che 116 prof. t. l. heise 1 che 116: general chemistry u chapter sixteen copyright © tyna l. heise...

CHE116Prof. T. L. Heise

1CHE 116: General Chemistry

CHAPTER SIXTEEN

Copyright © Tyna L. Heise 2001-2002

All Rights Reserved


2Acids and Bases: Review

Properties of Acidssour tastechange with litmus

Properties of Bases

bitter taste

change with litmus


3Acids and Bases: Review

1830 - scientists have recognized that all acids contain hydrogen, but not all hydrogen bearing compounds are acids

Svante Arrhenius - linked acid behavior with the presence of an H+ and base behavior with the presence of an OH-


4Bronsted-Lowry Acids and Bases

Arrhenius’ definition is useful, but restricts acid base reactions to aqueous conditions

Johannes Bronsted and Thomas Lowry proposed a more general definition which involves the transfer of an H+ ion from one molecule to another



H+ ion is simply a proton with no surrounding valence electron.Small particle interacts strongly with the nonbonding pairs of

water molecules to form hydrated hydrogen ionschemists use H+ and H3O+ interchangeably



Fig 16.1

Demonstrates the interconnections possible between hydrogenated water



Definitions:

Acid - any compound which transfers an H+ to another molecule

Base - any compound which accepts a transfer of an H+ from another molecule

* an acid and base always work together



Definitions:

Amphoteric - some substances can be an acid or a base depending on reaction

Conjugate Acid Base pairs - two compounds that differ only in the presence of an H+. The molecule with the extra H is the acid.



Fig 16.7, 16.8



Sample Exercise: Write the formula for the conjugate acid of each of the following:

HSO3-

F-

PO43-

CO




HSO3-

F-

PO43-

CO

Given a base, Given a base, bases accept bases accept HH++, so add an , so add an HH++ to each to each molecule.molecule.




H2SO3 HSO3-

F-

PO43-

CO





H2SO3 HSO3-

HF F-

PO43-

CO





H2SO3 HSO3-

HF F-

HPO42- PO4

3-

CO





H2SO3 HSO3-

HF F-

HPO42- PO4

3-

HCO+ CO




Sample Exercise: When lithium oxide, Li2O, is dissolved in water, the solution turns basic from the reaction of the oxide ion, O2-, with water. Write the reaction that occurs, and identify the conjugate acid base pairs.




O2- + H2O OH- + OH-




O2- + H2O OH- + OH-

acidacid base base

basebase acidacid



Relative Strengths of Acids and Bases:the more readily a substance donates an H+, the less readily it’s

conjugate base will accept one the more readily a substance accepts an H+, the less readily it’s

conjugate acid will donate onethe stronger one of the substances is, the weaker it’s conjugate



Relative Strengths of Acids and Bases:strong acids completely transfer their protons to water, leaving no

undissociated molecules weak acids are those that only partly dissociate in aqueous

solution and therefore exist in the solution as a mixture of acid molecules and component ions



Relative Strengths of Acids and Bases:negligible acids are those that have hydrogen but do not donate them

at all, their conjugate bases would be extremely strong, reacting with water to complete their octet and leaving OH- behind.

In every acid base reaction, the position of the equilibrium favors transfer of H+ from stronger side to weaker side



Fig. 16.4



Sample exercise: For each of the following reactions, use Fig. 16.4 to predict whether the equilibrium lies predominantly to the left or right:

a) PO43-(aq)+H2O(l)HPO4

2-(aq)+OH-(aq)

b) NH4+(aq)+OH-(aq) NH3(aq)+H2O(l)





2-(aq)+OH-(aq)

2 acids are: H2O and HPO42-

2 bases are: PO43- and OH-





2-(aq)+OH-(aq)



red indicates strength





2-(aq)+OH-(aq)



reverse reaction favored





2-(aq)+OH-(aq)



shifts left





2 acids are: NH4+ and H2O

2 bases are: NH3 and OH-







red indicates strength







favors forward reaction







shifts right


34The Autoionization of Water

One of the most important properties of water is its ability to ac as either a Bronsted acid or Bronsted base, depending on circumstances.One water molecule can donate a proton to another

water molecule Fig 16.10



The autoionization of water, although rapid and weak, does exist as an equilibrium, and therefore has an equilibrium constant expression:

Keq = [H3O+][OH-] [H2O]2

* because water is a liquid, it can be excluded from the equation...



