Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 1 of 58
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`Kowa Company, Ltd. et al. v. Aurobindo Pharma Limited et al.,
`Civil Action No. 14-CV-2497 (PAC) (and related cases)
`
`
`
`
`
`Exhibit 10 to Supplemental Declaration of
`Thomas R. Burns, dated June 10, 2015, in
`support of Defendants’ Joint Responsive
`Claim Construction Brief
`
`
`
`

`

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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 2 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 2 of 58
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`Steven S Zumdahl
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`MYLAN(Pitav) 037206
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`i,” Ir
`a, H,
`Casl:-cv-02758-PAC Document 69-5 Filed 06/10/15 Page30f 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 3 of 58
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`To my parents and to Eunice, Whitney, and Leslie.
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`Cover and title page photo:
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`
`
`V CONTETS
`‘ "
`‘
`n this chapter we reencounter two very important classes of compounds,
`acids and bases. We will explore their interactions and apply the funda— 14.1
`The Nature of Acids and
`mentals of chemical equilibria discussed in Chapter 13 to systems involv—
`Bases
`ing proton-transfer reactions.
`149 Am! Strength
`V
`Water as an Acrd and a
`Acid—base chemistry is important in a wide variety of everyday applica—
`Base
`tions. There are complex systems in our bodies that carefully control the
`The pH Scale
`acidity of our blood, since even small deviations may lead to serious illness
`Calculating the p
`
`and death. The same sensitivity is seen in other life forms. if you have ever
`Acid Solutions
`had tropical fish or goldfish, you know how important it is to monitor and
`CalFU'at'PSthe P Oi Weak
`.
`control the acidity of the water in the aquarium.
`AC'd SOIUUWS
`.
`.
`.
`.
`The pH of a Mixture of
`ACIds and bases are also important industrially. For example, the vast
`Weak Adds
`quantity of sulfuric acid manufactured in the United States each year is
`Percent Dissociation
`needed to produce fertilizers, polymers, steel, and many other materials
`Bases
`(see the Chemical Impact in Chapter 3).
`Polyprotic Acids
`Phosphoric Acid
`The influence of acids on living things has assumed special importance
`.
`i
`Sulfuric Add
`.
`.
`.
`in the United States, Canada, and Europe in recent years as a result of the
`Add_BaS€ Properties of Sans
`phenomenon of aCld rain. This problem is complex and has diplomath and
`Salts That produce Neutral
`economic overtones that make it all the more difficult to solve.
`Solutions
`Salts That Produce Basic
`Solutions
`Base Strength in Aqueous
`
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 4 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 4 of 58
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`CHAPTER FOURTEEN
`
`
`
`
`
`
`of Strong
`
`
`
`143
`14.4
`
`14‘5
`
`146
`14.7
`
`148
`
`1 4 ’1
`.
`
`.
`The Nature of ACldS and Bases
`
`Solution
`Sails That Produce Acidic
`Solutions
`
`14.9
`
`The Effect of Structure on
`.
`.
`Acid—Base Properties
`14.10 AC‘d_Bas€ Properties of
`OX1 deg
`The Lewis Acid-Base Model
`Strategy for Solving Acid—Base
`Problems—A Summary
`
`14.11
`14.19
`
`PURPOSE
`t
`.
`.
`.
`.
`.
`To discuss two models of a01ds and bases and to rela e equilibrium concepts to
`acid dissociation.
`
`
`
`
`Acids were first recognized as substances that taste sour. Vinegar tastes sour
`because it is a dilute solution of acetic acid; citric acid is responsible for the sour
`taste of a lemon. Bases, sometimes called alkalis, are characterized by their bitter
`
`
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 5 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 5 of 58
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`+
`..
`
`HOHiCl‘
`HOIiHCl
`
`H
`
`H
`
`The |abei on a bottie of concentrated, Note that the proton is transferred from the HCl molecule to the water molecule to
`hydrochloric acid.
`form H30+, which is called the hydronium ion.
`The general reaction that occurs when an acid is dissolved in water can best be
`represented as
`
`HA(oq) + H200) : H3O+(aq) + A’(aq)
`Acid
`Base
`Conjugate
`Conjugate
`acid
`base
`
`(14.1)
`
`taste and slippery feel. Commercial preparations for unclogging drains are highly
`basic.
