Chapter 6 Problems

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Chapter 6 Problems. 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,. 6-29. Distinguish between Lewis Acids/Bases & Bronsted-Lowry acids and bases. Give an example. 6-31. Why is the pH of water usually < 7? How can you prevent this from happening?. 6-39. - PowerPoint PPT Presentation

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Chapter 6 Problems

6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

6-29

Distinguish between Lewis Acids/Bases & Bronsted-Lowry acids and bases. Give an example.

6-31.

Why is the pH of water usually < 7? How can you prevent this from happening?

6-39.

The equilibrium constant for autoprotolysis of water is 1.0 x 10-14 at 25oC. What is the value of K for 4 H2O 4H+ + 4OH-

K = [H+]4 [OH]4

K = (1x10-7)4(1x10-7)4

K = 1 x 10-56

6-41

Use Le Chatelier’s principle and Kw in Table 6-1 to decide whether the autoprotolysis of water is exothermic or endothermic at 25oC 100oC 300oC

0

1E-12

2E-12

3E-12

4E-12

5E-12

6E-12

7E-12

0 100 200 300 400

Temp

K

6-42

Make a list of strong acid and strong bases. Memorize this list.

6-48 Which is the stronger acid?

Dichloracetic acid Chloroacetic acidKa = 8 x 10-2 Ka = 1.36 x 10-3

Stronger Base?

Hydrazine UreaKb = 1.1 x 10-6 Kb = 1.5 x 10-14

Chapter 8

ActivityActivity

HomeworkHomeworkChapter 8 - Chapter 8 - ActivityActivity

8.2, 8.3, 8.6, 8.9, 8.10, 8.12

8-1 Effect of Ionic Strength 8-1 Effect of Ionic Strength on Solubility of Saltson Solubility of Salts

Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.

Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18

A seemingly strange effectseemingly strange effect is observed when a salt such as KNO3 is added. As more KNO3 is added to the solution, more

solid dissolves until [Hg22+] increases to 1.0 x 10-6 M. Why?

1823

22 103.1]][[ IOHgK sp

IICCEE

somesome -- - --x-x +x+x +2x+2xsome-xsome-x +x+x +2x+2x

182 103.1]2][[ xxK sp7109.6][ x

Increased solubility

Why? LeChatelier’s Principle?

NO – not a product nor reactant Complex Ion?

No Hg2

2+ and IO3- do not form complexes

with K+ or NO3-.

How else?

The Explanation

Consider Hg22+ and the IO3

-

-2+

Electrostatic attraction

The Explanation

Consider Hg22+ and the IO3

-

2+

Electrostatic attraction

-

Hg2(IO3)2(s) The Precipitate!!The Precipitate!!

The Explanation

Consider Hg22+ and the IO3

-

-2+

Electrostatic attraction

Add KNO3

NO3-K+

The Explanation

Consider Hg22+ and the IO3

-

-2+

Add KNO3

K+K+

K+K+

K+K+

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

K+K+

K+K+

K+ K+

The Explanation

Consider Hg22+ and the IO3

-

2+

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

Hg22+ and IO3

- can’t getCLOSE ENOUGH to form Crystal latticeOr at least it is a lot “Harder” to form crystal lattice

-

K+ K+

K+K+

K+K+

K+K+

K+K+

K+ K+

The potassium hydrogen tartrate example

K+-O

O

OH

OH

O

OH

potassium hydrogen tartrate

Alright, what do we mean by Ionic strength?

Consider Hg22+ and the IO3

-

-2+

Add KNO3

K+K+

K+K+

K+K+

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

NO3-

K+K+

K+K+

K+ K+

Low Ionic StrengthHigher Ionic StrengthHigh Ionic Strength

Alright, what do we mean by Ionic strength?

Ionic strength is a measure of the total measure of the total concentration of ions in solutionconcentration of ions in solution.

Ionic strength is dependent on the number of ions in solution and their charge. Not dependent on the chemical nature of the

ions

Ionic strength () = ½ (c1z12+ c2z2

2 + …)Or Ionic strength (m) = ½ cizi

2

Examples Calculate the ionic strength of (a) 0.1 M

solution of KNO3 and (b) a 0.1 M solution of

Na2SO4 (c) a mixture containing 0.1 M KNO3

and 0.1 M Na2SO4.() = ½ (c1z1

2+ c2z22 + …)

Alright, that’s great but how does it affect the equilibrium constant?

A + B C + D

Activity = Ac = [C]c

AND

bB

baA

a

dD

dcC

c

bB

aA

dD

cC

BA

DC

AA

AAK

][][

][][

Relationship between activity coefficient and ionic strength

x

x

xz

3.31

51.0log

2

Debye-Huckel Equation

2 comments

= ionic strength of solution = activity coefficientZ = Charge on the species x = effective diameter of ion (nm)

(1)What happens to when approaches zero?(2)Most singly charged ions have an effective radius of about 0.3 nm

We generally don’t need to calculate – values are tabulated

Concept Test

List at least three properties of activity coefficients

• Dimensionless• Depends on size of the ions (ex. Hg2

2+ and IO3-)

• Depends on the Ionic Strength of the Solution (K+ & NO3

-)• Depends on the charge of the ions (ex. Hg2

2+ and IO3

-)• In dilute solutions, where ionic strength is minimal,

the activity coefficient -> 1, and has little effect on equilibrium constant

Activity coefficients are Activity coefficients are related to the hydrated related to the hydrated radius of atoms in radius of atoms in moleculesmolecules

Relationship between and

Back to our original problem

223

22

2

3223

22

][][ IOHgIOHgsp IOHgAAK

Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.

Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18

At low ionic strengths -> 1

11 11

Back to our original problem

223

22

2

3223

22

][][ IOHgIOHgsp IOHgAAK

Consider a saturated solution of Hg2(IO3)2 in ‘pure water’. Calculate the concentration of mercurous ions.

Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18

In 0.1 M KNO3 - how much Hg22+ will be dissolved?

Back to our original problem

223

22

2

3223

22

][][ IOHgIOHgsp IOHgAAK

Consider a saturated solution of Hg2(IO3)2 in 0.1 M KNO0.1 M KNO33.. Calculate the concentration of mercurous ions.

Hg2(IO3)2(s) Hg22+ + 2IO3

- Ksp=1.3x10-18 22

322

2

3322

775.0][355.0][ IOIOHgsp IOHgAAK

222 775.0]2[355.0][3

22

xxAAK IOHgsp

)4(213122.0103.1 318 xx x 65 101.1

Consider a saturated solution of Hg2(IO3)2 Calculate the concentration of mercurous ions in: In Pure water = 6.9 x 106.9 x 10-7-7

in 0.1 M KNO0.1 M KNO33 = 1.1 = 1.155 x 10 x 10-6-6

pH revisited

Definition of pH

pH = -log AH

or pH = - log [H+] H

pH of pure water

H2O (l) H+ (aq) + OH-

(aq)Kw =1.0 x 10-14

Kw = AH AOH

Kw = [H+] H [OH-] OH

pH of pure water

H2O (l) H+ (aq) + OH-

(aq)Kw =1.0 x 10-14

Kw = AH AOH

Kw = [H+] H [OH-] OH

Kw = x2 x = 1.0 x 10-7 M

11

- -ICE

+x +x+x +x

pH of pure water

x = 1.0 x 10-7 M

Therefore [H+] = 1.0 x 10-7 M

pH = -log AH

= -log [H+] H

= - log [1.0 x 10-7]

= 7.00

1

pH of pure water containing salt

Calculate the pH of pure water containing 0.10 M KCl at 25oC.