Titration of strong acid with strong base

1
Monoprotic Acid - Base Systems

• Titration of strong acid with strong base
• Example titrate 50.00 mL 0.1000 M HCl with
  0.1000 M NaOH
• At what volume of titrant has the equivalence
  point been reached?
• HCl NaOH NaCl H2O 1 mol NaOH ? 1
  mol HCl
• equivalence point is 50.00 mL
  added NaOH
• Equilibrium H2O H OH- Kw
  HOH- 1.00 x 10-14
• Charge balance equation H Na OH-
  Cl-
• At 0.00 mL added NaOH
• Na 0.0000 M and OH- ltlt H
• H OH- Cl- .1000 pH 1.00
• At 10.00 mL added NaOH
• From the charge balance equation H OH-
  Cl- - Na
• H Cl- - Na CHCl

2
Monoprotic Acid - Base Systems

• Titration of strong acid with strong base
• Example titrate 50.00 mL 0.1000 M HCl with
  0.1000 M NaOH
• At 49.90 mL added NaOH
• If one is closer to 50.00 mL, H may not be ltlt
  OH-
• Charge balance equation H Na OH-
  Cl-
• At 50.00 mL added NaOH equivalence point
• Na Cl-
• from the charge balance equation H Na
  OH- Cl-
• H OH-
• from Kw

3
Monoprotic Acid - Base Systems

• Titration of strong acid with strong base
• Example titrate 50.00 mL 0.1000 M HCl with
  0.1000 M NaOH
• At 50.10 mL added NaOH
• H Na OH- Cl-
• OH- Na - Cl- CNaOH
• Examine the change in pH for the 0.20 mL between
  49.90 mL and 50.10 mL added NaOH

4
Monoprotic Acid - Base Systems

• Titration of strong acid with strong base
• Compare this result to similar calculations for
  0.00100 M reagents
• At 49.90 mL added 0.00100 M NaOH to 50.00 mL
  0.00100 M HCl
• From the charge balance equation H OH-
  Cl- - Na
• At 50.10 mL add 0.00100 M NaOH
• H Na OH- Cl-
• OH- Na - Cl-

5
Monoprotic Acid - Base Systems

• Titration of strong acid with strong base
• Titration curves - a plot of a concentration
  related variable vs. volume of titrant
• Two types of titration curves - see Fig. 10-2,
  FAC7 p. 192
• Sigmoidal curves are produced from power function
  calculations or functions in which the power
  function of a reactant is related to another
  variable such as cell potential.
• Sigmoidal curves show how rapidly the power
  function changes with volume of titrant and are
  useful in evaluating indicator selection
• Linear segment curves are easy to determine from
  instrumental results in which a measurement is
  directly proportional to the concentration of a
  titration reactant and the titration reaction is
  complete far from the equivalence point.
• Significant figures in power functions use two
  significant figures
• Often the difference in mmoles of reactant is
  calculated to two sig. figs. Near the
  equivalence point in a titration
• See 50.00 mL 0.1000 M HCl 49.90 mL 0.1000 NaOH
• Uncertainties in the measurement of power
  functions due to instrumental limits

6
Monoprotic Acid - Base Systems
7
Monoprotic Acid - Base Systems

• Titration curves for HCl with NaOH
• A 50.00 mL 0.0500 M HCl with 0.100 M NaOH
• B 50.00 mL of 0.000500 M HCl with 0.00100 M NaOH

8
Monoprotic Acid - Base Systems

• Titration of weak acid
• Example titrate 50.00 mL 0.1000 M HAc with
  0.1000 M NaOH
• Initial point no added NaOH
• Equilibria
• HAc H2O Ac- H3O
• 2H2O H3O OH- Kw
  H3OOH-
• Mass balance equation
• CHAc HAc Ac-
• Charge balance equation
• H3O Ac- OH-
• One way to solve the problem
  another way to solve the problem

9
Monoprotic Acid - Base Systems

• Titration of weak acid
• Example titrate 50.00 mL 0.1000 M HAc with
  0.1000 M NaOH
• 10.00 mL NaOH
• Converted 1.000 mmol HAc to NaAc leaving 4.000
  mmol HAc
• Mass balance equation CHAc CNaAc HAc
  Ac-
• Charge balance equation Na H3O Ac-
  OH-
• CNaAc
  H3O Ac- OH-
• rearranging Ac- CNaAc
  H3O - OH-
• Combining the MB and CB equations
• CHAc - H3O Ac- OH- HAc Ac-
• HAc CHAc - H3O OH-
• Generally H3O - OH- ltlt CNaAc so Ac-
  CNaAc
• - H3O OH- ltlt CHAc so HAc
  CHAc

10
Monoprotic Acid - Base Systems

• Titration of weak acid
• Example titrate 50.00 mL 0.1000 M HAc with
  0.1000 M NaOH
• Equivalence point 50.00 mL added NaOH
• All the HAc has been converted to NaAc
• Equilibria Ac- H2O HAc OH-
• 2H2O
  H3O OH-
• Mass balance equation CNaAc Ac-
  HAc
• Charge balance equation Na H3O OH-
  Ac-
• rearranging Ac- CNaAC
  H3O - OH-
• substituting into MB equation CNaAc CNaAC
  H3O - OH- HAc
• HAc OH- - H3O
• Substituting into Kb equation

11
Monoprotic Acid - Base Systems

• Titration of weak acid
• Example titrate 50.00 mL 0.1000 M HAc with
  0.1000 M NaOH
• Beyond the equivalence point
• The OH- from the NaOH suppresses the dissociation
  of Ac-
• This part of the titration curve looks just like
  that for a strong acid titration

12
Monoprotic Acid - Base Systems

• Titration of weak acid
• The effect of concentration on some of the
  calculations
• Titrate 50.00 mL of 0.001000 M HAc with 0.001000
  M NaOH
• 10.00 mL added NaOH
• Converted 0.01000 mmol HAc to NaAc leaving
  0.04000 mmol HAc
• Mass balance equation CHAc CNaAc HAc
  Ac-
• Charge balance equation Na H3O Ac-
  OH-
• CNaAc
  H3O Ac- OH-
• rearranging Ac- CNaAc
  H3O - OH-
• Combining the MB and CB equations
• CHAc - H3O Ac- OH- HAc Ac-
• HAc CHAc - H3O OH-
• since H ? 10-4, cant ignore H3O relative
  to CHAc or CNaAc
• but the solution is acidic, so OH- ltlt CHAc -
  H3O

13
Monoprotic Acid - Base Systems

• Titration of weak acid
• The effect of concentration on some of the
  calculations
• Titrate 50.00 mL of 0.001000 M HAc with 0.001000
  M NaOH
• HAc CHAc - H3O OH-
• Ac- CNaAc H3O - OH-
• Substituting into the Ka equation

14
Monoprotic Acid - Base Systems

• Titration curves for acetic acid with NaOH
• A 0.1000 M HAc with 0.1000 M NaOH
• B 0.001000 M HAc with 0.00100 M NaOH