Keq[H2O]2 = [H3O+][OH-]

Kw = [H3O+][OH-] = 1.0 x 10-14

* this equation is not only applicable to water, but to all aqueous solutions, and it is upon this fact that the pH scale was built.



Sample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:

a) [H+] = 2 x 10-5

b) [OH-] = 3 x 10-9

c) [OH-] = 1 x 10-7




a) [H+] = 2 x 10-5 then [OH-] must equal 1.0 x 10-14 which is 2 x10-5

[OH-] = 5.0 x 10-10




a) [H+] = 2 x 10-5 then [OH-] must equal 1.0 x 10-14 which is 2 x10-5

[OH-] = 5.0 x 10-10

[H+] > [OH-] so acidic


40The Autoionization of WaterSample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:

b) [OH-] = 3 x 10-9 then [H+] must equal 1.0 x 10-14 which is 3 x10-9

[H+] = 3.3 x 10-6


41The Autoionization of WaterSample exercise: Indicate whether each of the following solutions is neutral, acidic, or basic:

b) [OH-] = 3 x 10-9 then [H+] must equal 1.0 x 10-14 which is 3 x10-9

[H+] = 3.3 x 10-6

[H+] > [OH-] so acidic




c) [OH-] = 1 x 10-7 then [H+] must equal 1.0 x 10-14 which is 1 x10-7

[H+] = 1.0 x 10-7




c) [OH-] = 1 x 10-7 then [H+] must equal 1.0 x 10-14 which is 1 x10-7

[H+] = 1.0 x 10-7

[H+] = [OH-] so neutral


44The pH Scale

For convenience, we can use a logarithmic version of concentration to turn the very small concentrations of [H+] and [OH-] into whole numbers.

p(anything) = - log[anything]

p(H) = -log[H+]

p(OH) = - log[OH-]

** pH + pOH = 14


45The pH Scale


46The pH Scale

Common household products and their relative pH’s.


47The pH Scale

Sample exercise: In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH?


48The pH Scale


pH = -log[H+]


49The pH Scale


pH = -log[H+]

pH = -log[3.8 x 10-4]


50The pH Scale


pH = -log[H+]

pH = -log[3.8 x 10-4]

pH = 3.42


51The pH Scale

Sample exercise: A commonly available window-cleaning solution has a [H+] is 5.3 x 10-9 M. What is the pH?


52The pH Scale


pH = -log[H+]


53The pH Scale


pH = -log[H+]

pH = -log[5.3 x 10-9]


54The pH Scale


pH = -log[H+]

pH = -log[5.3 x 10-9]

pH = 8.28


55The pH Scale

Sample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+].


56The pH Scale


pH = -log[H+]


57The pH Scale


pH = -log[H+]

9.18 = -log[H+]


58The pH Scale


pH = -log[H+]

9.18 = -log[H+]

10-9.18 = [H+]


59The pH ScaleSample exercise: A solution formed by dissolving an antacid tablet has a pH of 9.18.

Calculate [H+].

pH = -log[H+]

9.18 = -log[H+]

10-9.18 = [H+]

6.61 x 10-10 = [H+]


60The pH Scale

Measuring pH: a pH can be measured quickly and accurately using a pH meter.A pair of electrodes connected to a meter capable of measuring small voltagesa voltage which varies with pH is generated when the electrodes are placed in

a solutioncalibrated to give pH


61The pH Scale

Measuring pH: a pH can be measured quickly and accurately using a pH meter.Electrodes come in a variety of shapes and sizesa set of electrodes exists that can be placed inside a human cellacid base indicators can be used, but are much less precise


62The pH Scale

Fig 16.7


63Strong Acids and Bases

Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions. Strong Acids HCl

HBr HImonoprotic HNO3 HClO3

HClO4 H2SO4

diprotic



Strong acids and bases are strong electrolytes, existing in aqueous solution entirely as ions.

