For Advanced
`High School
`
`Chemistry
`CHEMISTRY
`
`Revised Eighth Edition
`
`1 of 7
`
`IPR2020-01045
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`

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`Chemistry
`
`The Central Science
`Eighth Revised Edition
`
`Theodore L. Brown
`University of Illinois at Urbana-Champaign
`
`H. Eugene LeMay,Jr.
`
`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
`
`
`
`
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`Ml35
`
`3 2
`
`Editor: John Challice
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`
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`Upper Saddle River, NJ 07458
`
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`reproduced, in any form or by any means,
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`Printed in the United States of America
`098 76543 2
`1
`
`ISBN 0O-13-Ob1142-5
`
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`
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`
`16.7 / Weak Bases
`
`615
`
`
`
`4.0 X 10°
`4,2 x 10™
`
`x” = (0.0037)(4.3 x 10°”) = 1.6 x 10?
`Solving forx,wehave
`x = [H*] = [HCO,-] = V1.6 x 10°? = 4.0 x 10-5 M
`The small valueof x indicates that our simplifying assumption was justified. The
`eH is therefore
`
`pH = —log [H*] = —log (4.0 x 1075) = 4.40
`If we were asked to salve for [CO;*], we would needto use K,». Let’s illustrate
`at calculation. Using the values of [HCO,”] and [H*] calculated above, and setting
`0°] = y, we have the following initial and equilibrium concentration values:
`HCO,"(aq) — H*(aq)
`+
` CO,*(aq)
`
`
`iply as “lars it proton. 7
`
`
`
`‘Sssuming that y is small compared to 4.0 x 107, we have
`zation)
`
`
`<10™. [H*]ICO2-]_(4.0 x 10-\y) ia,
`
`‘ fonthal
`Ko="Taco,]| + goxm= -°™*®
`
`'S
`mm
`ce
`i
`
`factor of 17
`y = 5.6 x 10°" M = [CO;*7]
`
`
`
`olyprotic The valuecalculated foryis indeed very small in comparison to 4.0 x 10~°, showi
`
`polyp
`ery
`p
`y
`wing
`Set our assumption wasjustified. It also showsthat the ionization of HCO,” is neg-
`le in comparison to that of HCO, as far as production of H™ is concerned. How-
`
`ex.
`it is the only source of CO,?", which has a very low concentration in the solution.
`Our calculations thustell us that in a solution of carbon dioxide in water most of
`CO, is in the form of CO, or HCO, a small fraction ionizes to form H* and HCO,”,
`
`37 M. Thee
`»d an even smaller fraction ionizes to give CO”.
`
`i TICE EXERCISE
`
`culate the pH and concentration of oxalate ion, [C,0,7"], ina 0.020 M solution of ox-
`acid, H,C,O,(see Table 16.3). Answers: pH = 1.80; [C,.07°] =64 x 10° M
`
`
`
`ats, Ka ané i
`».7 Weak Bases
`ican bea
`2a money
`
`he equili==
`s=y substances behave as weak bases in water. Such substances react with
`
`wer. removing protons from H,O,thereby forming the conjugate acid of the
`sand OH" ions:
`
`Weak base + H,O —— conjugate acid + OH™
`
`2
`most commonly encountered weak base is ammonia:
`ae
`[16.31]
`NH,(aq) + H,O(1) —=NH," (aq) + OH(aq)
`
`aie
`The equilibrium-constant expressionfor this reaction can be written as
`
`
`[NH,"][OH]aRarora16.32
`[NEGO]
`i602
`
`suse the concentration of water is essentially constant, the [H,O]term is in-
`0x 107
`»orated into the equilibrium constant, giving
`
`-
`mation
`+
`
`[NH,"][OH™]
`[16.33]
`=
`REO) =
`16.33
`K, = K{H,O] = ——.—_
`
`The constant K, is called the base-dissociation constant, by analogy with the
