1

Titration of a Weak Base with a Strong Acid

The same principles applied above are also applicable where we have:

1. Before addition of any acid, we have a solution of the weak base and calculation of the pH of the weak base should be performed as in previous sections.

2. After starting addition of the strong acid to the weak base, the salt of the weak base is formed. Therefore, a buffer solution results and you should consult previous lectures to find out how the pH of buffers is calculated.

2

3. At the equivalence point, the amount of strong acid is exactly equivalent to the weak base and thus there will be 100% conversion of the weak base to its salt.

The problem now is to calculate the pH of the salt solution.

4. After the equivalence point, we would have a solution of the salt with excess strong acid. The

presence of the excess acid suppresses the dissociation of the salt in water and the pH of the

solution controlled by the excess acid only.

Now, let us apply the abovementioned concepts on an actual titration of a weak base with a

strong acid.

3

Example

Find the pH of a 50 mL solution of 0.10 M NH3 (kb =

1.75x10-5) after addition of 0, 10, 25, 50, 60 and 100 mL of 0.10 M HCl.

Solution

1. After addition of 0 mL HCl

The solution is only 0.10 M in ammonia, therefore we have:

NH3 + H2O NH4+

+ OH-

4

Kb = [NH4+][OH-]/[NH3]

1.75*10-5 = x * x / (0.10 – x)

kb is very small that we can assume that 0.10>>x. We

then have:

1.75*10-5 = x2 / 0.10

x = 1.3x10-3 M

Relative error = (1.3x10-3 /0.10) x 100 = 1.3 %

The assumption is valid, therefore:

[OH-] = 1.3x10-3 M

pOH = 2.88

pH = 11.12

5

2. After addition of 10 mL HCl

A buffer will be formed from the base and its salt

Initial mmol NH3 = 0.10 x 50 = 5.0

mmol HCl added = 0.10 x 10 = 1.0

mmol NH3 left = 5.0 – 1.0 = 4.0

[NH3] = 4.0/60 M

mmol NH4+ formed = 1.0

[NH4+] = 1.0/60 M

NH3 + H2O NH4+

+ OH-

6

Kb = [NH4+][OH-]/[NH3]

1.75*10-5 = (1.0/60 + x) * x / (4.0/60 – x)

kb is very small that we can assume that 1.0/60

>>x. We then have:

1.75*10-5 = 1.0/60 x / 4.0/60

x = 7.0x10-5

Relative error = (7.0x10-5 /1.0/60) x 100 = 0.42 %

The assumption is valid, therefore:

[OH-] = 7.0x10-5 M

pOH = 4.15

pH = 9.85

7

3. After addition of 25 mL HCl

A buffer will be formed from the base and its salt

Initial mmol NH3 = 0.10 x 50 = 5.0

mmol HCl added = 0.10 x 25 = 2.5

mmol NH3 left = 5.0 – 2.5 = 2.5, [NH3] = 2.5/75 M

mmol NH4+ formed = 2.5, [NH4

+] = 2.5/75 M

NH3 + H2O NH4+

+ OH-

8

Kb = [NH4+][OH-]/[NH3]

1.75*10-5 = (2.5/75 + x) * x / (2.5/75 – x)

kb is very small that we can assume that 2.5/75

>>x. We then have:

1.75*10-5 = 2.5/75 x / 2.5/75

x = 1.75x10-5

Relative error = (1.75 x10-5 /2.5/75) x 100 = 0.052 %

The assumption is valid, therefore:

[OH-] = 1.75x10-5 M

pOH = 4.76

pH = 9.24

9

3. After addition of 50 mL HCl

A buffer will be formed from the base and its salt

Initial mmol NH3 = 0.10 x 50 = 5.0

mmol HCl added = 0.10 x 50 = 2.5

mmol NH3 left = 5.0 – 25.0 = 0

This is the equivalence point

mmol NH4+ formed = 5.0, [NH4

+] = 5.0/100 = 0.05 M

NH4+ H+ + NH3

10

Ka = 10-14/1.75x10-5 = 5.7x10-10

Ka = [H+][NH3]/[NH4+]

Ka = x * x / (0.05 – x)

Ka is very small. Assume 0.05 >> x

5.7*10-10 = x2/0.05

x = 5.33x10-6

Relative error = (5.33x10-6/0.05) x 100 = 0.011 %

The assumption is valid and the [H+] = 5.33x10-6 M

pH = 5.27

11

5. After addition of 60 mL HCl

Initial mmol of NH3 = 0.10 x 50 = 5.0

Mmol HCl added = 0.10 x 60 = 6.0

Mmol HCl excess = 6.0 – 5.0 = 1.0

[H+] = 1.0/110 M

pH = 2.04

6. After addition of 100 mL HCl

Initial mmol of NH3 = 0.10 x 50 = 5.0

Mmol HCl added = 0.10 x 100 = 10.0

Mmol HCl excess = 10.0 – 5.0 = 5.0

[H+] = 5.0/150 M

pH = 1.48

12

Titration of a Polyprotic Acid with a Strong BaseEach 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.

13

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.

14

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.

