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Chapter 15

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Chapter 15. Applications of Equilibrium. Common Ions. An ion that is present in both An acid and its conjugate base HNO 2 and NaNO 2 A base and its conjugate acid NH 3 and NH 4 Cl This common ion effects the position of the equilibrium and the pH. - PowerPoint PPT Presentation
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Chapter 15 Applications of Equilibrium
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Page 1: Chapter 15

Chapter 15

Applications of Equilibrium

Page 2: Chapter 15

Common Ions

• An ion that is present in both

– An acid and its conjugate base

• HNO2 and NaNO2

– A base and its conjugate acid

• NH3 and NH4Cl

• This common ion effects the position of the equilibrium and the pH

Page 3: Chapter 15

• How will the pH of a solution containing 0.10M HNO2 compare to a solution containing 0.10M HNO2 and 0.10M NaNO2? Why?

Page 4: Chapter 15

Points To Remember

• In common ion situations you MUST consider the initial concentrations of any common ions in solution before you begin the problem

• The larger Ka or Kb will govern the equilibrium

Page 5: Chapter 15

• Calculate the pH of a solution containing 0.10M HNO2 and 0.10M NaNO2? (Ka = 4.0x10-4)

Page 6: Chapter 15

Homework

• P. 774 #’s 21,23a&d, 34,36

Page 7: Chapter 15

Buffered Solutions

• Solutions that resists changes in pH when an acid or base is added

• Made of– Weak acid and its conjugate base– Weak base and its conjugate acid

• Made by adding both chemicals or reacting

Page 8: Chapter 15

Adding

.1M HF

NaF

Page 9: Chapter 15

Reacting

.1M HF

.1M NaOH

Page 10: Chapter 15

• Determine the pH of a solution made by dissolving NaHCO3 to 0.25M in 0.0233M H2CO3 (Ka = 4.3x10-7)

Page 11: Chapter 15

Henderson-Hasselbalch Equ.

• A way of calculating buffer problems

• ONLY USED 4 BUFFERS!!!!!!

Page 12: Chapter 15

HA H+ + A-

][

]][[

HA

AHKa

][

][log

Acid

BasepKapH

Page 13: Chapter 15

• Determine the pH of a solution made by dissolving NaHCO3 to 0.25M in 0.0233M H2CO3 (Ka = 4.3x10-7) Use H-H Equ.

Page 14: Chapter 15

How Do Buffers Work

• Strong acid or base is not allowed to build up in solution

• Strong acid is replaced by weak acid

• Strong base is replaced by weak base

– The strong acid or base reacts completely!!!

– Then equilibrium kicks in

Page 15: Chapter 15

Acidic Buffers

HF H+ + F-

Add NaOH

Add HCl

Page 16: Chapter 15

Basic Buffer

NH3 + H2O NH4

+ + OH-

Add NaOH

Add HCl

Page 17: Chapter 15

• Determine the pH after 0.20 grams of NaOH is added to 500.mL a solution made by dissolving NaHCO3 to 0.25M in 0.0233M H2CO3 (Ka = 4.3x10-7)

Page 18: Chapter 15

What Makes a Good Buffer?

• The best buffers have acids and bases at the same, large concentration

– Added acid or base has little effect on pH

– When the concentrations are the same pH=pKa

Page 19: Chapter 15

Homework

• P. 774 #’s 23d,31,37ab, 40, 42,46,48,49

Page 20: Chapter 15

Titrations

• The controlled addition of a chemical of known concentration to a chemical of unknown concentration

• Known Chemical – Standard or titrant

• Unknown Chemical - Analyte

Page 21: Chapter 15

• Titrations are carried out to the equivalence point– Point when mole ratio of chemicals is reached– Often an invisible point

• Marked by and indicator

• Indicator is a chemical that changes color in different pH’s

• Indicator color change marks the endpoint– Hopefully endpoint and equivalence point are

the same

Page 22: Chapter 15

Titration Curves

• Plotting of the pH of the titration solution as solution is added

• Used as a way of determining equivalence and Ka or Kb values

• Specific ranges for specific acid/base combinations – Strong/Strong or Strong/Weak

Page 23: Chapter 15

Titration Curve HCl vs NaOH

http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

Page 24: Chapter 15

Titration Curve Problems

• Remember the process. Do NOT memorize!

• Treat each step as a separate problem

– For Strong/Strong systems just stoichiometry

• For Strong/Weak First stoichiometry then equilibrium

Page 25: Chapter 15

Strong Acid / Strong Base

• Beginning – Just the pH of the chemical you have

• Before Equivalence – Stoichiometry

• Equivalence – pH is 7.00

• After Equivalence - Stoichiometry

Page 26: Chapter 15

• Strong Acid / Strong Base Titration

#54 page 776

80.0 mL of 0.100M Ba(OH)2 is titrated with 0.400M HCl

0.00 mL

20.0 mL

30.0 mL

40.0 mL

80.0 mL

Page 27: Chapter 15

Weak / Strong Titration

• Beginning – Just the pH of the chemical you have (Might need equilibrium)

• Before Equivalence – Stoichiometry then equilibrium

• Equivalence – pH is NOT 7 – Stoichiometry then equilibrium

• After Equivalence - Stoichiometry

Page 28: Chapter 15

Weak / Strong Combos

• Weak Acid / Strong Base

• Equivalence Point is ABOVE seven– Conjugate base of a weak acid controls pH

• Weak Base / Strong Acid

• Equivalence Point is BELOW seven– Conjugate acid of a weak base controls pH

• At halfway pH = pKa or pOH = pKb!!!!!!!!