`The first person to recognize the essential nature of acids and bases was Svante
`Arrhenius. Based on his experiments with electrolytes, Arrhenius postulated that
`acids produce hydrogen ions in aqueous solution, while bases produce hydroxide
`ions. At the time, the Arrhenius concept of acids and bases was a major step
`forward in quantifying acid—base chemistry, but this concept is limited because it
`applies only to aqueous solutions and allows for only one kind of base—the hydrox—
`ide ion. A more general definition of acids and bases was suggested by the Danish
`chemist Johannes Bronsted and the English chemist Thomas Lowry. In terms of the
`Bronsted.Lowry model, an acid is a proton (HT) donor, and a base is a proton
`acceptor. For example,when gaseous HCI dissolves in water, each HCl molecule
`donates a proton to a water molecule, and so qualifies as a Bronsted—Lowry acid.
`The molecule that accepts the proton in this case, water, is a Bronsted-Lowry base.
`To understand how water can act as a base, we need to remember that the
`oxygen of the water molecule has two unshared electron pairs, either of which can
`form a covalent bond with an H+ ion. When gaseous HCI dissolves, the following
`reaction occurs:
`
`
`
`This representation emphasizes the significant role of the polar water molecule in
`pulling the proton from the acid. Note that the conjugate base is everything that
`remains of the acid molecule after a proton is lost. The conjugate acid is formed
`when the proton is transferred to the base. A conjugate acid-base pair consists of
`two substances related to each other by the donating and accepting of a single
`proton. In Equation (14.1) there are two conjugate acid-base pairs: HA and AT, and
`1120 and H30+.
`It is important to note that Equation (14.1) really represents a competition for
`the proton between the two bases H20 and A'. If H20 is a much stronger base than
`A7 , that is, if H20 has a much greater affinity for H ' than does A‘ , the equilibrium
`position will be far to the right. Most of the acid dissolved will be in the ionized
`form. Conversely, if A‘ is a much stronger base than H20, the equilibrium position
`will lie far to the left. In this case most of the acid dissolved will be present at
`equilibrium as IIA.
`The equilibrium expression for the reaction given in Equation (14.1) is
`
`K = [H30+l[ATl : [H+l[A_]
`3
`[HA]
`[HA]
`
`(14.2)
`
`where Ka is called the acid dissociation constant. Both H30+(aq) and H+‘(aq) are
`commonly used to represent the hydrated proton. In this book we will often use
`simply H+, but you should remember that it is hydrated in aqueous solutions.
`
`
`
`Chapter Fourteen Acids and Bases
`
`MYLAN(Pitav) 037209
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 6 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 6 of 58
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`In Chapter 13 we saw that the concentration of a pure solid or a pure liquid is
`always omitted from the equilibrium expression. In a dilute solution we can assume
`that the concentration of liquid water remains essentially constant when an acid is
`dissolved. Thus the term [H20] is not included in Equation (14.2), and the equilib—
`rium expression for Ka has the same form as that for the simple dissociation
`
`HAW) : mag) i Mag)
`
`You should not forget, however, that water plays an important role in causing the
`acid to dissociate.
`
`Note that Ka is the equilibrium constant for the reaction in which a proton is
`removed from HA to form the conjugate base A’. We use Ka to represent only this
`type of reaction. Knowing this, you can write the Ka expression for any acid, even
`one that is totally unfamiliar to you. As you do Sample Exercise 14.1, focus on the
`definition of the reaction corresponding to Ka.
`
`sample Exercise 14.1 WWWWWW
`
`Write the simple dissociation reaction (omitting water) for each of the following
`acids:
`
`a. hydrochloric acid (HCl)
`b. acetic acid (HC2H302)
`
`c.
`
`d.
`
`the ammonium ion (NH4+)
`
`the anilinium ion (C6H5NH3‘)
`
`the hydrated aluminum(III) ion [Al(H20)6]3+
`e.
`Solution
`
`a. HCl(aq) : Inaq) + Cnaq)
`13- HC2H302(61Q) : H+(04) + C2H302_(061)
`c. NH4+(aq) : H+(aq) + NH3(aq)
`d. C6H5NH3+(aq) : H+(aq) + C5H5NH2(aq)
`e. Although this formula looks complicated, writing the reaction is simple if you
`concentrate on the meaning of Kn. Removing a proton, which can only come
`from one of the water molecules, leaves one OH‘ and five H20 molecules
`attached to the Al3+ ion. So the reaction is
`
`[AI(HZO)6]3 +(6161) : H+(aq) + [A1(H20)50Hl2+(aq)
`
`“-mwmmmcMWWW
`
`The Bronsted-Lowry model is not limited to aqucous solutions; it can be ex-
`tended to reactions in the gas phase. For example, we discussed the reaction be—
`tween gaseous hydrogen chloride and ammonia when we studied diffusion (Section
`5.7):
`
`NH3(g) + HCl(g) : NH4Cl(s)
`
`In this reaction a proton is donated by the hydrogen chloride to the ammonia, as
`shown by these Lewis structures:
`
`561
`
`
`
`
`14.1 The Nature of Acids and Bases.
