Remember: As a solution of a weak acid becomes more dilute, the % ionization increases.

Relate to LeChatlier's Principle

HC2H3O2 + H2O <--> C2H3O2- + H3O+

1. Get increasing relative amounts of water to acetic acid as [HC2H3O2] decreases, therefore shifts eq. to the right.

2. Conversely, get increasing relative amounts of hydronium ion as [HC2H3O2] decreases, therefore shifts eq. to the left.

A. Add a common ion: H+

HC2H3O2 + H2O <--> C2H3O2- + H3O+

Rewrite as:

HC2H3O2 <--> C2H3O2- + H+

1. When H+ is added, the system will shift to the left (decreased ionization of acetic acid).

2. Look at Ka

Ka = [H+][C2H3O2-]/[HC2H3O2]

With increased [H+], must increase [HC2H3O2] to keep fraction equal to Ka.

B. Add a different common ion: a solution of HC2H3O2 and NaC2H3O2

HC2H3O2 <--> C2H3O2- + H+

1. When C2H3O2- is added, the system will shift to the left (decreased ionization of acetic acid).

2. Look at Ka

Ka = [H+][C2H3O2-]/[HC2H3O2]

With increased [C2H3O2-], must increase [HC2H3O2] to keep fraction equal to Ka.

C. These results are referred to as the common ion effect: Addition of more of one component of an equilibrium will cause the system to shift to reestablish equilibrium.

1. If this is an ion, the percent ionization of the acid or base will decrease.

A. The solution of a weak acid and its common ion has special properties and is referred to as a buffer.

1. Buffer: an aqueous solution capable of maintaining approximately the same pH when small amounts of H+ or OH- are added.

2. How does this work? e.g. 0.20 M HC2H3O2 and 0.20 M NaC2H3O2

a. Initial pH of the solution?

Ka = 1.8 x 10-5 = [H+][C2H3O2-]/[HC2H3O2]

= x(0.20+x)/(0.20-x)

= x(0.20)/(0.20) ;

x = 1.8 x 10-5 M = [H+]

pH = -log(1.8 x 10-5) = 4.74

b. Test the buffer: add 0.010 moles NaOH to 1.00 L of buffer

Reaction of NaOH with what? HC2H3O2

pH of the buffer now?

Ka = 1.8 x 10-5 = [H+][C2H3O2-]/[HC2H3O2]

= x(0.21+x)/(0.19-x)

= x(0.21)/(0.19) ; x = 1.6 x 10-5 M = [H+]

pH = -log(1.6 x 10-5) = 4.79 => very close to the initial pH!

c. Compare with adding NaOH to pure water:

pH = 14.00 + log(0.010) = 14.00 - 2.00 = 12.00

d.The change in pH of the buffer: 0.05 ; the change in pH of water: 5.00

Big difference!

e. Reactions in solution:

1. The initial equilibrium between acetic acid and acetate

2. When base is added: it reacts with acetic acid, instead of being in solution, thus slightly adjusting ratio of acetate to acetic acid, instead of drastically changing pH ([H+]).

3. When acid is added: it reacts with acetate, instead of being in solution, thus slightly adjusting ratio of acetate to acetic acid, instead of drastically changing pH ([H+]).

4. Key to buffers: they must be solutions of

1. a weak acid and the salt of its conjugate base or
2. (ii) a weak base and the salt of its conjugate acid.

B. Simplifying buffer calculations: Henderson-Hasselbach (H-H) Equation

For the generic equilibrium of a weak acid: HA <--> H+ + A-

Ka = [H+][A-]/[HA] ; [H+] = Ka[HA]/[A-]

Take the negative log of the equation:

-log[H+] = -logKa -log([HA]/[A-])

pH = pKa - log([HA]/[A-])

Use the H-H equation to calculate:

initial pH of buffer

pH of buffer after addition of H+ or OH-

C. Guide to solving buffer problems

1. Initial pH of buffer: use H-H equation if 5% rule is followed

2. Challenged buffer:

a. write down equation of reaction of SB (or SA) with WA (or WB)

b. calculate change in amount of WA and CB (or WB and CA)

c. Use H-H equation to determine final pH

Purpose: determination of Ka of a WA or Kb of a WB

A. Experimental Procedure:

1. Place a known amount of the weak acid in water in a flask.

2. Add in a solution of SB of known concentration in small aliquots via buret.

3. Measure pH after each aliquot of SB.

4. Plot pH vs. volume of SB added.

B. Calculational Procedure:

1. Define titration - amount and concentration of WA in flask and concentration of SB in buret.

2. Calculate pH after each aliquot of SB.

3. Plot pH vs. volume of SB added.

C. Simple example: titration of a strong acid (50 mL of 0.10 M HCl) with a strong base (0.10 M NaOH)

1. Initial pH? pH = -log(0.10) = 1.00

2. pH after addition of 5.00 mL SB?