`
`-—cissociation constant, K,, for weak acids. The constant K, always refers to the equi-
`
`
`
`[COs(ah
`
`[16.30]
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`616=Chapter 16 / Acid-Base Equilibria
`
` ha Notice tha
`2
`;
`Ubrium-co1
`Lewis
`Conjugate
`expression
`Structure
`Acid
`H—N—H
`NH,"
`|H
`
`Equilibrium Reaction
`NH, + HO == NH,* + OH™
`
`K,
`1.8 x 107
`
`Base
`Ammonia
`(NH,)
`
`Pyridine
`
`(C5H5N)
`Hydroxylamine
`(H,NOH)
`
`Methylamine
`(NH;CH,)
`
`Hydrosulfide ion
`
`(HS")
`
`Carbonate ion
`(CO;*)
`Hypochlorite ion
`(ClO’)
`
`C) N:
`
`C;H;NH*
`
`C;H;N + H,O —C,H;NH* + OH™
`
`1.7 x 107
`
`H—N—GH
`|H
`
`H—N—CcH,
`|
`a
`ep
`[He
`‘O:
`|
`Pa,
`0:
`4
`[:i—G: 1
`
`2=
`
`H;NOH™*
`
`H,NOH + H,O == H;NOH* + OH-
`
`11x10"
`
`NH;CH;*
`
`NH,CH, + H,;O == NH,CH,* + OH™
`
`44x10"
`
`HLS
`
`HCO,”
`
`HCIO
`
`HS” +H,O——H,S + OH
`
`,
`
`1.8 X10
`
`CO,- + HO == HCO,- + OH™
`
`18x10"
`
`ClO- + HO ==HCIO + OH-
`
`3.3 x 107
`
`librium in which a base reacts with H,O toform the conjugate acid and OH”. Tab
`16.4 A lists the names, formulas, Lewis structures, equilibrium reactions, and v=-
`ues of K, for several weak bases in water. Appendix D includes a more extensive l=
`Notice that these bases contain one or more lone pairs ofelectrons. A lonepair is nex
`essary to form the bond with H*. Notice also that in the neutral moleculesthe lon:
`pairs are on nitrogen atoms andthat the other bases are anions derived from wes:
`acids.
`|
`
`SAMPLE EXERCISE 16.14
`
`Because K,
`compared -
`0.15 M.The
`
`}
`Notice that
`0.15 M. Th
`
`PRACTICE
`Which of tl
`Son: pyrid:
`
`Types of V
`
`Sow can
`s>le to beh,
`“rst catego
`~air ofelec
`dudingall
`These subs
`mines. In
`tha bon
`\H. with
`CH.NH,).
`‘orming an
`
`H—I
`
`4
`
`Calculate the concentration of OH” in a 0.15 M solution of NHs.
`Solution We use essentially the same procedure here as used in solving problems
`involving the ionization of weak acids. Thefirst step is to write the ionization reaction
`and the corresponding equilibrium-constant (K,) expression:
`NH4(aq) + H,0(!) = NH,“(aq) + OH™(aq)
`[NH,*][OH™]
`at
`The chemi:
`K,= ie, °
`-
`+
`Wethen tabulate the equilibrium concentrations involved in the equilibrium: _ .ese
`NH,(aq)
`+
`H,O() —=NH,t(aq) + OH(aq)
`weak acids
`nite, NaCl¢
`on is alwa
`Ge CIOi
`quently, th
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`5 of 7
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`IPR2020-01045
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`Equilibrium
`
`(0.15 — x) M
`
`clo~
`
`
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`
`
`16.7 / Weak Bases
`
`617
`
`(Notice that we ignore the concentration of H;O because it is not involved in the equi-
`librium-constant expression.) Inserting these quantities into the equilibrium-constant
`expression givesthe following:
`
`K,
`
`(x)(x)
`[NH,"][OH 7]
`=, eeee
`INH]
`i=<2
`Because K;, is small, we can neglect the small amount of NH;that reacts with water, as
`compared to the total NH; concentration; that is, we can neglect x in comparison to
`2.15 M. Then we have
`
`“5
`
`L7 x 107
`
`e—
`
`=18xi0°
`1.1 10%
`
`015 1.8x10
`x? = (0.15)(1.8 x 1075) = 2.7 x 1078
`x = [NH,*] = [OH”] = V2.7 x 10°° = 1.6 x 103M
`
`44% 10"
`
`1.8 x 107
`
`1.8 x 10™
`
`Notice that the value obtained for x is only about 1 percent of the NH; concentration,
`2.15 M.Therefore, our neglect of x in comparison with 0.15 is justified.