`
`i
`It
`i
`i‘
`H——N :[HI—Cl: F iH—an + :17
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 7 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 7 of 58
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`Note that this is not considered an acid—base reaction according to the Arrhenius
`concept.
`
`l
`
`Acid Strength
`
`PURPOSE
`
`
`
`
`
`
`To relate acid strength to the position of the dissociation equilibrium.
`To discuss the autoionization of water.
`
`7
`
`
`The strength of an acid is defined by the equilibrium position of its dissociation
`reaction:
`
`HA(aq) + H20(l) : H30+(aq) + ATaq)
`
`A strong acid is one for which this equilibrium lies far to the right. This means that
`almost all the original HA is dissociated at equilibrium [see Fig. l4.1(a)]. There is
`an important connection between the strength of an acid and that of its conjugate
`base. A strong acid yields a weak conjugate base—one that has a low affinity for a
`proton. A strong acid can also be described as an acid whose conjugate base is a
`much weaker base than water (see Fig. 14.2). In this case the water molecules win
`the competition for the H+ ions.
`Conversely, a weak acid is one for which the equilibrium lies far to the left.
`Most of the acid originally placed in the solution is still present as HA at equilib—
`rium. That is, a weak acid dissociates only to a very small extent in aqueous solu-
`tion [see Fig. 14.1(b)]. In contrast to a strong acid, a weak acid has a conjugate base
`After dissociation,
`at equilibrium
`H+ A‘
`
`Before dissociation
`HA
`
`Figure 14.1
`
`Graphical representation of the
`behavior of acids of different
`strengths in aqueous solution. (a) A
`strong acid is completely dissociated.
`(b) In contrast, only a small fraction
`of the molecules of a Weak acid are
`dissociated at equilibrium.
`
`HA
`
`
`
`(a)
`
`HA
`
`(b)
`
`m -
`
`Relative
`acid strength
`
`Relative
`conjugate
`base strength
`
`
`
`Very_ I
`ciakp: 3
`
`conjugate base strength for the dissociation
`
`Figure 14.2
`
`The relationship of acid strength and
`
`reaction
`
`HA(aq) l HQOU) :3 H30+(aq) + AKaq)
`Acid
`Conjugate
`base
`
`Chapter Fourteen Acids and Bases
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 8 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 8 of 58
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`W l
`
`
`
`.
`l
`
`means much less than
`means much greater than
`
`that is a much stronger base than water. In this case a water molecule is not very
`successful in pulling an H+ ion from the conjugate base. A weak acid yields a
`relatively strong conjugate base.
`The various ways of describing the strength of an acid are summarized in Table
`14.1.
`
`
`
`Various Ways to Describe Acid Strength
`
`l
`
`Property
`
`K3 value
`Position of the dissociation
`equilibrium
`Equilibrium concentration of H+
`compared to original
`concentration of HA
`
`Strong acid
`
`K3 is large
`Far t0 the right
`
`Weak acid
`
`.
`K3 is small
`FBI to the left
`
`[H+] z [HA](J
`
`[HT] < [HA]0
`
`Strength of Conjugate base
`compared to that of water
`
`A‘ much weaker
`base than H20
`
`A‘ much stronger
`base than H20
`
`Table 14.1
`
`The COIHIHOll strong acids are sulfuric acid (HZSO4(aq)), hydrochloric acid
`(HCl(aq)), nitric acid (HNO3)(aq), and perchloric acid (HClO4(aq)). Sulfuric acid
`is actually a diprotic acid, an acid having two acidic protons. The acid H2804 is a
`strong acid, virtually 100% dissociated in water:
`
`HZSO4(aq) —> l-l+(aq) + HSO4T(aq)
`
`but the HSO4" ion is a weak acid:
`
`lir’cmiilm‘ic acid can L‘rgtlt‘itlc ii bandied
`improperly.
`
`0
`
`(1
`
`HSO4T(aq) : H+(aq) + SO42”(aq)
`
`féull't‘n’ic mid
`
`_
`i
`
`:
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`i
`
`3
`%
`
`tl
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`3 is
`,ate
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`is a
`win
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`my
`Jlu-
`)ase
`
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`
`
`
`
`
`
`Most acids are oxyacids, in which the acidic proton is attached to an oxygen
`atom. The strong acids mentioned above, except hydrochloric acid, are typical
`examples. Many common weak acids, such as phosphoric acid (H3PO4), nitrous
`acid (HNOZ), and hypochlorous acid (HOCl), are also oxyacids. Organic acids,
`those with a carbon—atom backbone, commonly contain the carboxyl group:
`
`_C/0
`O—H
`
`\
`
`Acids of this type are usually weak. Examples are acetic acid (CH3COOH), often
`written HC2H302, and benzoic acid (CGHSCOOH).