[H+] = 0.0045 moles/0.055 L = 0.0818 M

pH = -log(0.0818) = 1.09

Repeat this for each aliquot of added NaOH:

3. Features to note on the titration curve:

a. equivalence point: inflection point in the curve;

steepest part of the curve;

corresponds to the point when a stoichiometric amount of base has been added to neutralize the acid.

moles SA = moles SB

b. shape of the curve: NOT a smooth increase in pH

D. Titration of weak acid (50.00 mL 0.100 M HC2H3O2) and strong base (0.100 M NaOH)

1. Initial pH?

Ka = [H+][C2H3O2-]/[HC2H3O2]

= (x)(x)/0.10-x

= x2/0.10 ; x = 1.3 x 10-3 M = [H+]

pH = -log(1.3 x 10-3) = 2.87

2. pH after addition of 5.00 mL SB? Beginning of the titration

NaOH will react with? acetic acid

total moles NaOH added: (0.100 M)(0.00500 L) = 5.00 x 10-4 moles

initial moles HC2H3O2: = (0.100 M)(0.0500 L) = 5.00 x 10-3 moles

[HC2H3O2] = (0.0045 moles)/(0.0055 L) = 0.0818 M

[C2H3O2-] = (0.0005 moles)/(0.0055 L) = 0.00909 M

H-H equation:

pH = pKa + log([A-]/[HA])

= -log(1.8 x 10-5) + log(0.00909/0.0818)

= 3.79

3. Half-equivalence point

pH after addition of 25.00 mL NaOH?

moles NaOH added = 1/2 initial moles acetic acid

NaOH will react with? acetic acid

total moles NaOH added: (0.100 M)(0.02500 L) = 0.0025 moles

initial moles HC2H3O2: = (0.100 M)(0.0500 L) = 0.0050 moles

[HC2H3O2] = (0.0025 moles)/(0.0075 L) = 0.034 M

[C2H3O2-] = (0.0025 moles)/(0.0075 L) = 0.034 M

H-H equation:

pH = pKa + log([A-]/[HA])

= -log(1.8 x 10-5) + log(0.034/0.034)

= 4.74

4. Equivalence point

pH after addition of 50.00 mL NaOH?

moles NaOH added = initial moles acetic acid

NaOH will react with? acetic acid

total moles NaOH added: (0.100 M)(0.0500 L) = 0.0050 moles

initial moles HC2H3O2: = (0.100 M)(0.0500 L) = 0.0050 moles

Operative equilibrium? What's in solution? Na+, C2H3O2-, H2O

=> treat as a sodium acetate solution

[C2H3O2-] = (0.005 moles)/(0.100 L) = 0.050 M

Kb = Kw/Ka = 1.0 x 10-14/1.8 x 10-5 = 5.6 x 10-10

= [HC2H3O2][OH-]/[C2H3O2-] = (x)(x)/0.05-x

= x2/0.05 ; x = 5.3 x 10-6 M = [OH-]

pOH = -log(5.3 x 10-6) = 5.28

pH = 14.00 - 5.28 = 8.72

5. Past the equivalence point

pH after addition of 51.00 mL of NaOH?

total moles NaOH: (0.100 M)(0.0510 L) = 0.0051 moles

initial moles HC2H3O2: (0.100 M)(0.050 L) = 0.0050 moles

moles excess OH-: 0.0051 - 0.0050 = 0.0001 moles

Operative equilibrium? What's in solution? Na+, C2H3O2-, H2O, OH-

=> treat as a sodium acetate solution

[C2H3O2-] = (0.005 moles)/(0.100 L) = 0.050 M

[OH-] = (0.0001 moles)/(0.100 L) = 0.001 M

Kb = Kw/Ka = 1.0 x 10-14/1.8 x 10-5 = 5.6 x 10-10

= [HC2H3O2][OH-]/[C2H3O2-] = (x)(0.001+x)/0.05-x

= x(0.001/0.05) ; x = 5.3 x 10-6 M = [HC2H3O2]

[OH-] = 0.001 + 5.3 x 10-6 Å 0.001 M

pOH = -log(0.001) = 3.00

pH = 14.00 - 3.00 = 11.00

Repeat for each aliquot of NaOH:

E. Titration of an acid weaker that acetic acid (0.100 M HOCl) with a strong base (0.050 M NaOH)

The initial pH is higher and the pH at equivalence is higher.

F. Titration of a weak base (0.100 M NH3) with a strong acid (0.100 M HCl)

The titration curve is the mirror image of that of WA-SB.

A. Often weak or organic acids with known Ka and color changes with deprotonation.

e.g. phenolphthalein

The acid form is colorless and the base form is pink.

Ka = 6 x 10-10 ; pKa = 9.2

B. Determining the pH range of the color change

The human eye can detect color changes of indicators when there is at least a factor of 10 difference in [HIn] and [In-].

Range of color change: 1/10 _ [In-]/[HIn] _ 10/1

Using the H-H equation:

pH = pKa + log([In-]/[HIn])

lower boundary: pH = pKa + log(1/10) = pKa - 1

upper boundary: pH = pKa + log(10/1) = pKa + 1

pH range of color change: pKa - 1 to pKa + 1

e.g. phenolphthalein

pH range is from (9.2 - 1 =) 8.2 to (9.2 + 1 =) 10.2