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19Ionic Equilibria: Part II
Buffers and Titration Curves
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Chapter Goals
1. The Common Ion Effect and Buffer Solutions2. Buffering Action3. Preparation of Buffer Solutions4. Acid-Base Indicators
Titration Curves5. Strong Acid/Strong Base Titration Curves6. Weak Acid/Strong Base Titration Curves7. Weak Acid/Weak Base Titration Curves8. Summary of Acid-Base Calculations
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The Common Ion Effect and Buffer Solutions• If a solution is made in which the same ion is produced by
two different compounds the common ion effect is exhibited.• Buffer solutions are solutions that resist changes in pH when
acids or bases are added to them.– Buffering is due to the common ion effect.
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The Common Ion Effect and Buffer Solutions• There are two common kinds of buffer solutions:1 Solutions made from a weak acid plus a soluble ionic
salt of the weak acid.2 Solutions made from a weak base plus a soluble
ionic salt of the weak base
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The Common Ion Effect and Buffer Solutions1. Solutions made of weak acids plus a soluble
ionic salt of the weak acid• One example of this type of buffer system is:
– The weak acid - acetic acid CH3COOH– The soluble ionic salt - sodium acetate NaCH3COO
acids. with reacts base) (aanion salt The
COOCH Na COOCHNa
H COOCH COOHCH
bases. with reacts acid weak The
-3
%1003
-33
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The Common Ion Effect and Buffer Solutions• Example 19-1: Calculate the concentration of H+and the pH of a
solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate.– This is another equilibrium problem with a starting concentration for both
the acid and anion.
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The Common Ion Effect and Buffer Solutions• Substitute the quantities determined in the
previous relationship into the ionization expression.
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The Common Ion Effect and Buffer Solutions• Apply the simplifying assumption to both the
numerator and denominator.
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The Common Ion Effect and Buffer Solutions
• This is a comparison of the acidity of a pure acetic acid solution and the buffer described in Example 19-1.
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The Common Ion Effect and Buffer Solutions
• Compare the acidity of a pure acetic acid solution and the buffer described in Example 19-1.
Solution [H+] pH
0.15 M CH3COOH 1.6 x 10-3 2.80
0.15 M CH3COOH &
0.15 M NaCH3COO buffer
1.8 x 10-5 4.74
[H+] is 89 times greater in pure acetic acid than in buffer solution.
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The Common Ion Effect and Buffer Solutions• The general expression for the ionization of a
weak monoprotic acid is:
• The generalized ionization constant expression for a weak acid is:
HA H+ + A-
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The Common Ion Effect and Buffer Solutions• If we solve the expression for [H+], this relationship
results:
• By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to:
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The Common Ion Effect and Buffer Solutions• The relationship developed in the previous slide is
valid for buffers containing a weak monoprotic acid and a soluble, ionic salt.
• If the salt’s cation is not univalent the relationship changes to:
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The Common Ion Effect and Buffer Solutions
• Simple rearrangement of this equation and application of algebra yields the
Henderson-Hasselbach equation.
The Henderson-Hasselbach equation is one method to calculate the pHof a buffer given the concentrations of the salt and acid.
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Weak Bases plus Salts of Weak Bases2. Buffers that contain a weak base plus the salt
of a weak base• One example of this buffer system is ammonia
plus ammonium nitrate.
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Weak Bases plus Salts of Weak Bases• Example 19-2: Calculate the concentration of
OH- and the pH of the solution that is 0.15 M in aqueous ammonia, NH3, and 0.30 M in ammonium nitrate, NH4NO3.
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Weak Bases plus Salts of Weak Bases
• Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.
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Weak Bases plus Salts of Weak Bases• A comparison of the aqueous ammonia
concentration to that of the buffer described above shows the buffering effect.
Solution [OH-] pH
0.15 M NH3 1.6 x 10-3 M 11.20
0.15 M NH3 &
0.15 M NH4NO3 buffer
9.0 x 10-6 M 8.95
The [OH-] in aqueous ammonia is 180 times greater than in the buffer.
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Weak Bases plus Salts of Weak Bases• We can derive a general relationship for buffer
solutions that contain a weak base plus a salt of a weak base similar to the acid buffer relationship.– The general ionization equation for weak bases is:
: B + H2O BH+ + OH-
where B represents a weak base
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Weak Bases plus Salts of Weak Bases• The general form of the ionization expression is:
• Solve for the [OH-]
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Weak Bases plus Salts of Weak Bases• For salts that have univalent ions:
• For salts that have divalent or trivalent ions:
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Weak Bases plus Salts of Weak Bases
• Simple rearrangement of this equation and application of algebra yields the
Henderson-Hasselbach equation.