`There are some important acids in which the acidic proton is attached to an
`atom other than oxygen. The most significant of these are the hydrohalic acids HX,
`where X represents a halogen atom.
`Table 14.2 contains a list of common monoprotic acids (those having one
`
`acidic proton) and their K,1 values. Note that the strong acids are not listed. When a
`strong acid molecule such as HCl, for example, is placed in water, the position of
`the dissociation equilibrium
`HCl(aq) : H+(aq) + Ci—(aq)
`
`0
`l
`
`/
`
`0
`
`{law-ll
`Wm my,
`
`o
`(,1r‘wretir ‘ ii
`
`{3
`gig/Mij “mg
`
`'
`
`14.2 Add Strength 3 563
`
`
`
`‘ M
`
`YLAN(Pitav) 037212
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 9 of 58
`ase 1:14-CV-02758-PAC Document 69-5 Filed 06/10/15 Page 9 of 58
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`
`
`Values of K8 for Some Common Monoprotic Acids
`
`6,
`M
`
`Formula
`H304'
`HC107
`11311—2002
`HF“
`HNO2
`HC2H302
`[A1(H20)613+
`HOCl
`HCN
`NH4+
`HOCGHS
`
`Name
`Hydrogen sulfate ion
`Chlorous acid
`Monochloracetic acid
`Hydrofluoric acid
`Nitrous acid
`Acetic acid
`Hydrated aluminum(lll) ion
`Hypochlorous acid
`Hydrocyanic acid
`Ammonium ion
`Phenol
`
`Value of Kfi
`1.2 x IOT2
`1.2 X 10“2
`1.35 x 10*3
`7.2 X 10’4
`4.0 x 10—4
`1.8 x 1d5
`1.4 x 10’5
`3.5 X 10’8
`6.2 x 10c10
`5.6 X 10“10
`1.6 x 10—10.
`
`v:
`go
`g
`g
`a
`§
`g
`
`<
`
`
`5
`
`dm€fi>/
`
`*The units of K3 are moi/L, but are customarily omitted.
`
`
`Table 14.2
`
`.
`lies so far to the right that [HCl] cannot be measured accurately. Th1s prevents an
`accurate calculation of Ka:
`
`K3:
`
`[H+][C1’]
`[HCl]
`\Very small and
`highly uncertain
`
`
`Sample Exercise 14. 2
`2,...
`Using Table 14.2, arrange the following species according to their strength as bases:
`H20, F: C1: N02: CN‘.
`Solution
`Remember that water is a stronger base than the conjugate base of a strong acid, but
`a weaker base than the conjugate base of a weak acid. This leads to the following
`order:
`
`Cl’ < H20 < conjugate bases of weak acids
`Weakest bases —————A Strongest bases
`
`We can order the remaining conjugate bases by recognizing that the strength of an
`acid is inversely related to the strength of its conjugate base. Since from Table 14.2
`we have
`
`Ka for HF > Ka for HNOZ > Ka for HCN
`
`the base strengths increase as follows:
`
`F‘ < NOZ‘ < CN—
`
`The combined order of increasing base strength is
`
`CIT < H20 <FT < N02 ' < CNw
`
`
`“I”
`
`
`
`
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`
`
`
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`
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`i
`
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`
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`
`i
`‘
`
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`
`1
`
`564
`
`Chapter Fourteen Acids and Bases
`
`
`
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`MYLAN(Pitav) 037213
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 10 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 10 of 58
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`
`
`Water as an Acid and a Base
`
`A substance is said to be amphoteric if it can behave either as an acid or as a base.
`Water is the most common amphoteric substance. We can see this clearly in the
`autoionization of water, which involves the transfer of a proton from one water
`molecule to another to produce a hydroxide ion and a hydronium ion:
`
`
`
`
` 5
`
`l
`
`is:
`
`)ut
`Hg
`
`an
`42
`
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`iron 1
`' m in“ “3
`
`565
`
`
`
`
`
`
`14.2 Acid Strength
`
`..//—H\
`'O'
`+
`H/ \H
`
`0'_'_ : (j
`H/
`[LI/19H]
`
`+ + iz'oflH]
`—
`
`I
`s-
`
`In this reaction one water molecule acts as an acid by furnishing a proton, and the
`other acts as a base by accepting the proton.
