`
`Brown LeMay Bursten
`
`CHEMISTRY
`
`Revised Eighth Edition
`
`
`
`1 of 7
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`IPR2020-01045
`Teva Ex. 1016
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`

`

`Chemistry
`
`The Central Science
`
`Eighth Revised Edition
`
`Theodore L. Brown
`
`University of Illinois at Urbana-Champaign
`
`H. Eugene LeMay, Ir.
`
`University of Nevada, Reno
`
`Bruce E. Bursten
`
`The Ohio State University
`
`With contributions by Julia R. Burdge, University of Akron
`
`PRENTICE HALL
`
`Upper Saddle River, New Jersey 07458
`
`
`
`
`
`2 0f 7
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`|PR2020-01045
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`Teva Ex. 1016
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`2 of 7
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`IPR2020-01045
`Teva Ex. 1016
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`

`

`. C
`3 .1 ‘
`-- 4
`z '1(
`""
`ii
`'
`
`Editor: John Challice
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`
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`Upper Saddle River, NJ 07458
`
`All rights reserved. No part of this book may be
`reproduced, in any form or by any means,
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`Printed in the United States of America
`10
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`ISBN El-lEI-EllalullIE-S
`
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`
`
`3 of 7
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`|PR2020-01045
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`Teva Ex. 1016
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`3 of 7
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`IPR2020-01045
`Teva Ex. 1016
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`

