Complex Acid Base Systems

1

• Complex Acid - Base Systems
• Mixture of a strong acid and a weak acid
• Example Titrate 50.00 mL of a solution that is
  0.1000 M in both HCl and HAc with 0.1000 M NaOH
• Initial pH no NaOH added
• HAc H2O Ac- H3O
• 2H2O H3O OH-
• Charge Balance H3O Cl- Ac- OH-
• ? ? ?
• H from H from H from
• HCl HAc H2O
• Mass Balance CHAc Hac Ac-
• Assumptions Since the solution is acidic, OH-
  ltlt Cl- Ac-
• The high H supresses Ac- from HAc
• H3O Cl- 0.1000 pH 1.00

2

• Complex Acid - Base Systems
• Mixture of a strong acid and a weak acid
• Example Titrate 50.00 mL of a solution that is
  0.1000 M in both HCl and
• HAc with 0.1000 M NaOH
• Check assumptions
• from Ka
• Add NaOH continue with similar calculations
  until most of HCl consumed
• In the initial stages of the titration the HCl is
  being consumed

3

• Complex Acid - Base Systems
• Mixture of a strong acid and a weak acid
• Example Titrate 50.00 mL of a solution that is
  0.1000 M in both HCl and
• HAc with 0.1000 M NaOH
• until most of the HCl is consumed and
  dissociation of the HAc is contributing
  significantly to the H 49.00 mL added NaOH
• HAc H2O Ac- H3O
• 2H2O H3O OH-
• Charge Balance H3O Cl- - Na Ac-
  OH-
• ?

4

• Complex Acid - Base Systems
• Mixture of a strong acid and a weak acid
• Example Titrate 50.00 mL of a solution that is
  0.1000 M in both HCl and
• HAc with 0.1000 M NaOH
• until most of the HCl is consumed and
  dissociation of the HAc is contributing
  significantly to the H 49.00 mL added NaOH
• Mass Balance CHAc Hac Ac-

5

• Complex Acid - Base Systems
• Mixture of a strong acid and a weak acid
• Example Titrate 50.00 mL of a solution that is
  0.1000 M in both HCl and
• HAc with 0.1000 M NaOH
• When all the HCl is consumed, Cl- Na and
  CHCl 0.0000
• Have a solution of 0.05000 M HAc
• The pH is the pH of a 0.05000 M HAc solution
• The rest of the titration curve is that for the
  titration of a 0.05000 M HAc solution
• Polyprotic acids and bases
• H3PO4, H2CO3. H2S, CO32-
• The calculation of the titration curve can be
  carried out rigorously but is not usually
  necessary
• If the ratio of successive Kas or Kbs is ?
  10-3, a general simplifying assumption can be
  made that the conjugate acid - base pair
  equilibrium can be considered as occuring
  between monoprotic species
• The only complication is the calculation of the
  pH of the solution of an amphiprotic substance
  at the equivalence point where they occur
• In the titration of H2CO3 with base or CO32- with
  acid, the first equivalence point occurs when
  all the H2CO3 or CO32- has been converted to HCO3-

6

• Complex Acid - Base Systems
• pH of a solution of an amphiprotic substance
• Equilibria
• H2O HCO3- CO32- H3O
• H2O HCO3- H2CO3 OH-
• 2H2O H3O OH-
• Mass balance CNaHCO3 HCO3- CO32-
  H2CO3
• Charge balance Na H3O 2CO32-
  HCO3- OH-
• Rearranging CB eqn. and noting Na CNaHCO3
• HCO3- CO32- H2CO3 2CO32- HCO3-
  OH-- H3O
• H2CO3 CO32-
  OH- - H3O

7

• Complex Acid - Base Systems
• pH of a solution of an amphiprotic substance
• From equilibrium constant expressions, find
  CO32- and H2CO3 in terms of HCO3-

For 0.10 M NaHCO3
8

• Complex Acid - Base Systems
• pH of a solution of a polyprotic acid
• If Ka1gtgtKa2gtgtKa3 and solution not too dilute, can
  assume only the 1st dissociation provides H
• Subsequent dissociations are suppressed by H
  from 1st dissociation
• End up with the usual equation for a monoprotic
  acid
• Example calculate the pH of a saturated solution
  of CO2 in water
• According to the CRC, a saturated solution of CO2
  contains 1.45 g/L ? 0.033 M
• Equilibria
• CO2(aq) H2O H2CO3(aq)
• H2CO3 H2O H3O HCO3-
• HCO3- H2O H3O CO32-

9

• Complex Acid - Base Systems
• pH of a solution of a polyprotic acid or base
• See example for pH of a solution of 0.040 M H2SO4
  , Feature 11-1, FAC7 p. 239