chapter 6 problems 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24 comments on “lab report & pop...

Chapter 6 Problems

6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24

Comments on “Lab Report & Pop Rocks”

6.6

From the equationsHOCl H+ + OCl- K = 3.0 x 10-8

HOCl + OBr- HOBr + OCl- K= 15

Find K for

HOBr H+ + OBr—

HOBr + OCl- HOCl + OBr-

Flip

K’ = 1/K= 1/15

K= ?K= 3.0 x 10-8/15

K= 0.2 x 10-8

6.9 The formation of Tetrafluorethlene

from its elements is highly exothermic. 2 F2 (g) + 2 C (s) F2C=CF2 (g)

(a) if a mixture of F2, graphite, and C2F4 is at equilibrium in a closed container, will the reaction go to the right or to the left if F2 is added?

(b) Rare bacteria … eat C2F4 and make teflon for cell walls. Will the reaction go to the right or to the left if these bacteria are added?

6.9

The formation of Tetrafluorethlene from its elements is highly exothermic. 2 F2 (g) + 2 C (s) F2C=CF2 (g)

(C) will the reaction go to the right or to the left if graphite is added?

(d) will reaction go left or right if container is crushed to one-eighths of original volume?

(e) Does “Q” get larger or smaller if vessel is Heated?

6-15. What concentration of Fe(CN)6

4- is in equilibrium with 1.0 uM Ag+ and Ag4Fe(CN)6 (s).

Ag4Fe(CN)6 4Ag+ + Fe(CN)64-

Ksp = [Ag+]4 [Fe(CN)64-]

8.5 x 10-45 = [1.0 x 10-6]4 [Fe(CN)64-]

[Fe(CN)64-] = 8.5 x 10-21 M = 8.5 zM

6-16.

Cu4(OH)6(SO4) 4 Cu2+ + 6OH- + SO42-

I’d first set up an ICE table:

Cu4(OH)6(SO4) 4 Cu2+ + 6 OH + SO42-

I Some - 1.0 x 10-6 M -C -x + 4x Fixed at 1.0 x 10-6 M + xE Some –x + 4x Fixed at 1.0 x 10-6 M + x

Ksp = [Cu2+]4 [OH-]6 [SO42-] = 2.3 x 10-69

Ksp = [4x]4 [1.0 x 10-6]6 [x] = 2.3 x 10-69

X = 9.75 x 10-8 M Cu2+ = 4x = 3.90 x 10-7 M

Chapter 6

Chemical Equilibrium

Chemical Equilibrium

Equilibrium Constant Solubility product (Ksp) Common Ion Effect Separation by precipitation Complex formation

Separation by Precipitation


Complete separation can mean a lot … we should define complete.

Complete means that the concentration of the less soluble material has decreased to 1 X 10-

6M or lower before the more soluble material begins to precipitate


EXAMPLE: Can Fe+3 and Mg+2 be separated quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH- concentrations is permissible.

Fe3+

Fe3+

Fe3+

Fe3+

Fe3+

Fe3+

Fe3+

Fe3+

Fe3+

Mg2+

Mg2+

Mg2+

Mg2+Mg2+

Mg2+Mg2+

Fe3+

Mg2+

Mg2+

Fe3+Fe3+

Mg2+

Mg2+

Add OHAdd OH--

Fe3+

Mg2+

Mg2+

Mg2+

Mg2+Mg2+

Mg2+Mg2+

Mg2+

Mg2+ Mg2+

Mg2+

Fe(OH)Fe(OH)33(s)(s)

What is the [OH-] when this happens

^

@ equilibrium

EXAMPLE: Separate Iron and Magnesium?

Ksp = [Fe+3][OH-]3 = 2 X 10-39

Ksp = [Mg+2][OH-]2 = 7.1 X 10-12

Assume [Fe+3]eq = 1.0 X 10-6M when “completely” precipitated.

What will be the [OH-] @ equilibrium required to reduce the [Fe+3] to [Fe+3] = 1.0 X 10-6M ?Ksp = [Fe+3][OH-]3 = 2 X 10-39


Ksp = [Fe+3][OH-]3 = 2 X 10-39

(1.0 X 10-6M)*[OH-]3 = 2 X 10-39

333 102][ OH

11103.1][ OH

Dealing with Mg2+

Find [OH-] to start precipitating Mg2+

• Conceptually – •Will assume a minimal amount of Mg2+ will precipitate and determine the respective concentration of OH-

Evaluate Q• If

•Q>K•Q<K•Q=K

“Left”“Right”“Equilibrium”

Fe3+

Mg2+

Mg2+

Mg2+

Mg2+Mg2+

Mg2+Mg2+

Mg2+

Mg2+ Mg2+

Mg2+

Fe(OH)Fe(OH)33(s)(s)

[OH-]̂@ equilibrium

Is this [OH-] (that is in solution) great enough to start precipitating Mg2+?

= 1.3 x 10-11


What [OH-] is required to begin the precipitation of Mg(OH)2?

[Mg+2] = 0.10 M

Ksp =

[OH-] = 8.4 X 10-6M

[Mg2+]eq = 0.09999999999999999 M

Really, Really close to 0.1 M

(0.10 M)[OH-]2 = 7.1 X 10-12


[OH-] to ‘completely’ remove Fe3+ = 1.3 X 10-11 M

[OH-] to start removing Mg2+ = 8.4 X 10-6M

“All” of the Iron will be precipitated b/f any of the magnesium starts to precipitate!!