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Buffering Action
• These movies show that buffer solutions resist changes in pH.
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Buffering Action
• Example 19-3: If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl.
1 Calculate the pH of the original buffer solution.
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Buffering Action
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Buffering Action
2 Next, calculate the concentration of all species after the addition of the gaseous HCl.– The HCl will react with some of the ammonia and
change the concentrations of the species.– This is another limiting reactant problem.
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Buffering Action
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Buffering Action
3 Using the concentrations of the salt and base and the Henderson-Hassselbach equation, the pH can be calculated.
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Buffering Action
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Buffering Action
4 Finally, calculate the change in pH.
-0.14=8.95-8.81=pH
pHpH pH originalnew
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Buffering Action
• Example 19-4: If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH.
You do it!You do it!
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Buffering Action
• pH of the original buffer solution is 8.95, from above.
1. First, calculate the concentration of all species after the addition of NaoH.
– NaOH will react with some of the ammonium chloride.– The limiting reactant is the NaOH.
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Buffering Action
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Buffering Action
2. Calculate the pH using the concentrations of the salt and base and the Henderson-Hasselbach equation.
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Buffering Action
3. Calculate the change in pH.
0.13=8.95-9.08=pH
pHpH =pH originalnew
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Buffering Action
• This table is a summary of examples 19-3 and 19-4.
• Notice that the pH changes only slightly in each case.
Original SolutionOriginal
pH
Acid or base
added
New pH
pH
1.00 L of solution containing
0.100 M NH3 and 0.200 M NH4Cl
8.95
0.020 mol NaOH
9.08 +0.13
0.020 mol HCl
8.81 -0.14
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Preparation of Buffer Solutions
• This movie shows how to prepare a buffer.
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Preparation of Buffer Solutions
• Example 19-5: Calculate the concentration of H+ and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions.
• Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction.
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Preparation of Buffer Solutions
• Sodium hydroxide and acetic acid react in a 1:1 mole ratio.
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Preparation of Buffer Solutions
• After the two solutions are mixed, the total volume of the solution is 300 mL (100 mL of NaOH + 200 mL of acetic acid).– The concentrations of the acid and base are:
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Preparation of Buffer Solutions
• Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.
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Preparation of Buffer Solutions
• For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH.
• Example 19-6:Calculate the number of moles of solid ammonium chloride, NH4Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15.– Because pH = 9.15, the pOH can be determined.
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Preparation of Buffer Solutions
• The appropriate equilibria representations are:
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Preparation of Buffer Solutions
• Substitute into the ionization constant expression (or Henderson-Hasselbach equation) for aqueous ammonia
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Preparation of Buffer Solutions
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Acid-Base Indicators
• The point in a titration at which chemically equivalent amounts of acid and base have reacted is called the equivalence point.
• The point in a titration at which a chemical indicator changes color is called the end point.
• A symbolic representation of the indicator’s color change at the end point is:
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Acid-Base Indicators
• The equilibrium constant expression for an indicator would be expressed as:
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Acid-Base Indicators
• If the preceding expression is rearranged the range over which the indicator changes color can be discerned.
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Acid-Base Indicators
Color change ranges of some acid-base indicators
Indicator
Color in acidic range pH range
Color in basic range
Methyl violet Yellow 0 - 2 Purple
Methyl orange Pink 3.1 – 4.4 Yellow
Litmus Red 4.7 – 8.2 Blue
Phenolphthalein Colorless 8.3 – 10.0 Red
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Titration CurvesStrong Acid/Strong Base Titration Curves
• These graphs are a plot of pH vs. volume of acid or base added in a titration.
• As an example, consider the titration of 100.0 mL of 0.100 M perchloric acid with 0.100 M potassium hydroxide.– In this case, we plot pH of the mixture vs. mL of KOH added.– Note that the reaction is a 1:1 mole ratio.
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Strong Acid/Strong Base Titration Curves• Before any KOH is added the pH of the HClO4
solution is 1.00.– Remember perchloric acid is a strong acid that
ionizes essentially 100%.
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Strong Acid/Strong Base Titration Curves• After a total of 20.0 mL 0.100 M KOH has been added
the pH of the reaction mixture is ___?
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Strong Acid/Strong Base Titration Curves• After a total of 50.0 mL of 0.100 M KOH has
been added the pH of the reaction mixture is ___?
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Strong Acid/Strong Base Titration Curves• After a total of 90.0 mL of 0.100 M KOH has been
added the pH of the reaction mixture is ____?
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Strong Acid/Strong Base Titration Curves• After a total of 100.0 mL of 0.100 M KOH has been
added the pH of the reaction mixture is ___?
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Strong Acid/Strong Base Titration Curves• We have calculated only a few points on the
titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.
