Titration of a Polyprotic Acid with a Strong Base
Each proton in a polyprotic acid is supposed to titrate separately. However, only those
protons which satisfy the empirical relation ka1 > 104 ka2 can result in an observable
break at the point of equivalence. For example, carbonic acid shows two breaks in the
titration curve. Each one corresponds to a specific proton of the acid. The method of
calculation of the pH is similar to that described above but initially for the first proton
then the second. Each equivalence point requires a separate indicator to visualize the
end point. There are few points to put in mind when dealing with problems of titration
of polyprotic acids with strong bases:
1. Before addition of any base, you only have the polyprotic acid solution and
thus calculation of the pH is straightforward as previously described.
2. When we start addition of base, the first proton is titrated and bicarbonate
will form. A buffer solution of carbonic acid and carbonate is formed and you
should refer to the section on such calculations.
3. When all the first proton is titrated, all carbonic acid is now converted to
bicarbonate (an amphoteric protonated salt) and calculation of the pH is
achieved using the appropriate root mean square equation.
4. Further addition of base starts titrating the second proton thus some
bicarbonate is converted to carbonate and a buffer is formed. Calculate the pH
of the resulting buffer in the same way as in step 2.
5. When enough base is added so that the titration of the second proton is
complete, all bicarbonate is converted to carbonate and this is the second
equivalence point. The pH is calculated for carbonate (unprotonated salt).
6. Addition of excess base will make the solution basic where this will suppress
the dissociation of carbonate. The hydrogen ion concentration is calculated from
the concentration of excess hydroxide.
CO32- / OHHCO3-/CO32-
pH
CO32-
H2CO3/HCO3HCO3-
mL Base
Example
Find the pH of a 50 mL solution of a 0.10 M H2CO3 after addition of 0, 25, 50, 75,
100, and 150 mL of 0.10 M NaOH. Ka1=4.3x10-7 and ka2 = 4.8x10-11.
Solution
1. After addition of 0 mL NaOH
We only have the carbonic acid solution and the pH calculation for such types of
solution was discussed earlier and can be worked as below:
H2CO3 = H+ + HCO3HCO3- = H+ + CO32-
ka1 = 4.3 x 10-7
ka2 = 4.8 x 10-11
Since ka1 is much greater than ka2, we can neglect the H+ from the second step and
therefore we have:
H2CO3 = H+ + HCO3-
ka1 = 4.3 x 10-7
Before Equilibrium
0.10
0
0
Equation
H2CO3
H+
HCO3-
At Equilibrium
0.10 - x
x
x
Ka1 = x * x/(0.10 – x)
Assume 0.10>>x since ka1 is small
4.3*10-7= x2/0.10
x = 2.1x10-4
Relative error = (2.1x10-4/0.10) x 100 = 0.21%
The assumption is valid and [H+] = 2.1x10-4 M
pH = 3.68
2. After addition of 25 mL NaOH
A buffer is formed from H2CO3 left and the formed HCO3Initial mmol H2CO3 = 0.10 x 50 = 5.0
Mmol NaOH added = 0.10 x 25 = 2.5
Mmol H2CO3 left = 5.0 – 2.5 = 2.5
[H2CO3] = 2.5/75
mmol HCO3- formed = 2.5
[HCO3-] = 2.5/75
H2CO3 = H+ + HCO3-
ka1 = 4.3 x 10-7
Before Equilibrium
2.5/75
0
2.5/75
Equation
H2CO3
H+
HCO3-
At Equilibrium
2.5/75 - x
x
2.5/75 + x
ka1 = x(2.5/75 + x)/(2.5/75 – x)
ka1 is very small and in presence of the common ion the dissociation will be
further suppressed. Therefore, assume 2.5/75>>x.
x = 4.3x10-7 M
Relative error = {4.3x10-7/(2.5/75)} x 100 = 0.0013%
The assumption is valid
[H+] = 4.3x10-7 M
pH = 6.37
3. After addition of 50 mL NaOH
Initial mmol H2CO3 = 0.10 x 50 = 5.0
Mmol NaOH added = 0.10 x 50 = 5.0
Mmol H2CO3 left = 5.0 – 5.0 = ??
