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Monoprotic Acid Base Systems

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... H2O H OH- Kw = [H ][OH-] = 1.00 x 10 ... to 50.00 mL, [H ] may not be [OH ... [H3O ] = [Ac-] [OH-] One way to solve the problem: another way to solve ... – PowerPoint PPT presentation

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Title: Monoprotic Acid Base Systems


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
  • 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

7
  • 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

8
  • 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

9
  • 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

10
  • 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

11
  • 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
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