additional aqueous equilibria chapter 16

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Additional Aqueous Equilibria CHAPTER 16

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Additional Aqueous Equilibria CHAPTER 16. I. Buffers. A. Definitions Buffer- solutions that resist changes in pH when acid or base are added to it. - they are composed of a weak acid and its conjugate base or a weak base and its conjugate acid . - PowerPoint PPT Presentation

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Page 1: Additional Aqueous  Equilibria CHAPTER 16

Additional Aqueous Equilibria

CHAPTER 16

Page 2: Additional Aqueous  Equilibria CHAPTER 16

I. Buffers

A. Definitions

Buffer- solutions that resist changes in pH when acid or base are added

to it.

- they are composed of a weak acid and its conjugate base or a weak base and its conjugate acid.

acetic acid / sodium acetate

carbonic acid / carbonate

Page 3: Additional Aqueous  Equilibria CHAPTER 16
Page 4: Additional Aqueous  Equilibria CHAPTER 16

B. Explanation of Buffers 1. Shifting of Equilibrium

For some weak acid and its conjugate base:

HA ⇌ H+ + A-

a. What happens if you add excess acid (H+)? b. What happens if you add excess base (OH-)?

** Best buffering capacity when approximately equal amounts of both acid and

conjugate base. Why strong acids and bases not good buffers?

Page 5: Additional Aqueous  Equilibria CHAPTER 16

2. Titration Curve For Buffer

Acetic acid / acetate (CH3COOH / CH3COO-)

Page 6: Additional Aqueous  Equilibria CHAPTER 16

3. Mathematical Relationship

a. Henderson-Hasselbalch Equation

For: HA H+ + A-

When [HA] = [A-], it is true that pH = pKa

Best buffering region = ±1 unit of pKa

log

+ -

a

-

a

[ ][ ]H A = K[HA]

[ ]ApH = + pK[HA]

Page 7: Additional Aqueous  Equilibria CHAPTER 16

C. Buffer Problem Solving

Similar to other equilibrium problems, except we will use the Henderson Hasselbalch Equation

Always utilizes assumption that we can neglect small amount of dissociation of weak acid or base!!!!!!!!

Different from Tro text, all our buffer problem solving will be done using the Henderson-Hasselbalch!!!!

logabase

pH pKacid

Page 8: Additional Aqueous  Equilibria CHAPTER 16

1. Determining the pH of a Buffer Solution

What is the pH of a buffered solution that is

0.015 M acetic acid and 0.045 M sodium acetate?

For acetic acid, Ka = 1.8 x 10-5 ?

Page 9: Additional Aqueous  Equilibria CHAPTER 16

2. Preparation of a Buffer Solutiona. Which acid-base pair should be used to

prepare a buffer pH 5.0?

Acetic acid / acetate Ka = 1.8 x 10-5

Ammonium / ammonia Ka = 5.6 x 10-10

b. A buffer solution pH 5.0 is to be prepared using acetic acid (Ka = 1.8 x 10-5) and sodium acetate . How many grams of sodium acetate (NaC2H3O2)

should you add to 250.0 mL of a 0.0500 M acetic acid (HC2H3O2) solution?

Page 10: Additional Aqueous  Equilibria CHAPTER 16

3. Addition of Acid or Base To A Buffer

A buffer is prepared by adding 0.30 mol of lactic acid and 0.40 mol of sodium lactate to sufficient water to make 2.00 L of buffer solution. The Ka

of lactic acid is 1.4 x 10-4. (Assume Ka small enough to neglect dissociation of weak acid).

a) Calculate the pH of the buffer.

b) Calculate the pH of the buffer after the addition

of 0.050 moles of NaOH.

Page 11: Additional Aqueous  Equilibria CHAPTER 16

When Adding Strong Acid or Base To Buffer:

1. Addition of strong acid to buffer → converts a stoichiometric amount of the base to the conjugate acid.

2. Addition of strong base to buffer → converts a stoichiometric amount of the acid to the conjugate base.

(Adding acid creates more acid, adding base creates more base)

Page 12: Additional Aqueous  Equilibria CHAPTER 16

CONCEPT QUESTION A 1.00 liter buffer solution is 0.10 M in HF and

0.050 M in NaF. Which action will destroy the buffer? Why?

a. addition of 0.050 mol HCl

b. addition of 0.050 mol NaOH

c. addition of 0.050 mol NaF

d. addition of 0.05 mol HF

Page 13: Additional Aqueous  Equilibria CHAPTER 16

D. Buffer Capacity

1. Definition- amount (moles) of acid or base required to change the

pH of a given volume of buffer by ±1 pH unit.

A measure of the ability of a buffer to resist changes in pH.

Addition of more acid or base will cause the buffer to fail, meaning inability to resist

changes in pH.

Page 14: Additional Aqueous  Equilibria CHAPTER 16

II. Acid-Base Titrations

A. Titration Process 1. Discuss Process (acid + base salt + water)

2. Equivalence Point: point in titration when stoichiometric amounts of titrant have

been added to neutralize acid or base titrated.

3. Endpoint: point where titration is actually ended because of indicator color change.

(Endpoint is used to visually determine equivalence pt.)

