ap chapter 17

AP Chapter 17Ionic Equilibria of Weak Electrolytes

pH and pOH scales

basicacidic

pH + pOH = 14.00

H+ = H3O+

pH = -log[H+]

• The pH scale is a logarithmic scale.

• This means that in order to change the pH by one unit there must be a tenfold change in the [H+].

• Find the pH of a solution with [H+] = 0.010M• Find the pH of a solution with [H+] = 0.10M

Example 17.1 21

• What is the pH of a solution of HCl with a concentration of 1.2 x 10-3M?

pH = 2.92

Example 17.2 Page 521

• Calculate the pH of:– 0.10 M solution of HNO3

– 0.10 M solution of CH3CO2H (1.3% ionized) pH = 2.89pH = 1.00


• Calculate the [H3O+] of a solution with a pH of 9.0

[H3O+] = 1.0 x 10-9


• Calculate the pH and pOH of a 0.0125 M KOH solution.

pOH = 1.903 pH = 12.097

Calculate the [OH-] of a solution with a pH of 5.56.

[OH-] = 3.63 x 10-9

What are the ion concentrations What are the ion concentrations in a 0.10M HCl solution?in a 0.10M HCl solution?

0.10M HCl

0.10M H+

0.10M Cl-

Strong electrolytes dissociate completely

What are the ion concentrations What are the ion concentrations in a 0.15 M Kin a 0.15 M K22SOSO44 solution? solution?

0.15M K2SO4

0.30M K+

0.15M SO42-

At equilibrium, a solution of acetic acid, [ CH3CO2H ] = 0.0788M and [H3O+] = [CH3CO2

-] = 0.0012M. What is the Ka of acetic acid?

Example 17.8 page 527


The pH of a 0.0516 M solution of nitrous acid, HNO2, is 2.34. What is the Ka?

Additional Ka values

Table contains more Ka values than are pictured here

Calculate the [H3O+], [CH3CO2-], and [ CH3CO2H ] in a 0.100 M

solution of acetic acid. What is the Ka = 1.8 x 10-5.


What else can you determine in the previous example?

• pH, pOH, [OH-]

• percent ionization Calculate this value.

Percent ionization = 1.3%

What is the percent ionization of a 0.25 M solution of trimethylamine, (CH3)3N, a weak base with a Kb = 7.4 x 10-5

1.7%

What is the pH of the trimethylamine solution?

[OH-] = 4.3 x 10-3

pOH = -log[4.3 x 10-3] = 2.37

pH = 14 – 2.37 = 11.63

[OH-] = 4.3 x 10-3

[H+] = 1 x 10-14 ÷ 4.3 x 10-3 = 2.3 x 10-12

pH = -log[2.3 x 10-12] = 11.63

Appendix G has additional values

Diprotic and Triprotic Acids

• A diprotic acid ionizes in two steps because it has two ionizable hydrogens.

• A triprotic acid ionizes in three steps because it has three ionizable hydrogens.

The Stepwise Dissociation of Phosphoric Acid.A triprotic acid.

H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+

(aq)

H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O+

(aq)

HPO42-

(aq) + H2O(l) PO43-

(aq) + H3O+(aq)

H3PO4 (aq) + 3 H2O(l) PO43-

(aq) + 3 H3O+(aq)

Each ionization step occurs to a lesser extent than one preceding it.

H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O+

(aq)

H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O+

(aq)

HPO42-

(aq) + H2O(l) PO43-

(aq) + H3O+(aq)

H3PO4 (aq) + 3 H2O(l) PO43-

(aq) + 3 H3O+(aq)

Ka = 7.5 x 10-3

Ka = 6.3 x 10-8

Ka = 3.6 x 10-13

Ka = ?

0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in 1.0L of solution. Calculate [H3O+].




We can work the problem using either Ka or Kb

Compare this answer to [H3O+] from problem 17.10 pp. 528.

How does this illustrates LeChatlier’s Principle.


CH3CO2H + H2O ↔ CH3CO2- + H3O+

[H3O+] = 1.3 x 10-3


[H3O+] = 3.6 x 10-6

Calculate the [H3O+], [CH3CO2-], and [ CH3CO2H ] in a 0.100 M

solution of acetic acid. What is the Ka = 1.8 x 10-5.

10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are mixed. Calculate the [OH-].

Common Ion Problems.What is the “common ion” in each example?


