1 19 Ionic Equilibria: Part II Buffers and Titration Curves

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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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• 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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• 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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• 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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• Sodium hydroxide and acetic acid react in a 1:1 mole ratio.

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• 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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• Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.

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• 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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• The appropriate equilibria representations are:

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• Substitute into the ionization constant expression (or Henderson-Hasselbach equation) for aqueous ammonia

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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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• The equilibrium constant expression for an indicator would be expressed as:

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• If the preceding expression is rearranged the range over which the indicator changes color can be discerned.

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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

-33

108.110.0

K

108.1COOHCH

COOCH HK

10.0

HCOOCHCOOHCH

x

xx

xMxMMx

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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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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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• 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

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1 19 Ionic Equilibria: Part II Buffers and Titration Curves