Page 15: Additional Aqueous  Equilibria CHAPTER 16
Page 16: Additional Aqueous  Equilibria CHAPTER 16

B. Explanation of Acid-Base Indicators

1. Weak organic acids which undergo color change when there is a shift from acid base at

complete neutralization.

HIn (aq) + H2O (l) ⇌ H3O+ (aq) + In-(aq)

acid base

one color another color

2. Different indicators change colors at different pH’s

Bromocresol green: pH 3.8 to 5.4

Phenolphthalein: pH 8.2 to 10

Page 17: Additional Aqueous  Equilibria CHAPTER 16

C. Titration of Strong Acid with Strong Base

Example: HCl + NaOH H2O + NaCl

Page 18: Additional Aqueous  Equilibria CHAPTER 16

Strong Acid-Strong Base Titration ProblemA 50.00 mL sample of 0.100 M HCl is titrated with a 0.100 M NaOH solution.

a. Determine the pH of the HCl solution before the addition of any NaOH solution.

a. Determine the pH after the addition of 25.00 mL

of the NaOH solution.

b. Determine the pH after the addition of 50.00 mL of the NaOH solution.

Page 19: Additional Aqueous  Equilibria CHAPTER 16

D. Titration of Weak Acid with Strong BaseExample: HC2H3O2 + NaOH C2H3O2

- Na+ + H2O

Page 20: Additional Aqueous  Equilibria CHAPTER 16

Weak Acid-Strong Base Titration Problem

A 50.00 mL sample of 0.100 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) is titrated with a 0.100 M NaOH solution.

a. Determine the pH of the solution before the addition of the NaOH solution.

b. Determine the pH after the addition of 50.00 mL of the NaOH solution.

Page 21: Additional Aqueous  Equilibria CHAPTER 16

E. Titration of Weak Base with Strong Acid

Example: NH3 + HCl NH4+ Cl-

1. Is the equivalence point less than or greater than pH 7? Explain you answer.

2. Draw a titration curve for the titration.

(titrate NH3 using HCl as the titrant)

Page 22: Additional Aqueous  Equilibria CHAPTER 16

F. Titration of Polyprotic Acid with Strong Base

Example: H3PO4

Page 23: Additional Aqueous  Equilibria CHAPTER 16

III. Solubility Equilibria

A. Equilibrium Expressions for Sparingly

Soluble Salts

1. Ksp : Solubility Product Constant

(Write like a normal Kc and recall it does not include solids in the equilibrium expression).

Meaning: idea of extent of solubility of salt

Larger Ksp Greater solubility of salt

Page 24: Additional Aqueous  Equilibria CHAPTER 16

Write the chemical equations and Ksp expressions for the slightly soluble salts:

AgCl

CaCO3

Ag3PO4

Mg(OH)2

Page 25: Additional Aqueous  Equilibria CHAPTER 16

2. Solubility: amount of substance dissolved based on grams / volume.

3. Molar Solubility: maximum amount of substance dissolved based on moles /

liter.

Molar solubility is not the same as Ksp !!!!!

The molar solubility of BaF2 is 6.27 x 10-3 M. The Ksp of BaF2 is 9.8 x 10-7. What is the concentration of F- ions in a BaF2 saturated solution?

Page 26: Additional Aqueous  Equilibria CHAPTER 16

B. Problem Solving

1. Calculating Ksp from Solubility Data

One liter of water is able to dissolve 2.15 x 10-3 mol of PbF2 at 250C.

What is the molar solubility of PbF2?

What is the Ksp for PbF2?

Page 27: Additional Aqueous  Equilibria CHAPTER 16

2. Calculating Molar Solubility From Ksp

Determine the molar solubility of barium phosphate, Ba3(PO4)2, given a Ksp value of

3.4 x 10-23.

What are the equilibrium concentrations of each ion in solution?

Page 28: Additional Aqueous  Equilibria CHAPTER 16

3. Common Ion Effect (Factor Affecting Solubility)

Ag2CO3 has a Ksp of 8.1 x 10-12 at 250C. The molar solubility of Ag2CO3 in water is 1.3 x 10-4 M. What is the molar solubility of Ag2CO3 in a 0.10 M NaCO3 solution. (NaCO3 is very soluble).

a. Qualitative Explanation – Le Chatelier’s Principle

Ag2CO3 (s) 2 Ag+ (aq) + CO32- (aq)

What is the effect of adding excess CO32- ?

What is the effect on the molar solubility of Ag2CO3 ?

Page 29: Additional Aqueous  Equilibria CHAPTER 16

b. Terminology Used:

Common Ion: the ion which is common to both salts being

considered.

Common Ion Effect: the solubility of a compound is always lowered

due to the addition of the common ion.

Now solving problem:

Page 30: Additional Aqueous  Equilibria CHAPTER 16

4. Predicting If Precipitation Occurs

Comparison of known Ksp to the calculated ion

product (Q) for a given mixture can serve to indicate

whether or not a precipitate will be formed.

Q

Ion Product > Ksp ppt. forms

Ion Product = Ksp ppt. forms

Ion Product < Ksp no ppt. forms

Page 31: Additional Aqueous  Equilibria CHAPTER 16

Will a precipitate form at equilibrium when 50.00 mL of 0.00100 M BaCl2 is added to

50.00 mL of 0.000100 M Na2SO4 ? The solubility product constant for barium sulfate is 1.1 x 10-10.