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Announcements Homework –

Chapter 4 8, 11, 13, 17, 19, 22

Chapter 6 6, 9, 14, 15

Exam Thursday

4-8

Meaning of Confidence interval? Is an interval around the

experimental mean that most likely contains the true mean ().

Homework

4.11

814.0x 403.0s

6

)03.0)(015.2(0.14 :confidence %90 4

8

88 02.00.14 :confidence %90

68 05.00.14 :confidence %95

Question 4-13.

4-13. A trainee in a medical lab will be released to work on her own when her results agree with those of an experienced worker at the 95% confidence interval. Results for a blood urea nitrogen analysis are shown ….

a) What does abbreviation dL refer to?dL = deciliter = 0.1 L = 100 mL

b) Should the trainee work alone?

Comparison of Means with Student’s t

Is there a significantIs there a significant difference?difference?

First you must ask, First you must ask, is there a significant difference in is there a significant difference in their standard deviations?their standard deviations?

21

2121

nn

nn

s

xxt

pooledcalculated

NONO YESYES

2

22

1

21

21

ns

ns

xxtcalculated

f-testf-test

4-13. dL = deciliter = 0.1 L = 100 mL

dLmgx /5.14 7 dLmgs /5.0 3 6n

dLmgx /9.13 5 dLmgs /4.0 2 5n

9

2

2

3 5.104.0

05.0

calcF Ftable = 6.26 No difference

Find spooled and t

484.0256

)4(42.0)5(53.0 22

pooleds

12.256

)5)(6(

484.0

9.135.14 57

calct

ttable = 2.262

No significant difference between two workers … Therefore trainee should be “Released”

Homework

4-17. If you measure a quantity four times and the standard deviation is 1.0 % of the average, can you be 90 % confident that the true value is within 1.2% of the measured average

%1.1x4

%)0.1)(353.2(x 8 Yes

Homework

4-19. Hydrocarbons in the cab of an automobile … Do the results differ at 95% CL? 99% CL?

3

2

01.18.29

0.30

calcF

3/0.304.31 mgx 32n3/8.299.52 mgx 32n

Ftable ~ 1.84 No DifferenceNo Difference

Find spooled and t

Homework

9.2923232

)31(8.29)31(0.30 22

pooleds

88.23232

)32)(32(

9.29

4.319.52

calct

The table gives t for 60 degrees of freedom, which is close to 62.ttable = 1.671 and 2.000 at the 90 and 95% CL, respectively.

The difference IS significant at both confidence levels.

4-22. Q-test, Is 216 rejectable? 192, 216, 202, 195, 204

50.0192216

204216

Q

Qtable = 0.64

Retain the “outlier” 216

Chapter 6

Chemical Equilibrium

Chemical Equilibrium Equilibrium Constant Equilibrium and Thermodynamics

Enthalpy Entropy Free Energy Le Chatelier’s Principle

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

Example

The equilibrium constant for the reaction

H2O H+ + OH-

Kw = 1.0 x 10-14

NH3 + H2O NH4+ + OH- KNH3 = 1.8 x 10-5

Find the Equilibrium constant for the following reaction

NH4+ NH3 + H+ K3 = ?

Equilibrium and Thermodynamics

A brief review …


enthalpy => Henthalpy change => H

exothermic vs. endothermicentropy => Sfree energy

Gibbs free energy => GGibbs free energy change => G


Go = Ho - TSo

Go = -RT ln (K)

K = e-(Go/RT)


The case of HCl

HCl H+ + Cl-

Ho = -74.83 x 103 J/molS0 = -130.4 kJ/mol

Go = Ho - TSo

Go = (-74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol)Go = -35.97 kJ/mol

K=?K=?


The case of HCl

HCl H+ + Cl-

Go = (-74.83 kJ/mol) – (298.15 K) (-130.4 kJ/mol)Go = -35.97 kJ/mol

K=?K=?

)15.298](314472.8[

/1097.35 3

KmolK

JmolJx

eK

61000.2 x

Predicting the direction in which an equilibrium will initially move

LeChatelier’s Principle and Reaction Quotient

Le Chatelier's Principle

If a stress, such as a change in concentration, pressure, temperature, etc., is applied to a system at equilibrium, the equilibrium will shift in such a way as to lessen the effect of the stress.