This is the first equivalence point
mmol HCO3- formed = 5.0
[HCO3-] = 5.0/100 = 0.05
Now the solution contains only the protonated salt. Calculation of the pH can be
done using the relation
[H+] = {(ka1kw + ka1ka2[HCO3-])/(ka1 + [HCO3-]}1/2
[H+] = {(4.3x10-7 * 10-14 + 4.3x10-7 * 4.8x10-11 * 0.0.05)/(4.3x10-7 + 0.0.05)}1/2
[H+] = 4.5x10-9 M
pH = 8.34
4. After addition of 75 mL NaOH
Here you should remember that 50 mL of the NaOH will be used in the titration
of the first proton. Therefore, it is as if we add 25 mL to the HCO3- solution. We
then have:
Initial mmol HCO3- = 5.0
Mmol NaOH added = 0.10 x 25 = 2.5
Mmol HCO3- left = 5.0 – 2.5 = 2.5
[HCO3-] = 2.5/125
mmol CO32- formed = 2.5
[CO32-] = 2.5/125
Once again we have a buffer solution from HCO3- and CO32-. The pH is calculated as
follows:
HCO3- = H+ + CO32-
ka2 = 4.8 x 10-11
Before Equilibrium
2.5/125
0
2.5/125
Equation
HCO3-
H+
CO32-
At Equilibrium
2.5/125 - x
x
2.5/125 + x
ka1 = x(2.5/125 + x)/(2.5/125 – x)
ka1 is very small and in presence of the common ion the dissociation will be
further suppressed. Therefore, assume 2.5/125>>x.
x = 4.8x10-11 M
Relative error = {4.8x10-11/(2.5/125)} x 100 =
The assumption is valid
[H+] = 4.8x10-11 M
pH = 10.32
5. After addition of 100 mL NaOH
At this point, all carbonic acid was converted into carbonate. The first 50 mL of
NaOH were consumed in converting H2CO3 to HCO3-. Therefore, as if we add 50
mL to HCO3- solution and we have:
Initial mmol HCO3- = 5.0
Mmol NaOH added = 0.10 x 50 = 5.0
Mmol HCO3- left = 5.0 – 5.0 = ??
This is the second equivalence point
mmol CO32- formed = 5.0
[CO32-] = 5.0/150
CO32- + H2O = HCO3- + OHThe equilibrium constant which corresponds to this equilibrium is kb where:
Kb = kw/ka2
We used ka2 since it is the equilibrium constant describing relation between
CO32- and HCO3-. However, in any equilibrium involving salts look at the highest
charge on any anion to find which ka to use.
Kb = 10-14/4.8x10-1
Kb = 2.1x10-4
Before Equilibrium
Equation
At Equilibrium
5.0/150
CO32- +
H2O
5.0/150 – x
0
HCO3-
0
OH-
X
x
Kb = x * x/(5.0/150 – x)
Assume 5.0/150 >> x
2.1x10-4 = x2/(5.0/150)
x = 2.6x10-3
Relative error = (2.6x10-3 /(5.0/150)) x 100 = 7.9%
Therefore, assumption is invalid and we have to use the quadratic equation. However,
I’ll accept the answer this time.
Therefore, [OH-] = 2.6x10-3 M
pOH = 2.58
pH = 14 – 2.58 = 11.42
6. After addition of 150 mL NaOH
At this point, all carbonic acid was converted into carbonate requiring 100 mL NaOH.
mmol NaOH excess = 0.1 x 50 = 5.0
[OH-] = 5.0/150
pOH = 1.48
pH = 14 – 1.48 = 12.52
Titration of a Polybasic base with a Strong Acid
This type of titration is similar to that described for the polyprotic acid but starting
from the base till we get to the acid at the end of the titration. An example is the
titration of carbonate or phosphate using a strong acid like HCl. Let us now consider
the calculation of pH of the base after addition of HCl as in the following example:
Example
Find the pH of a 50 mL solution of a 0.10 M Na3PO4 (ka1 = 1.1x10-2, ka2 = 7.5x10-8,
ka3 = 4.8x10-13) after addition of 0, 25, 50, 75, 100, 125, 150, and 175 mL of 0.10 M
HCl.