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`
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`16.7 J Weak Bases
`
`615
`
`
`
`4.0 X 1C" '
`
`4.2 X i-C'
`
`'
`
`x2 = (0.0037)(4.3 X 10‘?) = 1.6 X 10—9
`Solving for x, we have
`x = [W] = IHCO3’] = V1.5 x 10’9 = 4.0 x 10‘5M
`
`The small value of x indicates that our simplifying assumption was justified. The
`pH is therefore
`
`pH = —log [H+] = —log (4.0 x 10-5) = 4.40
`
`If we were asked to solve for [COSZ’], we would need to use Kfl. Let’s illustrate
`that calculation. Using the values of [HCOS‘] and [H+] calculated above, and setting
`:COf'] = y, we have the following initial and equilibrium concentration values:
`
`HCO3(sq) —‘
`
`H+(sq)
`
`+
`
`C032‘(mfi
`
`
`ht:
`
`
`[ply as "13:;
`it proton. .-
`
`zation)
`
`Assuming that y is small compared to 4.0 X 10'5,we have
`
`
`
`
`
`n
`
`IH"][C032']
`[4.0 x 105M).
`< 10H;
`KW: 4.0m =55”;
`30::
`
`_
`_
`as m
`y = 5-5 X 10 llM = [C03,2 ]
`factor of I?"
`
`The \alue calculated for y is indeed very smallin comparison to 4.0 x 105,showing
`POlypl‘ObC
`
`hit our assumption was justified. It also shows that the ionization of HC03' is neg-
`Italicin comparison to that of H2033 as far as production of H‘“15 concerned How-
`
`. 1 it'is the only source of (3032—, which has a very low concentration'in the solution.
`Our calculations thus teEl us that'm a solution of carbon dioxide'lI'l water most of
`
`C02 is in the form of CO; or HZCOW a small fraction ionizes to form H" and HCO3',
`37M. ”rhe-
`- - an even smaller fraction ionizes to give C03 ‘.
`
`”album '
`cncc EXERCISE
`“Mate the pH and concentration of oxalate ion, [C203], in a 0.020 M solution of ox-
`
`add, H2C204 (see Table 16.3). Answers: pH = 1.80; [(320421 = 6.4 X 10-5 M
`
`
`_
`. ad '
`_. -
`
`fiafie _
`_
`a a m-- . -
`be e-
`'
`-
`
`«‘r. removing protons from H20, thereby forming the conjugate acid of the
`- and 0H‘ ions:
`
`[comp
`
`Weak base + H20 = conjugate acid + OH‘
`most commonly encountered weak base is ammonia:
`
`[16.31]
`NH3(sq) + H20(1) = NI-Ifmq) + 0H*(aq}
`
`The equilibrium-constant expression for this reaction can be written as
`
`[NT-14+IIOH‘]
`
`K =— 16.32
`[NH3][H201
`I
`
`- .:se the concentration of water is essentially constant, the [H20] term is in-
`__0 x 10 .
`. stated into the equilibrium constant, giving
`
`_
`mation -'--
`+
`[NH4 ][OH 1
`K = K H O = ——
`1
`I
`t.
`[
`2
`1
`[NIL]
`
`inc constant Kb is called the base-dissociation constant, by analogy with the
`
`—.‘LSsociation constant, K“, for weak acids. The constant Kb almys rafters to the equi-
`
`
`
`Weak Bases
`
`-
`
`--
`
`substances behave as weak bases in water. Such substances react with
`
`0
`H M
`x M
`
`
`
`[16.30]
`
`1
`
`16.33
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`616
`
`Chapter 16 I Acid-Base Equilibria
`
`Base
`
` (Notice the.
`.
`.
`librium-cor
`Lawns
`Conjugate
`expression
`Structure
`Acid
`
`Equilibrium Reaction
`
`KI,
`
`Ammonia
`(NHsl
`
`Pyridine
`(CsHsN)
`
`H—N—H
`IH
`
`NH;
`
`NHa + H20 :NH; + OH‘
`
`1.8 x 10'
`
`:
`
`O N:
`
`C5H5NH'
`
`C5HSN + H20 :C5H5NH+ + OH“
`
`1.7 x in"
`
`Hydmxylamine
`(HZNOHJ
`
`H—N—QH
`lH
`
`H3NOH‘
`
`HZNOH + H20 2H3NOH‘ + OH‘
`
`1.1 X 1C '
`
`Methylarnine
`(NHZCHa)
`
`Hydrosulfide ion
`(HS‘)
`
`Carbonate ion
`(C032?
`
`Hypochlorite ion
`(CIO‘)
`
`H—N—CHa
`
`NH3CH3+
`
`NH2CH3 + H20: NH3CH3+ + 01-1"
`
`4.4 x 10'-
`
`H
`
`..
`[H-fiS-
`
`,
`
`3'
`
`:5:
`|
`__ /C.\\\¥ -.
`'9-
`-.
`
`HES
`
`HS" + H20 ——‘ H25 + OH‘
`
`_
`
`1.8 X it"
`
`HCOSF
`
`(3032‘ + H20 —‘—“ HCOS" + OH"
`
`1.8 X 11?"
`
`:(j—éj: _
`
`HCIO
`
`00* + H20 =HC10 + OH‘
`
`3.3 x it ‘
`
`lilm'um in which a base reacts with H20 toform the conjugate acid and OH‘. Tabs
`16.4 A lists the names, formulas, Lewis structures, equilibrium reactions, and v.=.-
`ues of Kb for several weak bases in water. Appendix D includes a more extensive is.
`Notice that thesebases contain one or more lone pairs of electrons. Alone pair is ne:-
`essary to form the bond with H”. Notice also that in the neutral molecules the lore
`pairs are on nitrogen atoms and that the other bases are anions derived from wee.
`acids
`
`I SAMPLE EXERCISE 16.14
`
`Calculate the concentration of OH" in a 0.15 M solution of NH3.
`
`Solution We use essentially the same procedure here as used in solving problems-
`involving the ionization of weak acids. The first step is to write the ionization reaction
`and the corresponding equilibrium-constant (K1,) expression:
`
`NH3(aq) + H200) ‘2‘ NHfQRq) + OH'(aq)
`N + 0H”
`Kt = l%3ll" = 1‘8 X 1075
`We then tabulate the equilibrium concentrations involved in the equilibrium:
`
`NHataq)
`
`+
`
`H200) : NH4+(aq) + OHfiaq)
`
`
`
`Because Kb
`compared '
`0.15 M. The
`
`Notice that
`
`0.15 M. Th1
`
`PRACTICE
`
`“hick of tl
`Sort: pyrid'
`
`TYPES 01‘ V
`. .
`,
`:10“ can it
`Able to belt.
`
`fist catego
`:air of 91.3
`finding all
`73959 subs
`.m'nefi In
`. 3th a hen
`\' H: with
`:H3NH2).
`refining an
`
`H—I
`
`the chemi:
`CH3NH3+.
`The se-
`
`wreak acids
`are, NaClC
`DH is alwa
`:"'.e 00‘ it
`
`gently, thu
`
`C10"
`
`
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`5 0f 7
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`|PR2020-01045
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`5 of 7
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`IPR2020-01045
`Teva Ex. 1016
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`