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Weak Acid/Strong Base Titration Curves• As an example, consider the titration of 100.0 mL of
0.100 M acetic acid, CH3 COOH, (a weak acid) with 0.100 M KOH (a strong base).– The acid and base react in a 1:1 mole ratio.
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Weak Acid/Strong Base Titration Curves
• Before the equivalence point is reached, both CH3COOH and KCH3COO are present in solution forming a buffer.– The KOH reacts with CH3COOH to form KCH3COO.
• A weak acid plus the salt of a weak acid form a buffer.• Hypothesize how the buffer production will effect the titration curve.
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Weak Acid/Strong Base Titration Curves1. Determine the pH of the acetic acid solution
before the titration is begun.• Same technique as used in Chapter 18.
5
a
5
3
-3
+
a
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108.110.0
K
108.1COOHCH
COOCH HK
10.0
HCOOCHCOOHCH
x
xx
xMxMMx
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Weak Acid/Strong Base Titration Curves
89.2pH 101.3H
101.3= 108.1
applied. becan assumption gsimplifyin The
108.110.0
K
108.1COOHCH
COOCH HK
10.0
HCOOCHCOOHCH
3-
3-62
5a
5
3
-3
+
a
-33
xx
x
xx
MxMxMx
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Weak Acid/Strong Base Titration Curves• After a total of 20.0 mL of KOH solution has
been added, the pH is: KOH + CH COOH K CH COO H O
Initial: 2.00 mmol 10.0 mmol
Chg. due to rxn:-2.00 mmol - 2.00 mmol + 2.00 mmol
After rxn: 0.00 mmol 8.00 mmol 2.00 mmol
8.0 mmol120 mL
2.0 mmol120 mL
3+
3-
2
CH COOH
CH COO
3
3-
M M
M M
0 067
0 017
.
.
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Weak Acid/Strong Base Titration Curves
Ka 1.8 10 5 H CH3COO
CH3COOH
H 1.8 10 5 CH3COOH CH3COO
H 1.8 10 5 0.067
0.0177.110 5M
pH 4.15
• Similarly for all other cases before the equivalence point is reached.
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Weak Acid/Strong Base Titration Curves
• At the equivalence point, the solution is 0.500 M in KCH3COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution.– This is a solvolysis process as discussed in
Chapter 18.– Both processes make the solution basic.
• The solution cannot have a pH=7.00 at equivalence point.
• Let us calculate the pH at the equivalence point.
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Weak Acid/Strong Base Titration Curves1. Set up the equilibrium reaction:
KOH + CH COOH K CH COO H O
Initial: 10.0 mmol 10.0 mmol
Chg. due to rxn:-10.0 mmol -10.0 mmol +10.0 mmol
After rxn: 0.0 mmol 0.0 mmol 10.0 mmol
3+
3-
2
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Weak Acid/Strong Base Titration Curves2. Determine the concentration of the salt in
solution.
COOCH 0500.00500.0
0500.0mL 200
mmol 10.0=
3COOKCH
COOKCH
3
3
MMM
MM
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Weak Acid/Strong Base Titration Curves3. Perform a hydrolysis calculation for the
potassium acetate in solution.
8.72pH5.28pOH
OH1027.5108.2
106.50500.00500.0
=K
106.5COOCH
OHCOOHCH=K
0500.0
H COOHCHOHCOOCH
6112
10
2
b
10
3
-
3
b
323
xx
x
x
xx
xMxMMx
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Weak Acid/Strong Base Titration Curves4. After the equivalence point is reached, the pH
is determined by the excess KOH just as in the strong acid/strong base example.
11.68=pH and 2.32pOH
108.4OH
108.4mL 210
mmol 0.1
mmol 10.00 mmol 0.00 mmol 1.00 :rxnAfter
mmol 10.00+ mmol 10.0- mmol -10.0:rxn todue Chg.
mmol 10.0 mmol 11.0 :Initial
OHCOOCHKCOOHCH + KOH
3
KOH3
KOH
2-
3+
3
M
MM
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Weak Acid/Strong Base Titration Curves• We have calculated only a few points on the
titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.
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Strong Acid/Weak BaseTitration Curves• Titration curves for Strong Acid/Weak
Base Titration Curves look similar to Strong Base/Weak Acid Titration Curves but they are inverted.
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Weak Acid/Weak BaseTitration Curves• Weak Acid/Weak Base Titration curves
have very short vertical sections.
• The solution is buffered both before and after the equivalence point.
• Visual indicators cannot be used.
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Group Question
• Blood is slightly basic, having a pH of 7.35 to 7.45. What chemical species causes our blood to be basic? How does our body regulate the pH of blood?
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19Ionic Equilibria: Part II
Buffers and Titration Curves