Chapter 12 AcidBase Titrations

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Chapter 12 Acid-Base Titrations

• Strong Acid Strong Base Titrations
• Weak Acid Strong Base
• Weak Base Strong Acid

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Example titration of HCl with NaOH

• Shape of the titration curve Conceptually this
  is very similar to the precipitation reaction
  curve we discussed in Chapter 7.
• Lets start with the MBE and CBE.
• CBE H Na Cl- OH-
• MBE Na OH- neglecting Kw.
• Cl- H

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Consider Na and Cl- throughout Titration
Substituting back into the CBE
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CBE H Na Cl- OH-
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Before Equivalence Point

• All of OH- consumed by excess H

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At the equivalence point

• All of H is now consumed

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After the equivalence point

• Excess OH-

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At any point on the titration curve, the fraction
of H titrated (?) with HO- is
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Example calculate the pH of the titration of
50.00 mL of 0.100 M HCl with 0.100 M NaOH
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b) 50.00 mL of NaOH added
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c) 60.00 mL of NaOH added
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Titration curve strong acid titrated with strong
base

• Note that only for the strong base strong acid
  case will the equivalence point pH be 7.00.

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Titration of a strong base (e.g. NaOH) with a
strong acid (e.g. HCl) will appear as
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Location of the equivalence point.

• Acid-Base indicators respond to pH changes based
  on color changes. These indicators are all weak
  acids themselves.
• Phenolphthalein p 240.
• P2- 2H H2P pKa ? 8
• pink colorless
• The titration error occurs because the end point
  is ?pH 8 while the true equivalence point is pH
  7.

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Remember when using visual indicators you must

• Use as little as possible since they are acids or
  bases and will react with your analyte.
• Pick the appropriate indicator. Based on expected
  equivalence point pH. See Table 12-4.
• We will skip a quantitative discussion of
  titration error.

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Titration of a Weak Acid (HA) with a Strong Base
(NaOH).

• Similar considerations are at work here, well
  treat this case less rigorously. Consider the
  following example
• 25.00 mL of 0.150 M CH3COOH (Ka 1.75e-5) is
  titrated with 0.200 M NaOH.
• Calculate the pH of the titration solution at

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a) 00.00 mL

• The starting pH is simply,
• HA H A-
• 0.150 x x x
• 1.75e-5 x2/0.150-x
• x 1.62e-3 H
• pH 2.793 this is the starting pH of this
  experiment.

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b) 10.00 mL

• First we must consider the volume of the
  equivalence point to (Ve) know where we are on
  the titration curve
• Ve 25.00 mL 0.150 M HA 1/0.200 M NaOH
  18.75 mL
• _at_10.00 mL we are before the titration eq. point.
• Initial moles HA 25.00 mL 0.150 M HA 3.75
  mmol
• Added OH- 10.00 mL 0.200 M OH- 2.00 mmol
• HA OH- ? H2O A-
• 3.75 mmol 2.00 mmol
• -2.00 mmol -2.00 mmol 2.00 mmol
• 1.75 mmol 0 2.00 mol

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b) 10.00 mL continued

• note that MBE is HA F HA A-
• or in terms of amount (3.75 2.00 1.75) mmol
• ?3.75 1.75 mmol 2.00 mmol
• HA 1.75 mmol/(35.00 mL) 5.00e-2 M
• A- 2.00 mmol/35.00 mL 5.71e-2 M

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b) 10.00 mL continued

• Essentially what we have is a buffer system.
• Ka 1.75e-5 HA-/HA
• H5.71e-2/5.00e-2
• H 1.53e-5
• pH 4.815

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c) 18.75 mL

• The volume of the equivalence point to (Ve) is
• Ve 25.00 mL 0.150 M HA 1/0.200 M NaOH
  18.75 mL
• Note that HA molinitial A- mol eq. pt.
• 25.00 mL 0.150 M HA 3.75 mmol HA or A-
• Total volume 18.75 25.00 mL 43.75 mL
• A- 3.75 mmol/43.75 mL 8.57e-2 M

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pH at equivalence point

• All of HA is consumed left with A- which
  undergoes hydrolysis.
• A- H2O HA OH-
• 8.57e-2 -- 0 0
• -x x x
• Kb Kw/Ka 5.71e-10 x2/8.57e-2 x
• x 7.00e-6 M
• pOH 5.155
• pH 8.845 note that it is not 7.00.