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`
`
`16.7 ,1 Weak Bases
`
`617
`
`(Notice that we ignore the concentration of H20 because it is not involved in the equi-
`librium-constant expression.) Inserting these quantifies into the equilibrium-constant
`expression gives the following:
`
`Ke=
`
`—— =— = 1. x
`[NHflIOH—l
`(35106)
`[NI-I3]
`0.15 — x
`8
`
`1°
`
`#5
`
`Because Kb is small, we can neglect the small amount of NH3 that reacts with water, as
`compared to the total NH3 concentration; that is, we can neglect x in comparison to
`3.15 M. Then we have
`
`x2
`0.15
`
`= 1.8 X 10—5
`
`x2 = (0.15)(1.8 x 10“) = 2.7 x 10'6
`
`x = [NI-14+] = [0H1 = +2.7 >< 10'6 = 1.6 x 10'3M
`
`Notice that the value obtained for x is only about 1 percent of the NH3 concentration,
`115 M. Therefore, our neglect of x in comparison with 0.15 is justified.
`
`PRACTICE EXERCISE
`
`Which of the following compounds should produce the highest pH as a 0.05 M solu-
`a'on: pyridine, methylan'u'ne, or nitrous acid? Answer: methylarnine
`
`157le"
`
`1.1 X li‘
`
`'
`
`4.4 X 11""
`
`1.8X1C .
`
`1.8x1L'
`
`'
`
`3.3xh‘"
`
`Types of Weak Bases
`
`d OH’. Table
`ions, and vs.-
`extensive is:
`
`no pair is nec-
`:ules the lane
`to! from west
`
`ng problems
`Ltion reactia'.
`
`lilibriurn:
`
`0H‘(flq)
`
`l-Iow can we recognize from a chemical formula whether a molecule or ion is
`3:1e to behave as a weak base? Weak bases fall into two general categories. The
`-_:st category contains neutral substances that have an atom with a nonbonding
`:air of electrons that can serve as a proton acceptor. Most of these bases, in-
`;xiding all the uncharged bases listed in Table 16.4, contain a nitrogen atom.
`Tnese substances include ammonia and a related class of compounds called
`amines. In organic amines, one or more of the N—H bonds in NH3 is replaced
`. 1th a bond between N and C. Thus, the replacement of one NfH bond in
`\E—I3 with a N—CH3 bond gives methylamine, NHZCHS (usually written
`:HgNHz). Like NH3, amines can extract a proton from a water molecule by
`firming an additional N—H bond, as shown here for methylamine:
`
`H
`
`..
`|
`H—lTJ—crgoq) + H200) = H—INII—CHS (sq) + 01-11%)
`H
`H
`
`[16.34]
`
`The chemical formula for the conjugate acid of methylamjne is usually written
`:HN‘HJ.
`The second general category of weak bases is composed of the anions of
`seak acids. Consider, for example, an aqueous solution of sodium hypochlo—
`:te, NaClO. This salt dissolves in water to give NaJr and C10" ions. The Na+
`ca is always a spectator ion in acid—base reactions.
`ISection 4.3) However,
`:e C10“ ion is the conjugate base of a weak acid, hypochlorous acid. Conse-
`qaently, the C10‘ ion acts as a weak base in water:
`
`Clo—(as) + H200) =HCIO(aq) + oer-(m)
`
`K, = 3.3 X 10"? [16.35]
`
`‘ Acids and Bases
`simulation
`
`‘6m7
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`|PR2020-01045
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`Teva EX. 1016
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`6 of 7
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`Teva Ex. 1016
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`