Stresses – Adding or removing reactants or products Changing system equilibrium temperature Changing pressure (depends on how the

change is accomplished

Consider

6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)

Predict in which direction the equilibrium moves as a result of the following stress:

Increasing [CO2]

Equilibrium moves Right

Consider

6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)


Increasing [O2]

Equilibrium moves Left

Consider

6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)


Decreasing [H2O]

Equilibrium moves Left

Consider

6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)


Removing C6H12O6(s)

NO CHANGE

62

62

62

1 ][][

]0[

OHCOK

K does not depend on concentration of solidsolid C6H12O6

Consider

6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)


Compressing the system

System shifts towards the direction which occupies the smallest volume. Fewest moles of gas.

Equilibrium moves Right

Consider

6CO2 (g) + 6 H2O(g) C6H12O6(s) + 6O2(g)


Increasing system temperature

System is endothermic … heat must go into the system (think of it as a reactant)

Equilibrium moves RightH = + 2816 kJ

Consider this

CoCl2 (g) Co (g) + Cl2(g)

When [COCl2] is 3.5 x 10-3 M, [CO] is 1.1 x 10-5 M, and [Cl2] is 3.25 x 10-6M is the system at equilibrium?

Q= Reaction quotient

K=2.19 x 10-10

2

2 ]][[

CoCl

ClCoQ

83

65

1002.1105.3

]1025.3][101.1[

xx

xx

Compare Q and K

Q = 1.02 x 10-8

K = 2.19 x 10-10

System is not at equilibrium, if it were the ratio would be 2.19x10-10

When

Q>K TOO MUCH PRODUCT TO BE AT EQUILIRBIUMEquilibrium moves to the left

Q<K TOO MUCH REACTANT TO BE AT EQUILIRBIUMEquilibrium moves to the Right

Q=K System is at EquilibriumSystem is at Equilibrium

Solubility Product

Introduction to Ksp

Solubility Product

solubility-productthe product of the solubilities

solubility-product constant => Ksp

constant that is equal to the solubilities of the ions produced when a substance dissolves

Solubility Product

In General:AxBy <=> xA+y + yB-x

[A+y]x [B-x]y

K = ------------ [AxBy]

[AxBy] K = Ksp = [A+y]x [B-x]y

Solubility Product

For silver sulfateAg2SO4 (s) <=> 2 Ag+

(aq) + SO4-

2(aq)

Ksp = [Ag+]2[SO4-2]

Solubility of a Precipitatein Pure WaterEXAMPLE: How many grams of AgCl

(fw = 143.32) can be dissolved in 100. mL of water at 25oC?AgCl <=> Ag+ + Cl-

Ksp = [Ag+][Cl-] = 1.82 X 10-10 (Appen. F)

let x = molar solubility = [Ag+] = [Cl-]

EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC?

AgCl(s) Ag+ (aq) + Cl- (aq)

Initial Some - -

Change -x +x +x

Equilibrium -x +x +x

(x)(x) = Ksp = [Ag+][Cl-] = 1.82 X 10-10

x = 1.35 X 10-5M

EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC?

How many grams is that in 100 ml?

# grams = (M.W.) (Volume) (Molarity) = 143.32 g mol-1 (.100 L) (1.35 x 10-5 mol

L-1) = 1.93X10-4 g = 0.193 mg

x = 1.35 X 10-5M

The Common Ion Effect


common ion effect a salt will be less soluble if one of

its constituent ions is already present in the solution


EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3.

Ag2CO3 <=> 2 Ag+ + CO3-2

Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12


Ag2CO3 <=> 2 Ag+ + CO3-2

Initial Solid - 0.0200M

Change -x +2x +x

Equilibrium Solid +2x 0.0200+x

Ksp=(2x)2(0.0200M + x) = 8.1 X 10-12

Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12

4x2(0.0200M + x) = 8.1 X 10-12


4x2(0.0200M + x) = 8.1 X 10-12

no exact solution to a 3rd order equation, need to make some approximationfirst, assume the X is very small

compared to 0.0200 M

4X2(0.0200M) = 8.1 X 10-12

4X2(0.0200M) = 8.1 X 10-12

X= 1.0 X 10-5 M


X = 1.0 X 10-5 M(1.3 X 10-4 M in pure water)

Second check assumption

[CO3-2] = 0.0200 M + X ~ 0.0200 M

0.0200 M + 0.00001M ~ 0.0200M

Assumption is ok!

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.

Two competing reactionsFe(OH)3(s) Fe3+ + 3OH-

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

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 quantitative separation requires that the concentration of the less soluble material to have decreased to < 1 X 10-6M before the more soluble material begins to precipitate.


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

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

Assume [Fe+3] = 1.0 X 10-6M

What will be the [OH-] 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

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

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


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


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

[Mg+2] = 0.10 M

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

[OH-] = 8.4 X 10-6M


[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

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