`Autoionization can occur in other liquids besides water. For example, in liquid
`ammonia the autoionization reaction is
`
`
`
`
`
`2. /.
`
`+
`
`\a*2:/
`
`\72
`
`+
`
`I:\ /E: l
`
`The autoionization reaction for water
`
`ZHZOU) : H30+(aq) + OH‘(aq)
`
`leads to the equilibrium expression
`Kw : [Hsouiom = [H'J[OH’l
`where KW, called the ion—product constant (or the dissociation constant), always
`refers to the autoionization of water.
`Experiment shows that at 25°C
`[H+] = [OH‘] : 1.0 X 10'7 M
`
`which means that at 25°C
`
`KW = [H+][OH‘] : (1.0 x 10*7 mol/L)(1.0 >< 10 7 mol/L)
`= 1.0 x 10'14 molZ/L2
`.
`.
`.
`The units are customarily omitted.
`It is important to recognize the meaning of KW. In any aqueous solution at
`25°C, no matter what it contains, the product of [H ‘ ] and [OH’] must always equal
`1.0 X too”. There are three possible situations:
`l. A neutral solution, where [H'] = [0H2].
`2. An acidic solution, where [H+] > [0H2].
`3. A basic solution, where [OH’] > [H+].
`
`In each case, however, at 25°C
`
`M
`
`Kw : [H+][OH"] = 1.0 X 10514
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`Sample Exercise 14.3
`Calculate [H+] of [OH’] as required for each of the following solutions at 25°C,
`and state whether the solution is neutral, acidic, or basic.
`
`3. 1.0 ><1OC5 M OH’
`
`Solution
`Kw : [H+][OHC] = 1.0 X 10!”. Since [OH’] is 1.0 X 10*5 M, solving for [H+]
`gives
`
`+ _
`[H ]
`
`1.0 x1044 _ 1.0 ><1tr14
`_
`[0H’]
`1.0 x 10*5
`
`— 1.0 x
`
`e9 M
`
`10
`
`Since [OH’] > [H+], the solution is basic.
`
`b. 1.0 ><10’7 M OHi
`
`Solution
`
`{\s in part a, solving for [H+] gives
`
`1.0 ><10*14
`1.0 x1044
`+ _
`~
`—1.0><10’7M
`[OHS]
`1.0 X 10*7
`[H ]
`
`Since [H ‘] = [OHS], the solution is neutral.
`
`c. 10.0 M H+.
`
`Solution
`
`Solving for [OH’] gives
`
`1.0 ><10‘14 _ 1.0 x 10—14
`e 1.0x 10—15M
`[H+]
`10.0
`
`[OH—i
`
`Since [W] > [OH’], the solution is acidic,
`
`
`
`Since KW is an equilibrium constant, it varies with temperature. The effect of
`temperature is considered in Sample Exercise 14.4.
`
`
`sample Exercise 14.4 W_____
`
`At 60°C the value of KW is l X 10-13.
`
`3. Using Le Chatelier’s principle, predict whether the reaction
`
`2HZO(Z) : H30+(aq) + OH (aq)
`
`is exothermic or endothermic.
`
`b. Calculate [H+] and [OH‘] in a neutral solution at 60°C.
`
`Solution
`21. KW increases from 1 >< 1tr14 at 25°C to 1 x 10‘13 at 60°C. Le Chatelier’s
`principle states that if a system at equilibrium is heated, it will adjust to con—
`sume energy. Since the value of KW increases with temperature, we must think
`of energy as a reactant, and so the process must be endothermic.
`
`
`
`
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`566
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`Chapter Fourteen Acids and Bases
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 12 of 58
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`"
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`pll wait: as
`wine-Sm: Malawi am
`
`b. At 60°C
`
`For a neutral solution
`
`[H+l[OH‘l = 1 X 10‘13
`
`lH+J = [OH—l =W : 3 X 10#7 M
`
`
`1 i The pH Scale
`
`PURPOSE
`
`D To define pH, pOH, and pK and to introduce general methods for solving
`acid-base problems.
`
`Because [H+] in an aqueous solution is typically quite small, the pH scale
`provides a convenient way to represent solution acidity. The pH is a log scale based
`on 10 where
`
`Thus for a solution where
`
`pH = —10g[H“l
`
`lH+l = 1.0 X 10"7 M
`
`pH = —(~7.00) = 7.00
`
`
`
`At this point we need to discuss significant figures for logarithms. The rule is
`that the number of decimal places in the log is equal to the number of significant
`figures in the original number. Thus
`
`hypoxia :1
`
`has:
`
`.1 mice, at
`
`tog».