15

16

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

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+ + HCO3- ka1 = 4.3 x 10-7

HCO3- H+ + CO3

2- ka2 = 4.8 x 10-11

17

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

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

18

After addition of 25 mL NaOH

A buffer is formed from H2CO3 left and the formed

HCO3-

Initial 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 M

mmol HCO3- formed = 2.5

[HCO3-] = 2.5/75 M

H2CO3 H+ + HCO3- ka1 = 4.3 x 10-7

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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 MRelative error = {4.3x10-7/(2.5/75)} x 100 = 0.0013%

The assumption is valid[H+] = 4.3x10-7 M

pH = 6.37

20

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 = 0

This is the first equivalence point

mmol HCO3- formed = 5.0

[HCO3-] = 5.0/100 = 0.05 M

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

21

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 M

mmol CO32- formed = 2.5

[CO32-] = 2.5/125 M

Once again we have a buffer solution from HCO3- and CO3

2-.

The pH is calculated as follows:

HCO3- H+ + CO3

2- ka2 = 4.8 x 10-11

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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 = V. small

The assumption is valid, [H+] = 4.8x10-11 M

pH = 10.32

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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 M

24

CO32- + H2O = HCO3

- + OH- 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 = 10-14/4.8x10-13 = 2.1x10-4

25

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

26

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-] = 5.0/200

pOH = 1.60

pH = 14.00 – 1.60 = 12.40

27

Titration of a Polybasic base with a Strong Acid

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

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PO43- + H2O HPO4

2- + OH- kb= kw/ka3

We used ka3 since it is the equilibrium constant describing

relation between PO43- and HPO4

2-. 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 = 0.020

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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

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2. After addition of 25 mL HCl

A buffer starts forming from phosphate remaining and the hydrogen phosphate produced from the reaction.

PO43- + H+ HPO4

2-

Initial 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 M

mmol HPO42- formed = 2.5

[HPO42-] = 2.5/75 M

Now we look at any dissociation equilibrium equation containing both species. This can be obtained from the

relation from ka3, for example

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HPO42- PO4

3- + H+

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 M

pH = 12.32

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3. After addition of 50 mL HCl

At this point, all PO43- will be converted to HPO4

2-

Initial 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 + [HPO4

2‑]}1/2

[H+] = 2.3x10-10

pH = 9.65

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4. After addition of 75 mL HCl

A second buffer is formed where we have

HPO42- + H+ H2PO4

-

You should understand that 50 mL were consumed in the conversion of PO4

3- to HPO42-, thus 25 mL only were

added to HPO42-

Initial 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

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H2PO4- H+ + HPO4

2-

Before Equilibrium2.5/1252.5/1250

EquationH2PO42-HPO4

2-H+

At Equilibrium2.5/125 – x2.5/125 + xx

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

It is clear that the relative error will be exceedingly small and the assumption is, for sure, valid

[H+] = 7.5x10-8 MpH = 7.12

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5. After addition of 100 mL HCl

50 mL of HCl were consumed in converting PO43- into HPO4

2-

Initial mmol HPO42- = 0.10 x 50 = 5.0

mmol H+ added = 0.10 x 50 = 5.0

mmol HPO42- left = 5.0 – 5.0 = 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+] = 2.5x10-5 M

pH = 4.6

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6. After addition of 125 mL HCl

H2PO4- + H+ H3PO4

50 mL were consumed in converting PO43- to HPO4

2- and 50

mL were consumed in converting HPO42- into H2PO4

-,

therefore as if we add 25 mL to H2PO4-

Initial 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 M

mmol H3PO4 formed = 2.5

[H3PO4] = 2.5/175 M

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:

37

H3PO4 H+ + H2PO4-

Before Equilibrium2.5/1752.5/1750

EquationH3PO4H2PO4-H+

At Equilibrium2.5/175 – x2.5/175 + xx

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 MpH = 2.29

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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 = 0

This is the third equivalence point

mmol H3PO4 formed = 5.0

[H3PO4] = 5/200 = 0.025

39

Therefore, we only have the acid in solution and calculation of the pH is done as follows:

H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2

H2PO4- H+ + HPO4

2- ka2 = 7.5 x 10-8

HPO42- H+ + PO4

3- 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:

40

H3PO4 H+ + H2PO4- ka1 = 1.1 x 10-2

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 = 68%

The assumption is invalid according to the criteria we set at 5% and thus we have to use the quadratic equation.

41

After addition of 175 mL HCl

50 mL were consumed in converting PO43- to HPO4

2-, 50 mL

were consumed in converting HPO42- into H2PO4

-, and

50 mL HCl were consumed in converting H2PO4- to

H3PO4, therefore, 25 mL of excess HCl are added

mmol H+ excess = 0.10 x 25 = 2.5

[H+]excess = 2.5/225 = 0.011 M

[H+] = [H+]excess + [H+]H3PO4

42

Since at this point both H+ from excess HCl and phosphoric acid contribute to the overall [H+]

We can calculate the [H+] from ka1:

Ka1 = x(2.5/225 + x)/(5.0/225 – x)

Since ka1 is very small (!!!), assume 2.5/225 >> x

1.1x10-2 = x(2.5/225)/(5.0/225)

x = 2.2x10-2 M

Relative error = {2.2x10-2/(2.5/225)} x 100 = 198%

It is clear that the relative error is very large and the assumption is, for sure, invalid and we should use the

quadratic equation.