Calculating the pH of a solution made up entirely of ions means the [H+] is proportional to [acid]



An aqueous solution of HNO3 has a pH of 2.66. What is the concentration of the acid?




pH = -log[H+]




pH = -log[H+] 2.66 = -log[H+]




pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+]

2.2 x 10-3 = [H+]




pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+] 2.2 x 10-3 = [H+] 2.2 x 10-3 M H+ 1 mole HNO3 1 mole H+




pH = -log[H+] 2.66 = -log[H+] 10-2.66 = [H+]2.2 x 10-3 = [H+] 2.2 x 10-3 M H+ 1 mole HNO3

1 mole H+ 2.2 x 10-3 HNO3



The most soluble common bases are the ionic hydroxides of the alkali and alkaline earth metals.

Due to the complete dissociation of the base into its ion components makes the pH calculation equally straightforward



Sample exercise: What is the concentration of a solution of KOH for which the pH is 11.89?




pH = -log[H+]




pH = -log[H+]11.89 = -log[H+]




pH = -log[H+] 11.89 = -log[H+]10-11.89 = [H+]




pH = -log[H+]11.89 = -log[H+] 10-11.89 = [H+]

1.29 x 10-12 = [H+]




pH = -log[H+] 11.89 = -log[H+]10-11.89 = [H+] 1.29 x 10-12 = [H+]1.0 x 10-14 = [OH-] 1.29 x 10-12




pH = -log[H+] 11.89 = -log[H+]10-11.89 = [H+] 1.29 x 10-12 = [H+]1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12




1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12

7.8 x 10-3 M OH- 1 mol KOH 1 mol OH-




1.0 x 10-14 = [OH-] = 7.8 x 10-3 1.29 x 10-12

7.8 x 10-3 M OH- 1 mol KOH 1 mol OH- 7.8 x 10-3 M KOH


81Weak Acids

Most acids are weak acids and only partially dissociate in aqueous solution.

The extent to which a weak acid dissociates can be expressed using an equilibrium constant for the ionization reaction HX + H2O H3O+ + X- Ka = [H3O+][X-] [HX] *the larger the Ka the stronger the acid


82Weak Acids

Calculating Ka from pH: a much more complicated calculation is required for the determinations of weak acids, in many cases, due to the extremely small magnitude of the values, some simpler approximations can be made.


83Weak Acids

Sample exercise: Niacin, one of the B vitamins, has the following molecular structure: C O H O N

A 0.020 M solution of niacin has a pH of 3.26

a) What % of the acid is ionized in this solution?

b) What is the acid-dissociation constant?


84Weak Acids

Sample exercise: A 0.020 M solution of niacin has a pH of 3.26

a) What % of the acid is ionized in this solution? pH = -log[H+] 3.26 = -log[H+] e-3.26 = [H+] 5.5 x 10-4 = [H+]


85Weak Acids


a) What % of the acid is ionized in this solution?C6H4NOOH C6H4NOO- + H+

initial 0.020 0 0

change -5.5 x 10-4 +5.5 x 10-4 +5.5 x 10-4

equil 0.01945 5.5 x 10-4 5.5 x 10-4


86Weak Acids


a) What % of the acid is ionized in this solution?% = part/total x 100

= 5.5 x 10-4/0.01945 x 100

= 2.8%


87Weak Acids


b) What is the acid-dissociation constant?

C6H4NOOH C6H4NOO- + H+

Ka = [C6H4NOO- ][H+] [C6H4NOOH] = (5.5 x 10-4 )(5.5 x 10-4 ) 0.01945 = 1.55 x10-5


88Weak Acids

Using Ka to calculate pH:

Using the value of Ka and knowing the initial concentration of the weak acid, we can calculate the concentration of H+(aq).