^̂

@ equilibrium@ equilibrium


Q vs. K

Ksp = [Mg2+][OH-]2 = 7.1 X 10-12

Q = [0.10 M ][1.3 x 10-11 ]2 = 1.69 x 10-23

Q<K Reaction will proceed to “Right”

Mg(OH)2(s) Mg2+ + 2OH-

Dealing with Mg2+

Find [OH-] to start precipitating Mg2+

• Conceptually – •Will assume a minimal amount of Mg2+ will precipitate and determine the respective concentration of OH-

Evaluate Q• If

•Q>K•Q<K•Q=K

“Left”“Right”“Equilibrium”

NO PPT

“Real Example” Consider a 1 liter solution that

contains 0.3 M Ca2+ and 0.5 M Ba2+. Can you separate the ions by adding

Sodium Carbonate? Sodium Chromate ? Sodium Fluoride? Sodium Hydroxide? Sodium Iodate? Sodium Oxylate?

An example

Consider Lead Iodide PbI2 (s) Pb2+ + 2I- Ksp

= 7.9 x 10-9

What should happen if I- is added to a solution?

Should the solubility go up or down?

Complex Ion Formation

Complex Formation

complex ions (also called coordination ions)

Lewis Acids and Basesacid => electron pair acceptor (metal)base => electron pair donor (ligand)

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

I- I-

Pb2+ I-

I-

I-

I-

I-

I-

Pb2+

Pb2+

I-

I- I-

I-

I- I-

Pb2+ I- I- Pb2+

I- I-

Pb2+

I-

I-

I- I-

I- I-

I-

I- I-

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

I- I-

Pb2+

I-

I-

I- I-

I- I-

I-

I- I-

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

I-

I- I-

I-

I- I-

I-

I-

Pb2+ I- I- Pb2+

I- I-

Pb2+

I-

I-

I-

I-

I- I-

I-

I-

I-

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

Pb2+ I- I- Pb2+

I-

I-

I-

I-

I- I-

I-

I-

Effects of Complex Ion Formation on Solubility

Consider the addition of I- to a solution of Pb+2 ions

Pb2+ + I- <=> PbI+

PbI+ + I- <=> PbI2 K2 = 1.4 x 101

PbI2 + I- <=> PbI3- K3 =5.9

PbI3+ I- <=> PbI42- K4 = 3.6

221 100.1

]][[

][x

IPb

PbIK

Effects of Complex Ion Formation on Solubility

Consider the addition of I- to a solution of Pb+2 ions

Pb2+ + I- <=> PbI+

PbI+ + I- <=> PbI2 K2 = 1.4 x 101

Pb2+ + 2I- <=> PbI2 K’ =?

221 100.1

]][[

][x

IPb

PbIK

Overall constants are designated with

This one is

Protic Acids and Bases

Section 6-7

Question

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

Can you think of a salt that when dissolved in water IS an acid or base?

Protic Acids and Bases - Salts

Consider Ammonium chloride Can ‘generally be thought of as the

product of an acid-base reaction.

NH4+Cl- (s) NH4

+ + Cl-

From general chemistry – single positive and single negative charges are STRONG ELECTROLYTES – they dissolve completely into ions in dilute aqueous solution

Protic Acids and Bases

Conjugate Acids and Bases in the B-L concept

CH3COOH + H2O CH3COO- + H3O+ acid + base <=> conjugate base + conjugate acid

conjugate base => what remains after a B-L acid donates its protonconjugate acid => what is formed when a B-L base accepts a proton

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

water at 250C.

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

H2O H+ + OH-

Initial liquid - -

Change -x +x +x

Equilibrium Liquid-x +x +x

KW=(X)(X) = 1.01 X 10-14

Kw = [H+][OH-] = 1.01 X 10-14

(X) = 1.00 X 10-7

Example

Concentration of OH-

if [H+] is 1.0 x 10-3 M @ 25 oC?

Kw = [H+][OH-]

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

1 x 10-11 = [OH-]

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

pH

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

Is there such a thing as Pure Water? In most labs the answer is NO Why?

A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value.

CO2 + H2O HCO3- + H+

6-9 Strengths of Acids and 6-9 Strengths of Acids and BasesBases

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). HCl as example

Strong Bronsted-Lowry Base

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

Example: NH2- (the amide ion)

Weak Bronsted-Lowry acid

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

Example? Use of double arrows! Said to reach

equilibrium.

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: NH3

Common Classes of Weak Acids and Bases

Weak Acids carboxylic acids ammonium ions

Weak Bases amines carboxylate anion

Weak Acids and Bases

HA H+ + A-

HA + H2O(l) H3O+ + A-

Ka

][

]][[

HA

AHKa

][

]][[ 3

HA

AOHKa

Ka’s ARE THE SAME

Weak Acids and Bases

B + H2O BH+ + OH-

][

]][[

B

OHBHKb

Kb

Relation Between KRelation Between Kaa and and KKbb

Relation between Ka and Kb

Consider Ammonia and its conjugate base.

][

]][[

4

33

NH

OHNHKb

][

]][[

3

4

NH

OHNHKa

NH3 + H2O NH4+ + OH-

Ka

NH4+ + H2O NH3 + H3O+

Kb

H2O + H2O OH- + H3O+

][

]][[

][

]][[

3

4

4

33

NH

OHNH

NH

OHNHK

][][ 3 OHOHKw

Example

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

Kw = Ka x Kb

a

wb K

KK

105

14

107.51075.1

100.1

bK

1st Insurance Problem

Challenge on page 120Challenge on page 120

chapter 6 problems 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24 comments on “lab report & pop...

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