Page 29: Chapter 15

Weak Base / Strong Acid Titration

http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

Page 30: Chapter 15

Weak Acid / Strong Base Titration

http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

Page 31: Chapter 15

• Strong Base/ Weak Acid Titration

#55 page 776

100.0 mL of 0.200M Acetic Acid is titrated with 0.100M KOH

0.00 mL

50.0 mL

100.0 mL

200.0 mL

250.0 mL

Page 32: Chapter 15

Homework

• P. 775 #’s 52,53,56

Page 33: Chapter 15

Which Indicator to Use?

• Choosing the appropriate indicator is important when doing titrations

• There are so many good ones to use

• So how does a chemist know???????

Page 34: Chapter 15

Which Indicator to Use?

• Indicators are weak acids and change color with varying pH

• Pick the indicator that changes a little above your calculated equivalence point.

• Pick indicator with pKa value close to equivalence point

• Change color +/- 1 from pKa

Page 35: Chapter 15
Page 36: Chapter 15

Example

• Choose an indicator for the titration of HCl with NH3 pH=5.36

Page 37: Chapter 15

Polyprotic Acids

• These acids have multiple equivalence points

• Carbonic acid has two

• Phosphoric acid has three

• Multiple step points on a titration curve

Page 38: Chapter 15

Diprotic Acid Titration

Page 39: Chapter 15

Molar Mass of an Acid

• You can find the molar mass of an acid by titration too

• Titrate a acid of known mass with a standard to determine number of moles

• Divided grams by moles– Molar Mass

Page 40: Chapter 15

Solubility Equilibrium

• Dissolving is an equilibrium process

• Chemicals that we say are insoluble are actually very slightly soluble– Meaning only a small amount of the solute

dissolves – Concentrations are very small

• 10-5M to 10-20M are common

Page 41: Chapter 15

Solubility Equilibrium

• Iron (II) Sulfide is “insoluble” by solubility rules

• However, it still dissolved to a small extent

• Consider the reaction

FeS(s) Fe+2 + S-2

• Since dissolving is an equilibrium process we can write an equilibrium expression

• Keq = [Fe+2] [S-2]

• FeS is omitted because it is a solid

Page 42: Chapter 15

Solubility Product Constant

• The previous equilibriums expression is a “special” type of equilibrium expression

• Ksp – solubility product constant

• Ksp = [Fe+2] [S-2]– Only used for slightly soluble salts– Never includes the reactants– Ksp’s are very small– Check out the table 15.4 page 753

Page 43: Chapter 15

Ksp Values

• Lead (II) Chloride 2.4X10-4

• Strontium Carbonate 3.8X10-9

• Nickel (II) Hydroxide 5.5X10-16

• Copper (II) Hydroxide 2.2X10-20

• Cadmium Sulfide 8.0X10-28

• Silver Sulfide 6.0X10-51

• And my personal favorite• Bismuth Sulfide 1.6x10-72

Page 44: Chapter 15

Ksp Values

• Q) What does a small Ksp value mean?

• A) Low concentration of ions• Small solubility

• Q) Is the compound with the smallest Ksp the least soluble?

• A) Not necessarily – There are different numbers of ions that changes

the expression

Page 45: Chapter 15

Solubility of Salts

• We can easily compare solubility for salts that have the same # of ions

• When there are the same # of ions the salt with the smallest Ksp is least soluble

• Which salt is least soluble, most soluble AgCl, FeS, BaSO4

– FeS 6.0x10-19 Least soluble– AgCl 1.6x10-10 Most soluble

– BaSO4 1.1x10-10 Middle

Page 46: Chapter 15

Calculating Solubility

• We can calculate solubility of salts and the concentration of the ions in solutions from Ksp

• Deal with saturated solutions– The salt has dissolved as much as it can

• Ksp has been reached

• In saturated solutions the concentrations of the ions are related to the mole ratio

Page 47: Chapter 15

Example• Calculate the solubility of silver chloride

Ksp=1.6x10-10

Page 48: Chapter 15

Example• Calculate the solubility of mercury (II) sulfide

and the concentration of each ion in a saturated solution Ksp=1.6x10-54

Page 49: Chapter 15

Example• Calculate the solubility of calcium phosphate

Ksp=1.3x10-32

Page 50: Chapter 15

Example• A saturated solution of Iron (III) Hydroxide

has a concentration of Fe+3 of 1.8x10-8. What is the Ksp of Iron (III) Hydroxide?

Page 51: Chapter 15

Common Ion Revisited

• Anyone ever have a barium shake for a scan?

• Contains barium sulfate.– Barium ions are very toxic to the body– Treated as calcium

• How can you have fewer barium ions in solution?

• Add Sodium Sulfate

Page 52: Chapter 15

• What is the concentration of barium ions in a saturated solution of barium sulfate?– Ksp = 1.1x10-10

• What is the concentration of barium ions in a when 0.10M Sodium Chloride is added to a saturated solution of barium sulfate

Page 53: Chapter 15

Precipitation

• If you combine Silver Nitrate with Sodium Chloride will a precipitate always form?

• Only if Ksp is exceeded

• Will a ppt. form if 2x10-5M Silver Nitrate is mixed with 3x10-3M Sodium Chloride?

• Ksp of AgCl is 1.6x10-10

• No


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