Example: Calculate the pH of a 0.30 M solution of acetic acid.

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)


89Weak Acids

Example: Calculate the pH of a 0.30 M solution of acetic acid.

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)

From Table 16.2, Ka = 1.8 x 10-5

Ka = 1.8 x 10-5 = [H+][C2H3O2-] [HC2H3O2]

set up data table of concentrations involved...


90Weak Acids

Ka = 1.8 x 10-5 = [H+][C2H3O2-] [HC2H3O2]

set up data table of concentrations involved…

[HC2H3O2] [H+] [C2H3O2-]

Initial 0.30 M 0 0

Change -x +x +x

Equilibrium 0.30 - x x x


91Weak Acids

Input concentrations in formula

Ka = 1.8 x 10-5 = [x][x] [0.30 -x]

This will lead to a quadratic equation, but we can simplify the problem a bit. All weak acids dissociate so little that we can assume the initial concentration of the acid remains essentially the same, that means we can rewrite the formula to read...


92Weak AcidsKa = 1.8 x 10-5 = [x][x] [0.30]

1.8 x 10-5 (0.30) = x2

1.8 x 10-5 (0.30) = x

0.0023 = x

0.0023 = [H+] pH = -log [H+]

= -log[0.0023]

= 2.64


93Weak Acids

Sample Exercise: The Ka for niacin is 1.5 x 10-5. What is the pH of a 0.010 M solution of niacin?


94Weak Acids



Ka = 1.5 x10-5 = [C6H4NOO-][H+] [C6H4NOOH]


95Weak Acids


C6H4NOOH C6H4NOO- H+

initial 0.010 0 0 change -x +x +x equilibrium 0.010 -x x x


96Weak Acids



Ka = 1.5 x10-5 = [x][x] [0.010]


97Weak Acids


1.5 x10-5 [0.010] = x2

1.5 x10-5 [0.010] = x

3.87 x 10-4 = x


98Weak Acids


3.87 x 10-4 = x

3.87 x 10-4 = [H+] pH = -log[H+] pH = -log[3.87 x 10-4] pH = 3.41


99Weak Acids

The results seen in the two previous examples are typical of weak acids. The lower number of ions produced during the partial dissociation causes less electrical conductivity and a slower reaction rate with metals.

Percent ionization is a good way to discover the actual conductivity, however...


100Weak Acids

As the concentration of a weak acid increases, the % ionized decreases.


101Weak Acids

As the concentration of a weak acid increases, the % ionized decreases.


102Weak Acids

Sample Exercise: Calculate the % of niacin molecules ionized in

a) the previous exerciseb) a 1.0 x 10-3 M solution


103Weak Acids


a) the previous exercise

Our approximation was good so% = part x 100 total

= 3.87 x 10-4 x 100 = 3.9% 0.010


104Weak Acids


b) a 1.0 x 10-3 M solution

Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3]

1.5 x10-5 (1.0 x 10-3) = x2

1.2 x 10-4


105Weak Acids



Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3] but, 1.2 x 10-3 x 100

1.5 x10-5 (1.0 x 10-3) = x2 1.0 x 10-3

1.2 x 10-3 is greater than 5% so use quadratic...


106Weak Acids



Ka = 1.5 x10-5 = [x][x] [1.0 x 10-3 - x]

-1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0


107Weak Acids



-1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0

x = -b ± b2 - 4ac 2a

x =


108Weak Acids



-1.5 x 10-5(1.0 x10-3) + 1.5 x 10-5x + x2 = 0

x = -b ± b2 - 4ac 2a

x = 1.1 x 10-4 or -1.3 x 10-4


109Weak Acids



x = 1.1 x 10-4

1.1 x 10-4 x 1001.0 x 10-3


110Weak Acids



x = 1.1 x 10-4

1.1 x 10-4 x 100 = 11%1.0 x 10-3


111Weak Acids

Polyprotic Acids: many acids have more than one ionizable H atom.

H2SO3(aq) H+(aq) + HSO3-(aq) Ka1

HSO3-(aq) H+(aq) + SO3

-2(aq) Ka2

The Ka are labeled according to which proton is dissociating.