• 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are mixed. Calculate the [OH-].


• 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are dissolved in 1.0L of solution. Calculate [H3O+].

Buffers are also examples of the common ion effect since they are mixtures where both substances produce the same ion.

Both of the solution mixtures above are buffers.

Buffers

• A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

• Buffer solutions have the property that the pH of the solution changes very little when a small amount of acid or base is added to it.

Demonstration: Buffered vs. Non-buffered solutions

Why are these solutions buffers?

Example 17.20 page 549• 10.0ml of 0.10M HCl and 25.0ml of 0.10M NH3 are

mixed. Calculate the [OH-].

Example 17.19 page 548• 0.10 mol CH3CO2H and 0.50 mol NaCH3CO2 are

dissolved in 1.0L of solution. Calculate [H3O+].

How buffers work

• Weak acids and weak bases tend to remain in high concentrations when added to water because by definition they do not ionize much in water since they are weak.

• However, they are very likely to react with any added strong base or strong acid.

Why are weak acids/bases used to create buffers?

Remember this problem

• Calculate the pH of:– 0.10 M solution of HNO3 – 0.10 M solution of CH3CO2H (1.3% ionized)

• What would the [CH3CO2H] have to be for it to have the same pH as the HNO3 assuming the 1.3% ionization factor does not change?

0.10M = [CH3CO2H](0.013)

[CH3CO2H] = 7.7 M

pH = 2.89pH = 1.00

Why are weak acids/bases used to create buffers?

• That’s right 7.7 M [CH3CO2H] vs. 0.10M have the same pH.

• In other words the [CH3CO2H] is 77 times greater [HNO3].

• It takes much more base to change the pH of a weak acid solution because there is a large reservoir of undissociated weak acid.

Weak Acid (HA) and its conjugate base (A-) buffer

Adding a strong base to the buffer• If a strong base is added to a buffer, the weak acid will give up its H+ in

order to transform the base (OH-) into water (H2O) and the conjugate base:

• HA + OH- → A- + H2O.

• Since the added OH- is consumed by this reaction, the pH will change only slightly.

Adding a strong acid to the buffer• If a strong acid is added to a buffer, the weak base will react with the

H+ from the strong acid to form the weak acid, HA: • H+ + A- → HA • The H+ gets absorbed by the A- instead of reacting with water to form

H3O+ (H+), so the pH changes only slightly.

An effective buffer requires relatively equal amounts of weak acid and conjugate base.

Basic Buffers

• Note that the same ideas hold true for weak bases, (B), and their conjugate acids (BH+).

For the most effective buffers Ka = [H3O+]

[HA]

]][AO[H K

-

a

3

HA + H2O ↔ A- + H3O+

Consider acetic acid

You want to make the most “effective” buffer you can using acetic acid. What would the [H3O+] be?

Calculate the pH of this buffer.

I want to make a buffer with a pH of 3.14. Which acid should I use?


Ka = [H3O+]

[H3O+] = antilog (-pH)

Ka = antilog (-3.14) = 7.2 x 10-4 HF


Formic Acid (HCO2H)

Ka = [H3O+]

[H3O+] = antilog (-pH)

Ka = antilog (-3.14) = 7.2 x 10-4

For the most effective buffers Kb = [OH-]

Consider ammonia

For the most effective buffers Kb = [OH-]

[B]]][BH[OH

Kb

B + H2O ↔ BH+ + OH-

I want to make a buffer with a pH of 10.64. Which base should I use?

I want to make a buffer with a pH of 10.64. Which base should I use?

Kb = [OH-]

Kb = antilog (-pOH)

Kb = antilog (-3.36) = 4.4 x 10-4

Methylamine (CH3NH2)

Effective Buffers

• The previous formulas only apply when the concentrations of weak acid and conjugate base or weak base and conjugate acid are equal.

• However it is important to note that you do not have to have equal concentrations to have a buffer.

Henderson – Hasselbach Equation

[HA]

][A pK pH

-

a log

Henderson – Hasselbach Equation

[B]][BH

log pK pOH b

I want to make a buffer using acetic acid. What chemicals should I get from the stock room?

• Describe how you would make the buffer?

• What would the pH of the buffer be?

I have no acetate salts. Can I still make the buffer using acetic acid? Explain.

I want to make a buffer using ammonia. What chemicals should I get from the stock room?

• Describe how you would make the buffer?

• What would the pH of the buffer be?