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d) 25.00 mL

• We are past the end point so we should expect a
  region in which we have excess OH-.
• Mol excess OH- mol OH- added initial mol HA
• 25.00 mL 0.200 M NaOH 5.00 mmol OH-
• Mol excess OH- 5.00 mmol 3.75 mmol 1.25
  mmol
• OH- 1.25 mmol / 50.00 mL 2.50e-2 M
• pOH 1.602
• pH 12.398

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Fig 12-2
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Role of Ka and HAinitial Figure 12-3

• This makes the titration of low concentrations
  and/or high pKa acids difficult.

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Titration of a weak base with a strong acid.

• Consider the titration of NH3 with HCl.
• Rxn NH3 HCl ? NH4 Cl-
• Example 50.00 mL of 0.222 M NH3 (Ka NH4
  5.5e-10) is titrated with 0.155 M HCl.

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a) 0.00 mL

• NH3 H2O NH4 OH-
• 0.222 -- 0 0
• -x x x
• Kb 1.00e-14 / 5.5e-10 x2 / 0.222-x
• x OH- 2.01e-3 M
• pH 11.30

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b) 50.00 mL 0.155 M HCl added

• First calculate volume of equivalence point Ve
• Ve 50.00 mL 0.222 M NH3 1/0.155 M
• 71.61 mL HCl solution
• We can see that we are before the end point, i.e.
  excess NH3.
• Initial mol NH3 50.00 mL 0.222 M 11.1 mmol
• Added mol HCl 50.00 mL 0.155 M 7.75 mmol
• Note that ? mol a/ mol b 7.75/11.1 0.698

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pH at 50.00 mL 0.155 M HCl added

• Excess NH3 11.1 7.75 mmol 3.35 mmol
• NH3 3.35 mmol / 100.00 mL 3.35e-2 M
• must calculate NH4
• NH3 HCl ? NH4 Cl-
• 11.1 7.75 0 0
• -7.75 -7.75 7.75 7.75
• mol NH4 7.75 mmol
• NH4 7.75 mmol / 100.00 mL 7.75e-2 M

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This a pH region where both NH3 and NH4 are
present

• Ka HNH3 / NH4
• 5.5e-10 H 3.35e-2 M / 7.75e-2 M
• H 1.27e-9
• pH 8.90

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pH pKa at ½ Ve added

• What happens at 35.805 mL (½ 71.61 mL) of 0.155 M
  HCl added?
• Initial mol NH3 50.00 mL 0.222 M 11.1 mmol
• Added mol HCl 35.805 mL 0.155 M 5.55 mmol
• Mol NH3 left 11.1 5.55 mmol 5.55 mmol
• Ka HNH3 / NH4
• H 5.55mmol / 5.55 mmol (vol. of solution
  cancel)
• pKa pH 9.26

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c) pH at the equivalence point, Ve, 71.61 mL of
0.155 M HCl added

• NH3 11.1 mmol / 121.61 mL 9.13e-2 M or ?
  1
• Here all of NH3 is consumed and we are left with
• NH4 NH3 H
• 9.12e-2M 0 0
• -x x x
• 5.5e-10 x2 / 9.12e-2 x
• x 7.08e-6 M
• pH 5.15

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d) 80.00 mL of 0.155 M HCl.

• Initial mol NH3 11.1 mmol
• Added mol H 80.00 mL 0.155 M 12.4 mmol
• ? 12.4/11.1 1.12
• Excess H 12.4 11.1 mmol 1.3 mmol
• H 1.3 mmol / 130.00 mL 0.010 M
• pH 2.00

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Summary of titration of 50.00 mL of 0.222 M NH3
with 0.155 M HCl.

• Vol of HCl added pH
• a) 0.00 mL 11.30
• 35.81 mL 9.26
• b) 50.00 mL 8.90
• c) 71.61 mL 5.15
• d) 80.00 mL 2.00

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Buffer region
Excess HCl
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Diprotic systems.

