AcidBase Titrations

1
Acid-Base Titrations

• Neutralization Reactions H OH ? H2O
• Equivalence Point Condition attained when
• no. mols H no. mols OH
• Recall that (Volume, L)x(Molarity, mol/L)
  mols
• Therefore, Macid Vacid Mbase Vbase
• at the equivalence point.

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• Titration of a Strong Acid with a Strong Base
• Ex. Titration of 50.0 mL of 0.200 M HNO3 with
  0.100 M NaOH
• (The acid is in the Erlenmeyer flask, and the
  NaOH is added from the burette.)
• a) The initial pH of the acid solution is equal
  to log(0.200) 0.70
• After 20.0 mL of NaOH have been added
• (20.0 mL)(0.100 M) 2.00 mmol NaOH
• (50.0 mL)(0.200 M) 10.00 mmol HNO3

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• Do the stoichiometry problem
• HNO3 NaOH ? NaNO3
  H2O
• Init. 10.00 mmol 2.00 mmol
  0
• Change 2.00 2.00
  2.00
• Final 8.00 0
  2.00
• Recall that salts formed from strong acids and
  strong bases (e.g., NaNO3) are neutral.
• We have (8.00 mmol / 70.0 mL) 0.114 M HNO3
• And pH log(0.114) 0.94

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• At the equivalence point
• (50.0 mL acid)(0.200 M) (Vbase)(0.100 M)
• and Vbase 100.0 mL
• Important! The volumes of acid and base are not
  necessarily equal. The numbers of moles are
  equal.
• HNO3 NaOH ? NaNO3
  H2O
• I 10.0 10.0
  0
• C 10.0 10.0
  10.0
• F 0 0
  10.0 mmol
• Since NaNO3 is a neutral salt, the pH 7.00

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• After 110.0 mL of NaOH has been added
• (110.0 mL)(0.100 M) 11.00 mmol NaOH
• Since 10.00 mmol of NaOH were used to neutralize
  the acid, 1.00 mmol remains, along with the 10.00
  mmol of NaNO3.
• (1.00 mmol / 160.0 mL) 0.00625 M NaOH
• The strong base is completely dissociated.
• pOH log(0.00625) 2.20
• pH 14.00 2.20 11.80

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Figure 15.1The pH Curve for the Titration of
50.0 mL of 0.200 M HNO3 with 0.100 M NaOH
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• Titration of a Weak Acid with a Strong Base
• Ex. Titration of 50.0 mL of 0.100 M HC2H3O2 with
  0.100 M NaOH
• The initial acetic acid solution has a pH of
  2.87. (Verify this for yourself.)
• After the addition of 10.0 mL of NaOH
• (10.0 mL)(0.100 M) 1.00 mmol NaOH
• (50.0 mL)(0.100 M) 5.00 mmol HC2H3O2
• Now do the stoichiometry problem.

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• HC2H3O2 NaOH ? NaC2H3O2
  H2O
• I 5.00 1.00
  0
• C 1.00 1.00
  1.00
• F 4.00 mmol 0
  1.00 mmol
• This is a buffer solution.
• (4.00 mmol acid / 60.0 mL) 0.0667 M
• (1.00 mmol salt / 60.0 mL) 0.0167 M
• pH 4.74 log(0.0167 / 0.0667) 4.14

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• At the equivalence point
• Vbase (Vacid)(Macid) / Mbase
• 50.0 mL of NaOH needed
• (Note that the acid and base have the same
  molarity.)
• The 5.00 mmol of acid reacts with the 5.00 mmol
  of base to form 5.00 mmol of NaC2H3O2.
• The product is the salt of a weak acid, which
  behaves as a weak base in solution.
• Kb Kw / Ka 5.6 x 1010
• (5.00 mmol salt / 100.0 mL) 0.0500 M NaC2H3O2

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• NaC2H3O2 ? Na C2H3O2
• 0 0.0500
  0.0500
• C2H3O2 H2O ? HC2H3O2 OH
• I 0.0500
  0 0
• C x
  x x
• E 0.0500x
  x x
• X OH 5.3 x 106 M
• pOH 5.28 pH 8.72

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• After 60.0 mL of NaOH has been added
• (60.0 mL)(0.100 M) 6.00 mmol NaOH
• Since 5.00 mmol of NaOH were used up in
  neutralizing the acid, 1.00 mmol NaOH is left in
  addition to the 5.00 mmol NaC2H3O2 formed.
• We have a strong base and a weak base (the salt).
  
• Only the strong base need be considered.

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• (1.00 mmol NaOH) / (100.0 mL total) 0.0100 M
• Since NaOH is a strong base, OH 0.0100 M
• pOH log(0.0100) 2.00
• pH 14.00 2.00 12.00

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Figure 15.3 The pH Curve for the Titration of
50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH
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• Now consider the titration of 50.0 mL of 0.100 M
  HCN with 0.100 M NaOH
• The Ka of HCN is 6.2 x 1010
• The pH of the initial 0.100 M HCN solution is
  5.10.
• As before, 50.0 mL of NaOH is required to reach
  the equivalence point (5.00 mmol each of acid and
  base).
• At the equivalence point, we have 5.00 mmol of
  NaCN in a total volume of 100.0 mL 0.0500 M.

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• The Kb of CN 1.0 x 1014 / 6.2 x 1010
• 1.6 x 105
• CN H2O ? HCN OH
• X OH 8.9 x 104 M
• pOH 3.05
• pH 10.95

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• Conclusion As the acid gets weaker (smaller
  Ka), the pH at the equivalence point gets higher.
• We will see that this has an implication for the
  choice of an indicator.
• Another outcome is that the length of the
  vertical portion of the pH curve decreases as the
  acid gets weaker. This makes the indicator
  endpoint less and less sharp as the acid gets
  weaker.

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Figure 15.4The pH Curves for the Titrations of
50.0-mL Samples of 0.10 M Acids with Various Ka
Values with 0.10 M NaOH
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• Titration of a Weak Base with a Strong Acid
• ex. Titrate 100.0 mL of 0.050 M NH3 with 0.10 M
  HCl
• The reaction is
• NH3 HCl ? NH4Cl
• Since NH4Cl is the salt of a weak base, it is
  completely dissociated
• NH4Cl ? NH4 Cl
• And the ammonium ion acts as a weak acid
• NH4 ? H NH3

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Figure 15.5The pH Curve for the Titration of
100.0 mL of 0.050 M NH3 with 0.10 M HCI
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Acid-Base Indicators

• These are substances which have different colors
  depending upon the pH.
• They are weak acids which are symbolized HIn.
• HIn ? H In
• Where HIn and In have different colors.
• Indicators are chosen such that the color change
  occurs in the pH region of the equivalence point
  of a titration.

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Figure 15.6The Acid and Base Forms of the
Indicator Phenolphthalein
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• The ionization constant Kind is defined as
• Using logarithms we can derive
• The color change is apparent when the ratio of
  In / HIn is in the range 0.1 10. Thus,
  the color change is observed to take place over a
  range of 2 pH units.

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• It is important to match the pH of the
  equivalence point with the pH at which the
  indicator changes color.
• Equivalence point mols H mols OH
• Endpoint color of indicator changes
• The two do not necessarily coincide unless a
  correct choice of indicator has been made!

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Figure 15.8The Useful pH Ranges for Several
Common Indicators
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Figure 15.9The pH Curve for the Titration of
100.0 mL of 0.10 M HCI with 0.10 M NaOH
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Figure 15.10The pH Curve for the Titration of 50
mL of 0.1 M HC2H3O2 with 0.1 M NaOH