Solution
1. After addition of 0 mL HCl
At this point, we only have the solution of PO43- (an unprotonated salt) and we can
find the pH as follows
PO43- + H2O = HPO42- + OHThe equilibrium constant which corresponds to this equilibrium is kb where:
Kb = kw/ka3
We used ka3 since it is the equilibrium constant describing relation between PO43- and
HPO42-. However, in any equilibrium involving salts look at the highest charge on any
anion to find which ka to use.
Kb = 10-14/4.8x10-13
Kb = 0.020
Before Equilibrium
Equation
0.10
PO4 + H2O
0
HPO4-
0
OH-
At Equilibrium
0.10 –x
x
x
3-
Kb = x * x/0.10 – x
Assume 0.10 >> x
0.02 = x2/0.10
x = 0.045
Relative error = (0.045/0.10) x 100 = 45%
Therefore, assumption is invalid and we have to use the quadratic equation. If we
solve the quadratic equation we get:
X = 0.036
Therefore, [OH-] = 0.036 M
pOH = 1.44
pH = 14 – 1.44 = 12.56
2. After addition of 25 mL HCl
A buffer starts forming from phosphate remaining and the hydrogen phosphate
produced from the reaction.
PO43- + H+ = HPO42Initial mmol PO43- = 0.10 x 50 = 5.0
Mmol H+ added = 0.10 x 25 = 2.5
Mmol PO43- left = 5.0 – 2.5 = 2.5
[PO43-] = 2.5/75
mmol HPO42- formed = 2.5
[HPO42-] = 2.5/75
Now we look at any dissociation equilibrium equation containing both species. This
can be obtained from the relation from ka3, for example
HPO43- = PO43- + H+
Before Equilibrium
Equation
2.5/75
HPO42-
2.5/75
PO43-
0
H+
At Equilibrium
2.5/75 – x
2.5/75 + x
x
Ka3 = x(2.5/75 + x)/(2.5/75 – x)
Since ka3 is very small, assume 2.5/75 >> x
4.8x10-13 = x(2.5/75)/(2.5/75)
x = 4.8x10-13
It is clear that the relative error will be exceedingly small and the assumption is, for
sure, valid.
[H+] = 4.8x10-13
pH = 12.32
3. After addition of 50 mL HCl
At this point, all PO43- will be converted to HPO42Initial mmol PO43- = 0.10 x 50 = 5.0
Mmol H+ added = 0.10 x 50 = 5.0
Mmol PO43- left = 5.0 – 5.0 = ??
This is the first equivalence point
mmol HPO42- formed = 5.0
[HPO42-] = 5.0/100 = 0.05 M
This is a protonated salt with two charges where we should use ka2 and ka3, i.e.
the relation
[H+] = {(ka2kw + ka2ka3[HPO42-])/(ka2 + [HPO42-]}1/2
[H+] = 2.3x10-10
pH = 9.65
4. After addition of 75 mL HCl
A second buffer is formed where we have
HPO42- + H+ = H2PO42You should understand that 50 mL were consumed in the conversion of PO43- to
HPO42-, thus 25 mL only were added to HPO42Initial mmol HPO4- = 0.10 x 50 = 5.0
Mmol H+ added = 0.10 x 25 = 2.5
Mmol HPO42- left = 5.0 – 2.5 = 2.5
[HPO42-] = 2.5/125
mmol H2PO42- formed = 2.5
[H2PO4-] = 2.5/125
Calculation of the pH is dependent on the ratio of hydrogen phosphate to dihydrogen
phosphate and the best way to calculate the pH is to use the ka2 relation which
combines both terms
H2PO4- = H+ + HPO42Before Equilibrium
Equation
2.5/125
H2PO42-
2.5/125
HPO42-
0
H+
At Equilibrium
2.5/125 – x
2.5/125 + x
x
Ka2 = x(2.5/125 + x)/(2.5/125 – x)
Since ka3 is very small, assume 2.5/125 >> x
7.5x10-8 = x(2.5/125)/(2.5/125)
x = 7.5x10-8 M
It is clear that the relative error will be exceedingly small and the assumption is, for
sure, valid
[H+] = 7.5x10-8 M
pH = 7.12
5. After addition of 100 mL HCl
50 mL of HCl were consumed in converting PO43- into HPO42Initial mmol HPO42- = 0.10 x 50 = 5.0
Mmol H+ added = 0.10 x 50 = 5.0
Mmol HPO42- left = 5.0 – 5.0 = ??
