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

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Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
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Transcript of Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Page 1: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Chapter 6 Problems

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

Page 2: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Outline Equilibrium of Acids and Bases

Bronsted-Lowry Acids/Bases Define strong Define weak

pH of pure water at 25oC Define Ka and Kb Relationship b/w Ka and Kb

Chapter 8 – Activity Relationship with K

Page 3: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Acids and Bases & EquilibriumEquilibrium

Section 6-7

Page 4: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Strong Bronsted-Lowry Acid

A strong Bronsted-Lowry Acid is one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor).

Page 5: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Strong Bronsted-Lowry Base

Accepts protons from water molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added.

Page 6: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
Page 7: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Question

Can you think of a salt that when dissolved in water is not an acid nor a base?

Page 8: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Weak Bronsted-Lowry acid

One that DOES not donate all of its acidic protons to water molecules in aqueous solution.

Example?

Page 9: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Weak Bronsted-Lowry base

Does NOT accept an amount of protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added.

example:

Page 10: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Common Classes of Weak Acids and Bases

Weak Acids

Weak Bases

Page 11: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Question: Question: Calculate the Concentration of H+ and OH- in Pure

water at 250C.

Equilibrium and Equilibrium and WaterWater

Page 12: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C.

H2O H+ + OH-

KW=

Kw =

Page 13: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
Page 14: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C.

H2O H+ + OH-

KW=

Kw =

Page 15: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Example

What is the concentration of OH- in a solution of water that is 1.0 x 10-3 M in [H+] (@ 25 oC)?

Kw = [H+][OH-]

1.0 x 10-14 = [1 x 10-3][OH-]

1.0 x 10-11 = [OH-]

“From now on, assume the temperature to be 25oC unless otherwise stated.”

Page 16: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

pH

~ -3 -----> ~ +16pH + pOH = - log Kw = pKw = 14.00

Page 17: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Weak Acids and Bases

HA H+ + A-

HA + H2O(l) H3O+ + A-

Ka

Page 18: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Weak Acids and Bases

B + H2O BH+ + OH-Kb

Page 19: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Relation Between KRelation Between Kaa and and KKbb

Page 20: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Relation between Ka and Kb

Consider Ammonia and its conjugate acid.

NH3 + H2O NH4+ + OH-

Kb

NH4+ + H2O NH3 + H3O+

Ka

Page 21: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Example

The Ka for acetic acid is 1.75 x 10-5. Find Kb for its conjugate base.

Page 22: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Example Calculate the hydroxide ion concentration in

a 0.0100 M sodium hypochlorite solution.OCl- + H2O HOCl + OH-

The acid dissociation constant = 3.0 x 10-8

][

]][[

OCl

OHHOClKb

Page 23: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

1st Insurance Problem

Challenge on page 120Challenge on page 120

Page 24: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Chapter 8

ActivityActivity

Page 25: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Write out the equilibrium constant for the following expression

Fe3+ + SCN- Fe(SCN)2+

Q: What happens to K when we add, say KNO3 ?

]][[

])([3

2

SCNFe

SCNFeK

Page 26: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

K decreases when an inert salt is added!!! Why?

Keq

Page 27: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

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

Page 28: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Increased solubility

Why? LeChatelier’s Principle?

Complex Ion?

?

Page 29: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
Page 30: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

The potassium hydrogen tartrate example

K+-O

O

OH

OH

O

OH

potassium hydrogen tartrate

Page 31: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Alright, what do we mean by Ionic strength?

Ionic strength is dependent on the number of ions in solution and their charge.

Ionic strength () = ½ (c1z12+ c2z2

2 + …)

Or Ionic strength (m) = ½ cizi2

Page 32: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

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 + …)

Page 33: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

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

Activity = Ac = [C]c

AND

bB

baA

a

dD

dcC

c

bB

aA

dD

cC

BA

DC

AA

AAK

][][

][][

Page 34: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Relationship between activity 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

Anyway … we generally don’t need to calculate – can get it from a table

Page 35: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

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

Page 36: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Relationship between and

Page 37: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
Page 38: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
Page 39: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Back to our original problem

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

Page 40: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Back to our original problem

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?

Page 41: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,

Back to our original problem

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

Page 42: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,
Page 43: Chapter 6 Problems 6-29, 6-31, 6-39, 6.41, 6-42, 6-48,