dominant equilibrium reaction.
Suppose that x M of H3PO4 dissociates in water
H3PO4
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H+
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H2PO4-
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Initial
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5.0 Μ
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0
|
0
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Change
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-x M
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+x M
|
+x M
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Final
(at equilibrium)
|
(5.0 – x) M
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x M
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x M
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STEP 4: Substitute the equilibrium concentrations into the
expression for ka1 and solve for x.
Since the value of x is small (5.0 – x) ~ 5
ka1 = [H+] * [H2PO4-]
/ [H3PO4] = (x)*(x) / (5.0 – x) ~ x2/5 = 7.5 * 10-3
and therefore x ~ 1.9*10-1 M, [H+]=
[H2PO4-] = x = 0.19 M, pH = 0.72
The [H3PO4] = 5.0 – x = 4.8 M
The concentration of [HPO4-2] can be obtained from
(11):
a1 = [HPO4-2]
/ C = a3 * (ka1* ka2) / [H+]2 (11)
The only unknown in the above equation is a3. It can be
calculated from (9):
a3 = [H3PO4] / C = [ 1 + ka1
/ [H+] + (ka1* ka2)/ [H+]2
+ (ka1* ka2* ka3)/ [H+]3]-1
=
[ 1 + (7.5*10-3/ 0.19) + (7.5 * 10-3*
6.2 * 10-8)/(0.19)2 + (7.5 * 10-3* 6.2 * 10-8*
4.8 * 10-13)/(0.19)3]-1 = [ 1 + (0.039) + (1.288*10-8)
+ (3.258*10-20) ]-1 ~ 1 / 1.039 = 0.96
Therefore, a3
~ 0.96
(12)
Substituting in (11) gives:
a1 = [HPO4-2] / 5 = a3 * (ka1*
ka2) / [H+]2 = 0.96 * (7.5 * 10-3*
6.2 * 10-8) / (0.19)2 = 1.236 *10-8
Therefore, [HPO4-2]
= 5 * 1.236 * 10-8 = 6.18 * 10-8 M ~
6.2 * 10-8 M (13)
Similarly, [PO4-3]
can be calculated. [PO4-3] = 1.6 *10-19 M (14)
“Chemical Approach” (Approximations)
Proceed as above up to the calculation of [H+] = [H2PO4-]
= 0.19 M. The concentration of [HPO4-2] can be obtained from the
equilibrium constant expression and ka2 (dissociation of H2PO4-).
H2PO4- (aq) = H+(aq) + HPO4-2 (aq)
ka2 = 6.2 * 10-8 (15)
ka2 = [H+] * [HPO4-2 ] / [H2PO4-]
= 6.2 * 10-8 (16)
Substituting the values for [H+] = [H2PO4-]
in (16):
[HPO4-2 ] = ka2 = 6.2 * 10-8 M (13)
Similarly, [PO4-3] can be calculated from the
equilibrium constant expression and ka3 to give as above:
[PO4-3] = 1.6 *10-19 M (14)
The results show that the dissociation
of H2PO4- and HPO4-2 is
negligible relative to that of H3PO4. This is apparent
from the fact that [HPO4-2] is 6.2 * 10-8 M –
showing that only 6.2 * 10-8 mol/l of H2PO4-
has dissociated. The value of [PO4-3] shows that the
dissociation of HPO4-2 is even smaller.
For many polyprotic acids ka1 is much larger than
subsequent dissociation constants,
in which case the H+ in the solution comes almost entirely from the
first dissociation reaction. As long as successive ka values differ by a factor of 103 or more,
it is possible to obtain a satisfactory estimate of the pH of polyprotic acid
solution by treating the acids as if they were monoprotic, considering only ka1.
In a manner analogous to a triprotic
acid the following formulas can be derived for a diprotic acid of the general form H2A:
a2 =[H2A]
/ C = ([ 1 + ka1 / [H+] + (ka1* ka2)/
[H+]2)-1 (17)
a1= [HA-]
/ C = a2 * ka1 / [H+] (18)
ao = [A-]
/ C = a2 * (ka1 * ka2)/ [H+]2
(19)
Where ao , a1, a2 are the fraction of a
diprotic acid present as each of the
three species H2A, HA- and A-.
However, as for the triprotic
acid there are very few situations where the above exact relations are
used. The above formulas should not be memorized since equilibrium concentrations for the species involved can be derived
from equilibrium constant expressions
and the assumption that all [H+] in solution is generated by the
dissociation of H2A.
For another solved example regarding pH calculation of
a polyprotic acid please see the video below:
Relevant Posts
Solving Weak Acid and Weak Base pH problems
Weak Acid Weak Base pH calculation solved example
Weak Acids and Bases - Calculate the pH of a weak acid
Cubic Equation Calculator for Weak Acid-Base Equilibria
Self-ionization of water - Autoionization of water - The ion product of water (kw)
References
J-L.
Burgot “Ionic Equilibria in Analytical Chemistry”, Springer Science &
Business Media, 2012
J.N.
Butler “Ionic Equilibrium – Solubility
and pH calculations”, Wiley – Interscience, 1998
Clayden,
Greeves, Waren and Wothers “Organic Chemistry”, Oxford,
D.
Harvey, “Modern Analytical Chemistry”,
McGraw-Hill Companies Inc., 2000
Toratane
Munegumi, World J. of Chem. Education, 1.1, 12 (2013)
J.N.
Spencer et al., “Chemistry structure and dynamics”, 5th Edition, John Wiley
& Sons, Inc., 2012
L.
Cardellini, Chem. Educ. Res. Pract. Eur., 1, 151, (2000)
Key Terms
polyprotic acids, dissociation, dissociation constants, phosphoric
acid, concentration of the species as a function of [H+],
conjugate base, equilibrium constant expressions, equilibrium
concentrations, calculate the pH, monoprotic, ka, triprotic acid, polyprotic
acids and bases problems polyprotic acid and pka, polyprotic acid base
equilibria, polyprotic acid concentration calculation, diprotic
acid concentration calculation, polyprotic acids dissociation
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