Chapter 19 Ionic Equilibria in Aqueous...

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Chapter 19Ionic Equilibria in Aqueous

Systems

Buffered Solutions

Buffer solutions resist changes in pH when

limited amounts of acid or base are added.

Buffer solutions must contain significant

quantities of a weak acid and its conjugate

base (or weak base and its conjugate

acid).

Fig.

19.1

Fig.

19.2

Unbuffered Weak Acid

• Consider

HC2H3O2 + H2O < == > H3O+ + C2H3O2

-

• The hydronium ion and acetate ion

concentrations are equal and quite small.

Acetic Acid/Acetate Buffered

SystemLet’s consider a solution made up of significant

quantities of acetic acid and sodium acetate.

HC2H3O2 + H2O H3O+ + C2H3O2

-

What happens to the concentrations of the

acid and its conjugate base if additional

H+ or OH- are added?

Fig. 19.3

Fig. 19.4

Common Ion Effect

Buffered solutions contain either a weak acid and its conjugate base or a weak base and its conjugate acid.

• HF and F- is a buffer pair in which F- is called a common ion.

• NH3 and NH4

+ is a buffer pair in which NH4

+ Is a common ion.

The Common Ion Effect

Buffer solutions resist changes in pH when limited amounts of strong acid or base are added because……

Each component of the buffer pair has a role in reacting with excess H+ or OH- .

Consider a solution made up of acetic acid

and sodium acetate.

1. Write a chemical reaction for the

dissociation of acetic acid.

2. What is the common ion if a buffer were

prepared?

3. What if OH- were added to the solution?

4. What would happen to the pH?

5. What if H+ were added to the solution?

6. What would happen to the pH?

Let’s consider an unbuffered, then

buffered solution of benzoic acid

1. Determine the [H+] of 0.15 M benzoic acid, HC7H5O2. Ka = 6.3 X 10-5

2. Determine the [H+] of 0.15 M benzoic acid in which the benzoate concentration has been adjusted to 0.010 M using sodium benzoate.

What happened to the [H+]? Why?

Condensed Method ofCalculating

the pH of a Buffer Solution

Write the equilibrium expression for a weak

acid.

Take the negative log of both sides.

Rearrange the equation and substitute two

definitions.

What do you get?

Henderson-Hasselbalch

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

And

pOH = pKb + log [HB+]/[B]

Note: These equations are for buffer solutions ONLY.

Problem Solving

1. What is the pH of a buffer solution made

by mixing 1.00 L of 0.020 M HC7H5O2 with

3.00 L of 0.060 M NaC7H5O2 . Ka = 6.3 X

10-5

2. What is the [NH4+] in a buffer solution

having 0.20 M NH3 with a pH of 9.0? Kb =

1.8 X 10-5

An experiment requires you to prepare 50.00 mL of a

pH 9.56 buffer solution. You have 1.00 M solutions

of the following along with solutions of the

conjugate base or acid.

H2CO3 pKa = 6.38 NaHCO3 pKa = 10.32

NaH2PO4 pKa = 7.21 Na2HPO4 pKa = 12.44

NH4Cl pKa = 9.25

Which acid and its conjugate base pair would you

choose to use? What volume of each of the

conjugate pair solutions would you need to use to

obtain a pH of 9.56?

pH Resistance of Buffers

What happens in the following systems when

strong acid or base is added to the solution?

Buffer Solutions’ Resistance to

pH Changes• Consider the following buffered systems:

• H2 CO3 + H2 O < == > H3 O+ + HCO3

–

• NH3 + H2 O < == > NH4+ + OH-

• SHOW HOW ADDED H+ OR ADDED

OH- WOULD REACT WITH EACH

BUFFER PAIR.

Problem Solving with an Acidic

Buffer• Calculate the pH of a 100.0 mL of a buffer

containing 0.200M HCOOH and 0.150M

HCOO- . Ka for HCOOH = 1.8 x 10-4 .

• (1) Calculate the resulting pH if 10.0 mL

of 0.100M HCl(aq) were added.

• (2) What is the resulting pH if 5.00 mL of

0.100M NaOH(aq) is added to the original

buffer?

Problem Solving With a Basic

BufferCalculate the change in pH of a 50.0 mL

buffer solution containing 0.10 M NH3 and

0.10 M NH4+ if 5.0 mL of 0.10 M HCl are

added. Kb = 1.82 X 10-5

Fig. 19.7: Titration of a Strong Acid with a Strong

Base

Fig. 19.8: Titration of a Weak Acid with a Strong Base

Titration of 40.00 mL of 0.1000 M HPr with 0.1000 M NaOH

0

2

4

6

8

10

12

14

10 20 30 40 50 60 70 80

Volume of NaOH added (mL)

pH

Strong acid–strong base curve


0

2

4

6

8

10

12

14

10 20 30 40 50 60 70 80


pH

0

2

4

6

8

10

12

14

10 20 30 40 50 60 70 80


pH



0

2

4

6

8

10

12

14

10 20 30 40 50 60 70 80


pH


pH = 8.80 at

equivalence point


0

2

4

6

8

10

12

14

10 20 30 40 50 60 70 80


pH

Methyl red[HPr] = [Pr–]


pH = 8.80 at

equivalence point

pKa of HPr = 4.89


0

2

4

6

8

10

12

14

10 20 30 40 50 60 70 80


pH

Methyl red[HPr] = [Pr–]


Phenolphthalein

pH = 8.80 at

equivalence point


pKa of HPr = 4.89

Fig. 19.10: Titration of a Weak Polyprotic Acid

with a Strong Base

Fig. 19.9: Titration of a Weak Base with a Strong

Acid

Volume of HCl added (mL)