`
`r2 significant figures
`[HT] = 1.0 X 10’9 M
`
`pH = 9.00
`AV2 decimal places
`
`Similar log scales are used for representing other quantities, for example:
`
`pOH = —log[OH“]
`
`pK = —log K
`
`
`
`
`
`
`
`
`
`
`
`
`
`Since pH is a log scale based on 10, the pH changes by Ifor every power 0f10 m an "i
`change in [H+]. For example, a solution of pH 3 has an H4r concentration 10 times
`that of a solution of pH 4 and 100 times that of a solution of pH 5. Also note that
`because pH is defined as —log[H+], the pH decreases as [H+] increases. The pH it“: hit Warmer: tits
`scale and the pH values for several common substances are shown in Fig. 14.3.
`M'Kl rm?! “Wt-‘11:
`The pH of a solution is usually measured using a pH meter, an electronic device
`HA
`with a probe that can be inserted into a solution of unknown pH. The probe contains
`an acidic aqueous solution enclosed by a special glass membrane that allows migra—
`tion of H4r ions. If the unknown solution has a different pH from the solution in the
`probe, an electrical potential results, which is registered on the meter (see Fig.
`14.4).
`
`r
`
`to, 41;}; t
`
`3
`
`lt’l '
`
`t éna‘t‘cusm»,
`
`
`
`14.3 The pH Scale
`
`
`I:
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`567
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`MYLAN(Pitav) 037216
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 13 of 58
`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 13 of 58
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`Figure 14.4
`A typical pH meter. The probe at the
`right is placed in the solution With
`unknown pH. The difference between
`the [HT] in the solution sealed into
`the probe and the [H+] in the
`solution being analyzed is translated
`into an electrical potential arid
`registered on the meter as a pH
`reading.
`
`,
`333”
`
`14 .— 1 MNdOH
`
`Ammonia
`(Househdd Cleaner)
`
`4-— Sea Water
`
`
`
`Neutral Wz—wm
`
`Blood
`
`Pure water
`Milk
`
`Acidic
`
`Vinegar
`Lemon juice
`Stomach acid
`
`1 M HCl
`
`
`
`Si
`
`5"
`T1P
`St
`
`Si
`
`Tl
`
`W
`a1
`
`Ti
`la
`C
`
`0]
`
`a]
`
`11
`
`-
`
`a P
`
`t]
`'5
`p
`
`
`
`$ample Exercise 14.5
`o
`.
`_
`_
`Calculate pH and pOH for each of the followmg solutlons at 25 C;
`
`
`
`a. 1.0x 10'3MOH'
`
`b.1.0MH+.
`
`Figure 14.3
`The pH scale and pH values of some
`common substances.
`
`Solution
`
`3"
`
`b.
`
`
`d 1.0 X 10—14 _
`KW
`_
`[11+] _ [0H1 _ 1‘0 X 10—3 ‘ 1'0 X 10
`
`_11
`
`M
`
`pH = -10g[H+J = —1og(l.0 x 10*”) = 11.00
`
`[OH—1:
`
`pOH = —log[OH_] = flog(1.0 X 10'3) = 3.00
`
`1.0 x 10—14
`K
`: =————=1.0><10'14M
`[H ]
`1.0
`pH = —1og[H+] = -log(1.0) : 0
`pOH = —1og[0Hc] = —log(1.0 X 10’”) = 14.00
`
`m
`
`_,,,,,,,,,,,,,,,,,,,.1
`
`It is useful to consider the log form of the expression
`Kw = [H+][OH"J
`log KW = log[H+] -l- log[OH’]
`—log Kw = —1ogiH+] e log[OH_]
`pKW : pH -l— pOH
`
`That is,
`or
`Thus
`
`(14.3)
`
`Chapter Fourteen Acids and Bases
`
`
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`
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`568
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`1‘
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`p
`1;
`‘
`‘
`1
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`1
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 14 of 58
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 14 of 58
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`Since KW = 1.0 x 10—14,
`
`pKW = —log(1.0 x 10*”) = 14.00
`
`Thus, for any aqueous solution at 25°C, pH and pOH add up to 14.00.
`
`Sample Exercise 14.6
`
`pH + pOH = 14.00
`
`
`(14-4)
`
`The pH of a sample of human blood was measured to be 7.41 at 25°C. Calculate
`pOH, [11+], and [OH‘] for the sample.
