5.111 Lecture Summary #23 Monday, November 3, 2014
Reading for Today: Sections 12.4-12.6 in 5th ed. (4th ed: 11.4-11.6)
Reading Lecture #24: Section K
Topics: Acid-Base Titrations
I. Titration of Strong Acids and Strong Bases
II. Titration of Weak Acids/Strong Bases & Strong Acids/Weak Bases
ACID BASE TITRATIONS
An acid-base titration is the addition of a volume of base of known concentration to acid of
unknown concentration (or addition of acid to base).
This technique can be used to determine the of an acid or base.
I. TITRATION OF STRONG ACIDS AND STRONG BASES
A. Shapes of Curves and Some Definitions
In a titration of a strong acid with a strong base, or a strong base with a strong acid, the pH
changes slowly initially, changes rapidly through pH 7 at the equivalence point and then
changes slowly again.
Titration curves:
Strong Acid titrated with Strong Base Strong Based titrated with Strong Acid
Figure by MIT OpenCourseWare.
Equivalence (stoichiometric, S) point = theoretical volume at which moles of base (or acid)
added equals moles of acid (or base) that was originally present.
End point = experimentally measured volume at which the indicator changes color.
Endpoint should equivalence point.
pH indicators are weak acids or weak bases that have different colors based on different
chemical structures in acidic or basic environments. Anthocyanins are examples of natural
acid-base indicators.
1
, B. Calculating Points on a pH Curve for a Strong/Strong Acid-Base Titration
Example: a strong base (0.250 M NaOH) is titrated with a strong acid (0.340 M HCl)
1. Calculating the pH before the equivalence point when 5.00 mL of 0.340 M HCl (aq) is added to
25.00 mL of 0.250 M NaOH (aq)
a). Calculate moles of OH- present.
(Base is strong so moles of NaOH added moles of OH- formed.)
0.02500 L x 0.250 mol/L = 6.25 x 10-3 moles of OH- present
b). Calculate moles of H3O+ supplied by titrant.
(Acid is strong so moles HCl added moles of H3O+ formed.)
0.00500 L x 0.340 mol/L = 1.70 x 10-3 moles
c). Find the moles of OH- remaining after the reaction with H3O+ ions.
because stoichiometry is 1:1
6.25 x 10-3 moles - 1.70 x 10-3 moles = 4.55 x 10-3 mol of left
(Since all the moles of strong acid are gone, from this point on, it is a strong base in water
problem, and pH can be calculated from pOH, and pOH can be calculated from [OH-])
d). Calculate molarity of OH-
4.55 x 10-3 mol/ 0.03000 L = 0.152 mol/L (recall volume is now 5.00 mL + 25.00 mL)
e). Calculate pH
pOH = -log 0.152 = 0.818 pH = 14.00 -0.818 =
2. Calculating the volume of HCl needed to reach the equivalence point.
Initially 6.25 x 10-3 mol of OH- were present. At the equivalence point, m ol
of HCl will have been added (1:1 stoichiometry)
6.25 x 10-3 mol x 1L = 0.0184 L
0.340 mol
3. What is the pH at the equivalence point?
4. Calculate the pH after 1.00 mL of HCl has been added after equivalence point has been
reached. Note: this is a strong acid in water problem.
a). Find moles of H3O+ formed due to the 1.00 mL addition of HCl
(H3O+ formed = amount of HCl added, since strong acid)
0.340 mol/L x (0.00100 L) = 3.40 x 10-4 mol of H3O+
b). Calculate molarity of H3O+
c). Calculate pH. pH = -log (7.66 x 10-3) = 2.116
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Reading for Today: Sections 12.4-12.6 in 5th ed. (4th ed: 11.4-11.6)
Reading Lecture #24: Section K
Topics: Acid-Base Titrations
I. Titration of Strong Acids and Strong Bases
II. Titration of Weak Acids/Strong Bases & Strong Acids/Weak Bases
ACID BASE TITRATIONS
An acid-base titration is the addition of a volume of base of known concentration to acid of
unknown concentration (or addition of acid to base).
This technique can be used to determine the of an acid or base.
I. TITRATION OF STRONG ACIDS AND STRONG BASES
A. Shapes of Curves and Some Definitions
In a titration of a strong acid with a strong base, or a strong base with a strong acid, the pH
changes slowly initially, changes rapidly through pH 7 at the equivalence point and then
changes slowly again.
Titration curves:
Strong Acid titrated with Strong Base Strong Based titrated with Strong Acid
Figure by MIT OpenCourseWare.
Equivalence (stoichiometric, S) point = theoretical volume at which moles of base (or acid)
added equals moles of acid (or base) that was originally present.
End point = experimentally measured volume at which the indicator changes color.
Endpoint should equivalence point.
pH indicators are weak acids or weak bases that have different colors based on different
chemical structures in acidic or basic environments. Anthocyanins are examples of natural
acid-base indicators.
1
, B. Calculating Points on a pH Curve for a Strong/Strong Acid-Base Titration
Example: a strong base (0.250 M NaOH) is titrated with a strong acid (0.340 M HCl)
1. Calculating the pH before the equivalence point when 5.00 mL of 0.340 M HCl (aq) is added to
25.00 mL of 0.250 M NaOH (aq)
a). Calculate moles of OH- present.
(Base is strong so moles of NaOH added moles of OH- formed.)
0.02500 L x 0.250 mol/L = 6.25 x 10-3 moles of OH- present
b). Calculate moles of H3O+ supplied by titrant.
(Acid is strong so moles HCl added moles of H3O+ formed.)
0.00500 L x 0.340 mol/L = 1.70 x 10-3 moles
c). Find the moles of OH- remaining after the reaction with H3O+ ions.
because stoichiometry is 1:1
6.25 x 10-3 moles - 1.70 x 10-3 moles = 4.55 x 10-3 mol of left
(Since all the moles of strong acid are gone, from this point on, it is a strong base in water
problem, and pH can be calculated from pOH, and pOH can be calculated from [OH-])
d). Calculate molarity of OH-
4.55 x 10-3 mol/ 0.03000 L = 0.152 mol/L (recall volume is now 5.00 mL + 25.00 mL)
e). Calculate pH
pOH = -log 0.152 = 0.818 pH = 14.00 -0.818 =
2. Calculating the volume of HCl needed to reach the equivalence point.
Initially 6.25 x 10-3 mol of OH- were present. At the equivalence point, m ol
of HCl will have been added (1:1 stoichiometry)
6.25 x 10-3 mol x 1L = 0.0184 L
0.340 mol
3. What is the pH at the equivalence point?
4. Calculate the pH after 1.00 mL of HCl has been added after equivalence point has been
reached. Note: this is a strong acid in water problem.
a). Find moles of H3O+ formed due to the 1.00 mL addition of HCl
(H3O+ formed = amount of HCl added, since strong acid)
0.340 mol/L x (0.00100 L) = 3.40 x 10-4 mol of H3O+
b). Calculate molarity of H3O+
c). Calculate pH. pH = -log (7.66 x 10-3) = 2.116
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