I have no ammonium salts. Can I still make a buffer using ammonia? Explain.

Making Effective Buffers

1. Equal moles of a weak acid and a salt of its conjugate base.

2. Equal moles of a weak base and a salt of its conjugate acid.

3. A weak acid and half as many moles of strong base.

4. A weak base and half as many moles of strong acid.

5. “Use a pH meter”

pH meter

Calculate the pH of a buffer that is 0.10 M acetic acid and 0.10 M sodium acetate.

pH = 4.74


Buffer Capacity

• Buffer capacity: the amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly.

• Buffer capacity depends on the amount (moles) of the conjugate pair used to make the buffer.

• Buffer solutions have essentially lost their buffering capabilities when one component of the conjugate pair is about 10% or less of the other.

Buffers

• Which buffer has the greater capacity?

• Which buffer is more effective?

Buffer A:– 100mL of 0.1M CH3CO2H with 100 mL of 0.1M NaCH3CO2.

Buffer B: – 100mL of 1.0M CH3CO2H with 100 mL of 1.0M NaCH3CO2.

Calculate the [OH-] of a 0.050 M solution of NaCH3CO2

[OH-] = 5.3 x 10-6

Calculate the pH of a 0.050 M solution of NaCH3CO2

pH = 8.72

Calculate the percent reaction of a 0.050 M solution of NaCH3CO2

0.011%

Calculate the pH of a 0.10 M solution of AlCl3 (Ka = 1.4 x 10-5)

pH = 2.93

Titration

titrant

analyte

Titration Curves

• A titration curve is a plot of the pH against the volume of acid or base added in a titration.

• The equivalence point or endpoint for a titration is the point at which exactly enough of the titrant has been added to completely react with the analyte.

• In other words, at the equivalence point, the number of moles of titrant added corresponds exactly to the number of moles of substance being titrated, the analyte, according to the reaction stoichiometry.

Table 17.8 Page 568

How do we choose an appropriate indicator for a titration?

Indicators• An acid – base indicator is a substance that

indicates how acidic or basic a solution is using certain color changes.

• Indicators are normally weak acids that have a different color than their conjugate base.

HA + H2O ↔ H3O+ + A-

Indicators can also be weak bases that have a different color than their conjugate acid.

Indicators• Remember that the equivalence point of a titration is

where you have mixed the two substances in exact stoichiometric proportions.

• You obviously need to choose an indicator which changes color as close as possible to that equivalence point.

• The indicator should have a pKa value near the pH of the titration's endpoint.

• That varies from titration to titration.

pKa and pH range of Acid-Base Indicators

Which indicator(s) would be appropriate for the titrations below?

Indicators

Now that we have chosen an appropriate indicator for a titration. Let’s consider in

more detail what happens during the titration of different types of acids.

(a) (b)

(c) (d)

What species are present at half way to the equivalence point?


NaOH + HCl → NaCl + H2O

•At half way to the equivalence point the main species present are H+ (hydronium), Cl- and Na+.

•The solution is neutral.


NaOH + CH3CO2H → NaCH3CO2 + H2O

•At half way to the equivalence point the main species present are CH3CO2H, Na+, and CH3CO2

- and to a lesser extent H+ (hydronium).

•The solution is neutral.

The titration curve is produced when a 10.0 ml sample of HCl is titrated with 0.100M NaOH. What is

the concentration of the HCl solution.

0.25M HCl

The titration curve is produced when a 10.0 ml sample of CH3CO2H is titrated with 0.100M NaOH.

What is the concentration of the CH3CO2H solution.

0.25M CH3CO2H

Determine the Ka of the weak acid using the titration curve.

Diprotic Acid

• When titrating a diprotic acid with a strong base it is essentially like doing two titrations at once.

Titration curve of a diprotic acid

The titration curve shown above is for a diprotic acid such as H2SO4. This proves that polyprotic acids lose their protons in a stepwise manner.

H2SO4 → H+ + HSO4-

HSO4- → H+ + SO4

2-

How would you determine the Ka for each acid (H2SO4 and HSO4

-)?

How would you determine the Ka for each acid (H2SO4 and HSO4

-)?

http://z.about.com/d/chemistry/1/0/g/g/dipatitration.JPG

Why do you have to learn all this very challenging information?

This stuff is hard. Why do you have to learn it?

Why? Because I had to learn this crap and now its payback time!

ap chapter 17

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