`Solution
`
`Since pH + pOH = 14.00,
`
`pOH = 14.00 ‘ pH = 14.00 — 7.41 = 6.59
`
`To find [H+] we must go back to the definition of pH:
`
`pH = ~10g[H+]
`
`Thus
`
`7.41 = a-log[H+]
`
`or
`
`log[H+] = ~7.41
`
`We need to know the antilog of —7.41. As shown in Appendix 1.2, taking the
`antilog is the same as exponentiation, that is,
`
`antilog(n) = 10"
`
`There are different methods for carrying out the antilog operation on various calcu—
`lators. The most common are the
`key and the two—key
`sequence.
`Consult the user’s manual for your calculator to find out how to do the antilog
`operation.
`Since pH : —log[H+],
`
`pr = 10g[H+l
`
`and [H+J can be calculated by taking the antilog of —pH:
`
`[H+] = antilog(—pH)
`
`In the present case
`
`[H+] = antilog(—pH) = antilog(—7.41): 10—7-41 2 3.9 x 10—8
`
`Similarly, [OH’] = antilog(~pOH), and
`
`
`
`Now that we have considered all of the fundamental definitions relevant to
`
`acid-base solutions, we can proceed to a quantitative description of the equilibria
`present in these solutions. The main reason that acid-base problems seem difficult is
`that a typical aqueous solution contains many components so the problems tend to
`be complicated. However, you can deal with these problems successfully if you use
`the following general strategies:
`
`
`
`Think chemistry. Focus on the solution components and their reactions. It will
`almost always be possible to choose one reaction that is the most important.
`
`
`
`‘i
`
`i-f
`'3-It
`
`
`
`
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`14.3)
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`569
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`14.3 The pH Scale
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 15 of 58 Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 15 of 58
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`wi
`
`for
`
`ate
`
`Acid—base problems require a step—by—step approach.
`5 Be systematic.
`gh all acid—base problems are similar in many ways, impor-
`E Be flexible. Althou
`occur. Treat each problem as a separate entity. Do not try to
`tant differences do
`u have solved before. Look for both
`force a given problem into matching any yo
`ties and the differences.
`the similari
`he complete solution to a complicated problem cannot be seen
`Be patient. T
`n all its detail. Pick the problem apart into its workable steps.
`immediately i
`thin the problem for the solution, and let the problem
`Be confident. Look wi
`guide you. Assume that you can think it out. Do not rely on memorizing solu—
`tions to problems. In fact, memorizing solutions is usually detrimental because
`you tend to try to force a new problem to be the same as one you have seen
`before. Understand and think; don’t just memorize.
`
`
`
`
`14.4
`
`Calculating the pH of
`Strong Acid Solutions
`
`PURPOSE
`
`
`
`
`f solutions of strong acids.
`
`To demonstrate the systematic treatment 0
`When we deal with acid—base equilibria, it is essential to focus on the solution
`components and their chemistry. For example, what species are present in a 1.0 M
`solution of HCl? Since hydrochloric acid is a strong acid, we assume that it is
`completely dissociated. Thus, although the label on the bottle says 1.0 M HCl, the
`solution contains virtually no HCl molecules. Typically, container labels indicate
`the substance(s) used to make up the solution, but do not necessarily describe the
`solution components after dissolution. Thus a 1.0 M HCl solution contains H+ and
`CIT ions rather than HCl molecules.
`lutions is to determine which compo—
`The next step in dealing with aqueous so
`nents are significant and which can be ignored. We need to focus on the major
`species, those solution components present in relat‘vely large amounts. In 1.0 M
`HCl, for example, the major species are H+, C17, and H20. Since this is a very
`classed as a minor
`acidic solution, OH’ is present only in tiny amounts and is
`species. In attacking acid—base problems, the importance of writing the major spe-
`cies in the solution as the first step cannot be overemphasized. This single step is the
`key to solving these problems successfully.
`To illustrate the main ideas involved, let us calculate the pH of 1.0 M HCl. We
`first list the major species: HT, ClT, and H20. Since we want to calculate the pH,
`we will focus on those major species that can furnish H+. Obviously, we must
`consider HJr from the dissociation of HCl. However, H20 also furnishes H+ by
`autoionization, which is often represented by the simple dissociation reaction
`H20(l) :3 H+(aq) + OHTaq)
`+ ions? In pure water at 250C, [H+] is
`But is autoionization an important source of H
`10T7 M. In 1.0 M HCl solution, the water will produce even less than lOT7 M H+,
`since by Le Chatelier’s principle, the H+ from the dissociated HCl will drive the
`position of the water equilibrium to the left. Thus the amount of H+ contributed by
`water is negligible compared to the 1.0 M H" from the dissociation of HCl. There-
`
`‘
`
`:
`'
`
`l
`
`‘
`
`i
`3
`
`ll,
`
`‘
`
`;
`
`,‘
`
`‘
`
`‘_
`I
`
`i
`
` I
`
`l
`‘
`
`iiiwzys mite the major species
`
`WWW“ m m“ WWW”
`
`from the strong acid drives
`‘l'lie ill
`the equilibrium Hf!) :1 H ‘ + OH'
`to the lcll,
`
`Sr
`
`“Ff-:2
`
`Chapter Fourteen Acids and Bases
`
`
`
`
`
`
`
`570
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`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 16 of 58
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`Case 1:14-cv-02758—PAC Document 69-5 Filed 06/10/15 Page 16 of 58
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`
`
`
`a. Calculate the pH of 0.10 M HN03.