This is the second equivalence point
[H2PO4-] = 5.0/150 = 0.033 M
At this point, all HPO42- will be completely converted into H2PO4- which is a
protonated salt where the pH can be calculated from the relation
[H+] = {(ka1kw + ka1ka2[H2PO4-])/(ka1 + [H2PO4-]}1/2
[H+] = {ka1ka2}1/2
[H+] = 2.9x10-5 M
pH = 4.54
6. After addition of 125 mL HCl
H2PO4- + H+ = H3PO4
50 mL were consumed in converting PO43- to HPO42- and 50 mL were consumed
in converting HPO42- into H2PO4-, therefore as if we add 25 mL to H2PO4Initial mmol H2PO4- = 0.10 x 50 = 5.0
Mmol H+ added = 0.10 x 25 = 2.5
Mmol HPO42- left = 5.0 – 2.5 = 2.5
[HPO42-] = 2.5/175
mmol H3PO4 formed = 2.5
[H3PO4] = 2.5/175
This is a buffer formed from the acid and its conjugate base. The best way to calculate
the pH is to use the ka1 expression where:
H3PO4 = H+ + H2PO4-
Before Equilibrium
2.5/175
2.5/175
0
Equation
H3PO4
H2PO4-
H+
At Equilibrium
2.5/175 – x
2.5/175 + x
x
Ka1 = x(2.5/175 + x)/(2.5/175 – x)
Since ka1 is very small (!!!), assume 2.5/175 >> x
1.1x10-2 = x(2.5/175)/(2.5/175)
x = 1.1x10-2 M
Relative error = {1.1x10-2/(2.5/175)} x 100 = 77%
It is clear that the relative error is very large and the assumption is, for sure, invalid
and we should use the quadratic equation.
[H+] = 5.2x10-3 M
pH = 2.29
7. After addition of 150 mL HCl
At this point, all PO43- is converted into the acid
Initial mmol H2PO4- = 0.10 x 50 = 5.0
Mmol H+ added = 0.10 x 50 = 5.0
Mmol HPO42- left = 5.0 – 5.0 = ??
This is the third equivalence point
mmol H3PO4 formed = 5.0
[H3PO4] = 5/200 = 0.025
Therefore, we only have the acid in solution and calculation of the pH is done as
follows:
H3PO4 = H+ + H2PO4ka1 = 1.1 x 10-2
H2PO4- = H+ + HPO42ka2 = 7.5 x 10-8
23+
HPO4 = H + PO4
ka3 = 4.8 x 10-13
Since ka1 >> ka2 (ka1/ka2 > 104) the amount of H+ from the second and consecutive
equilibria is negligible if compared to that coming from the first equilibrium.
Therefore, we can say that we only have:
H3PO4 = H+ + H2PO4-
ka1 = 1.1 x 10-2
Before Equilibrium
0.025
0
0
Equation
H3PO4
H+
H2PO4-
At Equilibrium
0.025 - x
x
x
Ka1 = x * x/(0.025 – x)
Assume 0.025>>x since ka1 is small (!!!)
1.1*10-2 = x2/0.025
x = 0.017
Relative error = (0.017/0.025) x 100 = 6.6%
The assumption is invalid according to the criteria we set at 5% and thus we
have to use the quadratic equation. However, I'll accept this answer.
Therefore, [H+] = 0.017 M
pH = 1.78
7. After addition of 175 mL HCl
50 mL were consumed in converting PO43- to HPO42-, 50 mL were consumed in
converting HPO42- into H2PO4-, and 50 mL HCl were consumed in converting H2PO4to H3PO4, therefore, 25 mL of excess HCl are added
Mmol H+ excess = 0.10 x 25 = 2.5
[H+] = 2.5/225 = 0.011
pH = 1.95
It may be of interest to indicate that the abovementioned calculations are
theoretically, however when the equilibrium constant is exceedingly small it is
not possible to titrate that very weak acid or base. Thus, an end point may not be
observed for such a titration step.