`

`618
`
`Chapter 16 / Acid-Base Equilibria
`
`SAMPLE EXERCISE 16.15
`
`
`
`
`
`
`
`
`A solution is made by adding solid sodium hypochlorite, NaClO, to enough water In
`. make 2.00 L of solution. If the solution has a pH of 10.50, how many moles of Neat!
`were added to the water?
`
`Solution NaClO is an ionic compound consisting of Na+ and C10" ions. As such. 2
`' is a strong electrolyte that completely dissociates in solution into Na ’3 which is a spec-
`tator ion, and CIO‘ ion. which is a Weak base with Ki, : 3.3 X 10‘7 (Equation 16.35.
`We wish to determine the concentration of ClO’ in Solution that would genera!
`enough OH‘ ion to raise the pH to 10.50.
`We first calculate the concentration of 0H'(aq) at equilibrium. We can calculn
`IOH'] by using either Equation 16.14 or Equation 16.17; we will use the late:
`method here:
`
`pOH = 14.00 — pH = 14.00 — 10.50 = 3.50
`
`[OH] = 10‘350 = 3.2 x arm
`
`This concentration is high enough that we can assume that Equation 16.35 is the only
`source of OH'; that is, we can neglect any OH’ produced by the autoionization 3'
`H20. We now assume a value of x for the initial concentration of C10" and solve it
`= equilibrium problem in the usual way:
`
`(DH—(sq)
`+
`H200) =HCIO(aq)
`+
`C10'(aq)
`
`——
`
`-———
`amt—4,... — .
`
`
`
`
`We now use the expression for the base-dissociation constant to solve for x:
`
`K,
`
`= [HCIOHOH'] =
`[ClO']
`
`(3.2 X 10‘4)2
`x - 3.2 X 10—4
`
`= 3.3 X 10—7
`
`Thus,
`
`x _ (3.2 x 10-4)2
`3.3 x 10‘7
`
`+ (3.2 x 10'4) = 0.31 M
`
`We say that the solution is 0.31 M in NaClO, even though some of the C10’ ions
`have reacted with water. Because the solution is [1.31 M in NaClO and the total m3-
`. ume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was adds:
`to the water.
`
`PRACTICE EXERCISE
`
`A solution of NH;5 in water has a pH of 10.50. What is the molarity of the soluticr.‘
`' Answer: 0.0058 M
`
`16.8 Relationship Between K” and Kg,
`
`We’ve seen in a qualitative way that the stronger acids have the weaker corri—
`gate bases. The fact that this qualitative reiationship exists suggests that we nag:
`be able to find a quantitative relationship. Let’s explore this matter by consider
`ing the NH{( and NH3 conjugate acid—base pair. Each of these species rear:
`with water:
`
`Acr
`
`Many amine
`'fishy” odor
`.tvic (abseno
`matter. Two
`H;.\I{CH2)4I\
`known as ca
`
`Many d
`mphetamil
`amines, thes
`
`the acid salt .
`
`:)>~c
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`Erin of these
`
`New notice a
`112' 16.37 are
`
`Inch just the e
`
`To determ
`aided reactio:
`
`scmciples gov
`2‘27: reaction, i
`an: equilibrium
`
`
`
`NI-I4+(aq) =N'l-Ia(aq) + H"'(sq)
`[16.3w
` NT-l3(aq) + H200) =NH4+(aq) + OH‘(aq)
`[165"
`
`
`
`7 of 7
`
`|PR2020-01045
`
`Teva EX. 1016
`
`7 of 7
`
`IPR2020-01045
`Teva Ex. 1016
`
`