Titration of a Weak Base with a Strong Acid

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 D NH4 OH-

4

• Kb NH4OH-/NH3
• 1.7510-5 x x / (0.10 x)
• kb is very small that we can assume that 0.10gtgtx.
  We then have
• 1.7510-5 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 D NH4 OH-

6

• Kb NH4OH-/NH3
• 1.7510-5 (1.0/60 x) x / (4.0/60 x)
• kb is very small that we can assume that 1.0/60
  gtgtx. We then have
• 1.7510-5 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 D NH4 OH-

8

• Kb NH4OH-/NH3
• 1.7510-5 (2.5/75 x) x / (2.5/75 x)
• kb is very small that we can assume that 2.5/75
  gtgtx. We then have
• 1.7510-5 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 D H NH3

10

• Ka 10-14/1.75x10-5 5.7x10-10
• Ka HNH3/NH4
• Ka x x / (0.05 x)
• Ka is very small. Assume 0.05 gtgt x
• 5.710-10 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 Base
• Each proton in a polyprotic acid is supposed to
  titrate separately. However, only those protons
  which satisfy the empirical relation ka1 gt 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.

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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. Ka14.3x10-7 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 D H HCO3- ka1 4.3 x 10-7
• HCO3- D H CO32- 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 D H HCO3- ka1 4.3 x 10-7
• Ka1 x x/(0.10 x)
• Assume 0.10gtgtx since ka1 is small
• 4.310-7 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 D 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 2.5/75gtgtx. 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
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 ka1ka2HCO3-)/(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 CO32-. The pH is calculated as follows
• HCO3- D H CO32- 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 2.5/125gtgtx.
• 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

23

• 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 gtgt 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, Ill 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

28

• PO43- H2O D HPO42- OH- 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 0.020

29

• Kb x x/0.10 x
• Assume 0.10 gtgt 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 D HPO42-
• 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- D PO43- H
• Ka3 x(2.5/75 x)/(2.5/75 x)
• Since ka3 is very small, assume 2.5/75 gtgt 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

32

• 3. After addition of 50 mL HCl
• At this point, all PO43- will be converted to
  HPO42-
• 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 ka2ka3HPO42-)/(ka2
  HPO42-1/2
• 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 D H2PO4-
• You should understand that 50 mL were consumed in
  the conversion of PO43- 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- D H HPO42-
0 2.5/125 2.5/125 Before Equilibrium
H HPO42- H2PO42- Equation
x 2.5/125 x 2.5/125 x At Equilibrium
Ka2 x(2.5/125 x)/(2.5/125 x) Since ka3 is
very small, assume 2.5/125 gtgt 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 M pH 7.12
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• 5. After addition of 100 mL HCl
• 50 mL of HCl were consumed in converting PO43-
  into HPO42-
• 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 ka1ka2H2PO4-)/(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 D 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 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 D H H2PO4-

0 2.5/175 2.5/175 Before Equilibrium
H H2PO4- H3PO4 Equation
x 2.5/175 x 2.5/175 x At Equilibrium
Ka1 x(2.5/175 x)/(2.5/175 x) Since ka1 is
very small (!!!), assume 2.5/175 gtgt 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
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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

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• Therefore, we only have the acid in solution and
  calculation of the pH is done as follows
• H3PO4 D H H2PO4- ka1 1.1 x 10-2
• H2PO4- D H HPO42- ka2 7.5 x 10-8
• HPO42- D H PO43- ka3 4.8 x 10-13
• Since ka1 gtgt ka2 (ka1/ka2 gt 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 D H H2PO4- ka1 1.1 x 10-2
• Ka1 x x/(0.025 x)
• Assume 0.025gtgtx since ka1 is small (!!!)
• 1.110-2 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
  HPO42-, 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
• Hexcess 2.5/225 0.011 M
• H Hexcess HH3PO4

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 gtgt
  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.