10 20 30 40 50 60 70

0

80

14

12

10

8

6

4

2

pH

Titration of 40.00 mL of 0.1000 M NH3

with 0.1000 M HCl

10 20 30 40 50 60 70

0

80

14

12

10

8

6

4

2

pH



with 0.1000 M HCl

10 20 30 40 50 60 70

0

80

14

12

10

8

6

4

2

pH


pH = 5.27 at

equivalence

point


with 0.1000 M HCl

pH = 5.27 at

equivalence

point

10 20 30 40 50 60 70

0

80

14

12

10

8

6

4

2

pH



with 0.1000 M HCl

pH = 5.27 at

equivalence

point

10 20 30 40 50 60 70

0

80

14

12

10

8

6

4

2

pH


[NH3] = [NH4+]


with 0.1000 M HCl

Phenolphthalein

pKa of NH4+ = 9.25

10 20 30 40 50 60 70

0

80

14

12

10

8

6

4

2

pH


Phenolphthalein

Methyl red

[NH3] = [NH4+]

pH = 5.27 at

equivalence

point


with 0.1000 M HCl

pKa of NH4+ = 9.25

Solubility Equilibria of Relatively

Insoluble Salts

Solubility Product Constant (Ksp)

Fig. 19.12

Equilibrium Expressions for Solids

Write the equilibrium expression for the

previous slide.

Write the solubility product expression for

calcium phosphate.

Write the solubility product expression for

aluminum hydroxide.

Table

19.2

The larger the Ksp the more soluble the

compound.

Remember, K values are temperature

dependent.

Metal Sulfides

The sulfide ion is relatively unstable in water

and is the weak conjugate base of HS-.

Therefore two equilibrium reactions must

be considered. Let’s consider aluminum

sulfide.

Al2S3(s)2Al3+ + 3 S2-

3 S2- + 3 H2O(l)3 HS- + 3 OH-

Total Reaction

Ksp =

Calculating Ksp

If 0.133 mg of silver bromide dissolves in

1.00 L of solution. Determine the solubility

product constant for silver bromide.

An experiment finds the solubility of barium

fluoride to be 0.55 grams in 500. mL of

solution. Determine the Ksp for barium

fluoride.

Determining Molar Solubility

How can we determine how much of a

slightly soluble salt dissolves?

What is the molar solubility of Al(OH)3?

Ksp = 3.0 X 10-34

What is the molar solubility of Al(OH)3 in

0.0200 M barium hydroxide?

More Molar Solubility Problem

Solving

What is the molar solubility of aluminum

hydroxide in 0.0200 M aluminum nitrate?

Mercury (I) chloride (once used as a

laxative) called Calomel has a Ksp = 1.3 X

10-18. What is the solubility in grams/L?

The Effect of pH on Solubility

Consider lead (II) fluoride.

If strong acid is added what happens to equilibrium and the solubility of lead fluoride?

Compounds having anions of weak acids or hydroxide will be more soluble in acidic (low pH) solutions.

Consider lead (II) bromide and lead (II) hydroxide. If strong acid is added what happens to equilibrium and the solubility of lead bromide?

PbF2(s)Pb2+ + 2 F-

Predicting the Formation of a

Precipitate

Determining whether or not a ppt will form

you will need to compare Q with Ksp.

If Q > Ksp the rxn will proceed in reverse

If Q < Ksp the rxn will proceed forward

If Q = Ksp the rxn is at equilibrium

Problem

Ca3(PO4)2 forms kidney stones. If a urine

sample contains 1.0 X 10-3 M Ca2+ and 1.0

X 10-8 M PO43-, will a stone (ppt) form?

Ksp = 1.0 X 10-26.

If 400. mL of 0.50 M lead (II) nitrate are

mixed with 1600. mL of 2.50 X 10-2 M

sodium chloride will a ppt form?

Fractional Precipitation

This is a process to selectively separate 2 or more ions in solution.

In an experiment you have a solution containing 0.10 M Ba(NO3)2 and 0.10 M Sr(NO3)2 and you use K2CrO4 to titrate the solution. (a) Which one will ppt first? (b) What is the [Ba2+] when the SrCrO4 just begins to ppt? In other words, is the Ba2+

all gone? Ksp of BaCrO4 is 1.2 X 10-10 and Ksp of SrCrO4 is 3.5 X 10-4.

Fig.

19.17

Fig.

19.1

8

Another Fractional Precipitation

A solution contains 0.10 M lead (II) nitrate and 0.20 M strontium (II) nitrate. Using sodium sulfate to titrate the solution, (a) determine the first ppt, (b) the concentration of the first one to ppt when the second begins to ppt and (c) What percentage of the first remains when the second begins to ppt.

Ksp for lead (II) sulfate is 1.7 X 10-8 and Ksp

for strontium sulfate is 2.5 X 10-7.

Determination of a Good Titrant

Complex Ion Equilibria

A complex ion consists of a metal and two or

more anions or molecules covalently

bonded. These are complex ions are

Lewis adducts.

Tabl

e

19.4

Fig. 19.14

Problem

Write a chemical reaction for the formation

of the complex ion presented in the

previous slide.

Write the equilibrium expression for this

reaction.

Determining the Concentration of a

Complex Ion

If a chemist mixes 1.0 X 10-4 M silver nitrate with 1.0 M sodium cyanide, what is the concentration of (a) the silver complex ion (Ag(CN)2

- (b) the silver ion, and (c) the cyanide ion ? Kf = 5.6 X 1018.

What is the [Cu2+] in 0.00020 M Cu(NO3)2

that is also 0.900 M in NH3?

Kf of Cu(NH3)42+ is 4.8 X 1012.

Chapter 19 Ionic Equilibria in Aqueous...

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