• In cases where the polyprotic acid has well
  separated pKa vaules, each step can be treated
  individually.
• Example Consider the titration of 50.00 mL of
  0.050 M H2CO3 (Ka1 4.5e-7, Ka2 5.6e-11) with
  0.100 M NaOH.

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a) 0.00 mL

• The solution consists only of 0.050 M H2CO3.
• H2CO3 H HCO3-
• 0.050 M 0 0
• -x x x
• 4.5e-7 x2 / 0.050 x
• x 1.5e-4
• pH 3.82

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b) 10.00 mL 0.100 M NaOH solution

• Calculate the mol OH- added and initial moles of
  H2CO3.
• Initial mol H2CO3 50.00 mL 0.050 M 2.5 mmol
• Added mol OH- 10.00 mL 0.100 M 1.00 mmol
• Excess mol H2CO3 2.5 1.00 mmol 1.5 mmol
• H2CO3 1.5 mmol / 60.00 mL 2.5e-2 M
• Must realize that this is a buffer system (why?).

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Realizing that this is a buffer system we must
calculate HCO3-

• From an MBE consideration
• initial mol H2CO3 mol H2CO3 left mol HCO3-
  formed
• 2.5 mmol mol H2CO3 mol HCO3- 1.5 x
• x 1.0 mmol
• HCO3- 1.0 mmol / 60.00 mL 1.67e-2
• Ka1 4.5e-7 HHCO3- / H2CO3
• 4.5e-7 H1.67e-2 / 2.5e-2 note that volumes
  of solution cancel
• H 6.75e-7
• pH 6.17

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c) 25.00 ml added

• Initial mol H2CO3 50.00 mL 0.050 M 2.5 mmol
• Added mol OH- 25.00 mL 0.100 M 2.5 mmol
• This is the first equivalence point.
• HCO3- 2.5 mmol / 75.00 mL 0.033 M
• HCO3- CO32- H H HCO3- H2CO3

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d) 28.00 mL

• We should recognize that that we will have HCO3-
  and CO32-, which constitutes a buffer system.
• Initial mols HCO3- 2.5 mmol
• Mols OH- beyond 1st eq. pt.
• (28.00 25.00) mL 0.100 M 0.30 mmol
• mols HCO3- 2.5 0.3 mmol 2.2 mmol
• HCO3- 2.2 mmol / 78.00 mL 2.82e-2 M

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• MBE 2.5 mmol HCO3- CO32- 2.2 x
• x 0.30 mmol
• CO32- 0.3 / 78.00 mL 3.85e-3
• Ka2 5.6e-11 HCO32- / HCO3-
• H3.85e-3 / 2.82e-2
• H 4.10e-10
• pH 9.39

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e) 50.00 mL 0.100 M NaOH added

• Initial mol H2CO3 50.00 mL 0.050 M 2.5 mmol
• Added mol OH- 50.00 mL 0.100 M 5.0 mmol
• This is the second eq. pt.
• Sole species CO32- 2.5 mmol / 100.00 mL
  0.025 M
• CO32- H2O HCO3- OH-
• 0.025 M -- 0 0
• -x x x
• Kb 1.00e-14 / 5.6e-11 x2 / 0.025 x
• x 2.11e-3 pH 11.32

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f) 53.00 mL of 0.100 M NaOH

• We are 3.00 mL beyond the 2nd eq. pt.
• Excess OH- 3.00 mL 0.100 M 0.300 mmol
• OH- 0.300 mmol / 103.00 mL 2.91e-3 M
• pH 11.464

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Summary

• Vol titrant pH
• 0.00mL 3.82 excess H2CO3
• 10.00 6.17 1st buffer region
• 25.00 8.30
• 28.00 9.39 2nd buffer region
• 50.00 11.32
• 53.00 11.464 excess OH-
• Draw entire curve.

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• Read about the case of a weak base with a strong
  base in section 12-4 p232 of your text.
• Read about how to locate the equivalence point
  with a pH electrode in section 12-5 of your text.

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Location of end points by 1st and 2nd derivatives
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