-it is always easier to remove the first proton than the second


112Weak Acids


113Weak Acids

If Ka values differ by 103 or more, only use Ka1 to determine calculations.


114Weak Acids

Sample Exercise: Calculate the pH and concentration of oxalate ion, [C2O4

2-] in a 0.020 M solution of oxalic acid.


115Weak Acids



H2C2O4 HC2O4- + H+

Ka = 5.9 x 10-2 = [HC2O4- ][H+] [H2C2O4]


116Weak Acids



H2C2O4 HC2O4- + H+

strong acid so 100% dissociation

0.020 M H+


117Weak Acids



HC2O4- C2O4

-2 + H+

I 0.020 0 0.020 -x +x +x E 0.020 - x x 0.020 + x


118Weak Acids



Ka = 6.4 x 10-5 = [0.020 + x ][x] [0.020 - x]

use your assumption


119Weak Acids



Ka = 6.4 x 10-5 = [0.020 ][x] [0.020]

6.4 x 10-5 = x = [C2O42-]


120Weak Acids



[H+] = 0.020


121Weak Acids



pH = -log[H+]

pH = - log[0.020]

pH = 1.70


122Weak Bases

Many substances behave as weak bases in water. Such substances react with water, removing protons from water, leaving the OH- ion behind.

NH3 + H2O NH4+ + OH-

Kb = [NH4+][OH-] [NH3]

* Kb is the base dissociation constant utilizing the [OH-]


123Weak Bases

* Kb is the base dissociation constant utilizing the [OH-]bases must contain one or more lone pair to bond with the H+

from water.as before, the larger the Kb the stronger the basestronger base have low pOH, but high pH


124Weak Bases

Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?A) pyridine

B) methylamine

C) nitrous acid


125Weak Bases

Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?A) pyridine Kb = 1.7 x 10-9

B) methylamine Kb = 4.4 x 10-4

C) nitrous acid Kb = 2.2 x 10-11


126Weak Bases

Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? A) pyridine Kb = 1.7 x 10-9

Kb = 1.7 x 10-9 = [x][x] [0.05]

x = [OH-] = 9.2 x 10-6

pOH = 5.0 so pH = 9.0


127Weak Bases

Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? B) methylamine Kb = 4.4 x 10-4

Kb = 4.4 x 10-4 = [x][x] [0.05]

x = [OH-] = 4.6 x 10-3

pOH = 2.32 so pH = 11.68


128Weak Bases

Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution? C) nitrous acid Kb = 2.2 x 10-11

Kb = 2.2 x 10-11 = [x][x] [0.05]

x = [OH-] = 1.0 x 10-6

pOH = 5.97 so pH = 8.02


129Weak Bases

Sample exercise: Which of the following compounds should produce the highest pH as a 0.05 M solution?A) pyridine pH = 9.0

B) methylamine pH = 11.68

C) nitrous acid pH = 8.02


130Weak Bases

Identifying a Weak BaseNeutral substances that have an atom with a nonbonding pair

of electrons that can serve as a proton acceptor.Most of these are nitrogen atoms

Anions of weak acids


131Weak Bases

Sample exercise: A solution of NH3 in water has a pH of 10.50. What is the molarity of the solution?


132Weak Bases


pH = 10.50 so pOH = 3.50


133Weak Bases


pOH = 3.50NH3 + H20 NH4+ + OH-

[OH-] = 3.16 x 10-4


134Weak Bases


NH3 + H20 NH4+ + OH-

x 0 0 -3.16 x 10-4 +3.16 x 10-4 +3.16 x 10-4

x - 3.16 x 10-4 3.16 x 10-4 3.16 x 10-4


135Weak Bases


NH3 + H20 NH4+ + OH-

Kb = 1.8 x 10-5 = [3.16 x 10-4][3.16 x 10-4] [x - 3.16 x 10-4]


136Weak Bases


NH3 + H20 NH4+ + OH-

Kb = 1.8 x 10-5 = [3.16 x 10-4][3.16 x 10-4] [x - 3.16 x 10-4]

x = 0.0058 M


137Relationship Between Ka and Kb

When two reactions are added to give a third reaction, the equilibrium constant for the third reaction is equal to the product of the equilibrium constants for the two added reactants.