`
`Solution
`a. Since HN03 is a strong acid, the major species in solution are:
`H+, NO3‘, and H20
`
`The concentration of HNO3 is viltually zero, since the acid completely dissoci-
`ates in water. Also, [0H7] will be very small because the H+ ions from the acid
`will drive the equilibrium
`H200) : H+(aq) + 0H“(aq)
`to the left. That is, this is an acidic solution where [W] > [OI-1‘] and so [OH—1 <
`lOT7 M. The sources of H4r are
`
`‘
`
`l. W from HNo3 (0.10 it).
`2. H+ from H20
`
`i
`i
`i
`
`
`
`The number of H+ ions contributed by the autoionization of water will be very small
`compared to the 0.10 M contributed by the HN03 and can be neglected. Since the
`dissolved HN03 is the only important source of H+ ions in this solution,
`[H '] = 0.10 M and pH = —log(0.10)= 1.00
`
`b. Calculate the pH of 1.0 x 10“10 M HCl.
`Solution
`
`Normally, in an aqueous solution of HCl the major species are HT, CIT. and H20.
`However, in this case the amount of HCl in solution is so small that it has no effect;
`the only major species is H20. Thus the pH will be that of pure water, or pH :
`7.00.
`
`W ____________,m_mmmmmw_
`
`W 4 5 Calculating the pH of
`
`Weak Acid Solutions
`
`.
`
`.
`
`PURPOSE
`
`
`
`To demonstrate the systematic treatment of solutions of weak acids.
`
`3
`
`To show how to calculate percent dissociation.
`Since a weak acid dissolved in water can be viewed as a prototype of almost
`any equilibrium occurring in aqueous solution, we will proceed carefully and sys-
`tematically. Although some of the procedures we develop here may seem superflu—
`0118, they will become essential as the problems become more complicated. We will
`
`57’1
`
`
`
`
`
`
`
`14.5 Calculating the pH of Weak Acid Solutions
`
`
`
`MYLAN(Pitav) 037220
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`Ty to
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`
`seen
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`:ause
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`seen
`
`__
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`.0 M
`litthis
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`iicate
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`fth:
`an
`
`mpO-
`lajOI‘
`.0 M
`very
`ninor
`
`' spe—
`is the
`
`i. We
`
`,pH
`
`must
`+ by
`n
`
`[+1is
`'H+,
`e the
`ed by
`here-
`
`
`
`fore, we can say that [H+] in the solution is 1.0 M. The pH is then
`H = ~10 H+ : e
`=
`g[
`i
`
`log(l.0)
`
`0
`
`p
`561771016 Exercise 14.7 "1
`
`

`

`Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 17 of 58
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`“"“Case 1:14-cv-02758-PAC Document 69-5 Filed 06/10/15 Page 17 of 58
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`deVelop the necessary strategies by calculating the pH of a 1.00 M solution of HF
`(Ka : 7.2 x 10-4).
`The first step, as always, is to write the major species in the solution. From its
`ofluoric acid is a weak acid and will be dissoci—
`small K3 value, we know that hydr
`ated only to a slight extent. Thus , when we write the major species, the hydrofluoric
`acid will be represented in its dominant form, as HF. The major species in solution
`are: HF and H20.
`The next step (since this is a pH problem) is to decide which of the major
`species can furnish H+ ions. Actually, both major species can do so:
`HF(aq) :3 H+(aq) + F‘(aq)
`K2, 2 7.2 >< 10*4
`H200) :3 H F(m1) + OH‘(aq)
`Kw = 1.0 x tor14
`But in aqueous solutions typically one source of H+ can be singled out as dominant.
`By comparing Ka for HF to KW for H20, we see that hydrofluoric acid, although
`Weak,
`is still a much stronger acid than water. Thus we will assume that hydro—
`fluoric acid will be the dominant source of H" . We will ignore the tiny contribution
`by water.
`T