NH4+(aq) NH3(aq) + H+(aq)

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

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



NH4+(aq) NH3(aq) + H+(aq)

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

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

Ka x Kb = Kw

pKa + pKb = pKw = 14



Sample exercise: Which of the following anions has the largest base-dissociation constant? A) NO2

-

B) PO43-

C) N3-




- Ka = 4.5 x 10-4

Ka x Kb = 1.0 x 10-14

Kb = 1.0 x 10-14 = 2.2 x 10-11

4.5 x 10-4



Sample exercise: Which of the following anions has the largest base-dissociation constant? B) PO43- Ka = 4.2 x

10-13

Ka x Kb = 1.0 x 10-14

Kb = 1.0 x 10-14 = 2.4 x 10-2

4.2 x 10-13



Sample exercise: Which of the following anions has the largest base-dissociation constant? C) N3

- Ka = 1.9 x 10-5

Ka x Kb = 1.0 x 10-14

Kb = 1.0 x 10-14 = 5.2 x 10-10

1.9 x 10-5




- Kb = 2.2 x 10-11

B) PO43- Kb = 2.4 x 10-2

C) N3- Kb = 5.2 x 10-10



Sample exercise: The base quinoline has a pKa of 4.90. What is the base dissociation constant for quinoline?




pKa + pKb = 14

4.90 + x = 14

x = 9.1




pKb = 9.1

pKb = -log[Kb]

9.1 = -log[Kb] [Kb] = 7.9 x 10-10


147Acid Base Properties of Salt Soln’s

Salt solutions have the potential to be acidic or basic.Hydrolysis of a saltacid base properties are due to the behavior of their cations and anionsperform the necessary double replacement reaction and examine the

products using your strength rules



If a strong acid and strong base are produced, the resultant solution will be neutral.

If a strong acid and weak base are produced, the resultant solution will be acidic.

If a strong base and a weak acid are produced, the resultant solution will be basic.



If a weak acid and weak base are produced, the resultant solution will be dependent on the Ka values.



Sample exercise: In each of following, which salt will form the more acidic solution.

A) NaNO3, Fe(NO3)3

NaNO3 + H2O NaOH + HNO3

SB SA


151Acid Base Properties of Salt Soln’sSample exercise: In each of following, which salt will form the more acidic solution.

A) NaNO3, Fe(NO3)3

Fe(NO3)3 + 3H2O Fe(OH)3 + 3HNO3

WB SA



A) NaNO3, Fe(NO3)3

Fe(NO3)3 + 3H2O Fe(OH)3 + 3HNO3

WB SA



Sample exercise: In each of following, which salt will form the more acidic solution.

B) KBr, KBrO



B) KBr, KBrO

KBr + H2O KOH + HBr

SB SA



B) KBr, KBrO

KBrO + H2O KOH + HBrO

SB WA



B) KBr, KBrO


157

Acid-Base Behavior & Chem Structure

How does the chemical structure determine which of the behaviors will be exhibited?

Acidica molecule containing H will transfer a proton only if the H-X bond is polarized like

H -- X the stronger the bond the weaker the acid and vice versa


158



Acidicoxyacids exist when the H is attached to an oxygen bonded to a central atom if

the OH’s attached are equal in number to the O’s present, acid strength increases with electronegativity


159



Acidicoxyacids exist when the H is attached to an oxygen bonded to a central atom

the more O’s present compared to the OH’s, the more polarized the OH bond becomes and the stronger the acid is


160



Acidiccarboxylic acids exist when the functional group COOH is present the

strength also increases as the number of electronegative atoms in the molecule increase

che 116 prof. t. l. heise 1 che 116: general chemistry u chapter sixteen copyright © tyna l. heise...

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