`
`PRACTICE EXERCISE
`Whichof the following compounds should produce the highest pH as a 0.05 M solu-
`Son: pyridine, methylamine, or nitrous acid? Answer: methylamine
`
`33x 107
`
`Types of Weak Bases
`“owcan werecognize from a chemical formula whether a molecule or ion is
`sble to behave as a weak base? Weak basesfall into two general categories. The
`“est category contains neutral substances that have an atom with a nonbonding
`d OH. Ta>=
`»air of electrons that can serve as a proton acceptor. Mostof these bases, in-
`“uding all the uncharged baseslisted in Table 16.4, contain a nitrogen atom.
`ions, and va~
`extensive “=
`These substances include ammonia anda related class of compoundscalled
`emines. In organic amines, one or more of the N—H bondsin NH,is replaced
`ne pair is ne=
`cules the lor
`sth a bond betweenNand C. Thus, the replacement of one N—H bond in
`od from wea:
`“H, with a N—CH; bond gives methylamine, NH,CH;(usually written
`°H.NH,). Like NH, amines can extract a proton from a water molecule by
`‘ermingan additional N—H bond,as shownhere for methylamine:
`
`Acids and Bases
`simulation
`
`ng problems
`ition reaction
`
`iilibrium:
`
`OH(aq)
`
`H
`-
`|
`area +100 — re (ag) + OH-(aq)
`
`H
`
`H
`
`[16.34]
`
`The chemical formula for the conjugate acid of methylamine is usually written
`-H;NH;".
`The second general category of weak bases is composedof the anionsof
`weak acids. Consider, for example, an aqueous solution of sodium hypochlo-
`=te, NaClO. This salt dissolves in water to give Na* and ClO” ions. The Na’
`on is always a spectatorion in acid-basereactions.
`(Section 4.3) However,
`*e ClO” ion is the conjugate base of a weak acid, hypochlorous acid. Conse-
`guently, the ClO™ ion acts as a weak basein water:
`C1O~(aq) + H,O(1) == HC1O(aq) + OH~(aq)
`
`KK, = 3.3 X 10°? [16.35]
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`618
`
`Chapter 16 / Acid-Base Equilibria
`
`
`
`SAMPLE EXERCISE 16.15
`a ct
`A solution is made by adding solid sodium hypochlorite, NaClO, to enough water =
`| make 2.00Lof solution.If the solution has a pH of10.50, how many moles of NaC”
`“fany amine
`were addedto the water?
`“Sshy” odor
`»>ic (absenc
`matter. Two
`FLN(CH,),»
`“mown as ca
`Many d
`emphetamir
`mines, thes
`eenis readily
`sssulting prc
`»>>reviation
`woth hydrocl
`semetimes 1
`“eochloride.
`Se acid salt.
`
`Solution NaClOis an ionic compoundconsisting of Na* and C1O™ions. As such ~
`_ isa strong electrolyte that completely dissociatesin solution into Na*, whichis a spe
`tator ion, and CIO” ion, whichis a weak base with K, = 3.3 X 1077 (Equation 16.35
`We wish to determine the concentration of CIO” in solution that would geners=
`enough OHionto raise the pH to 10.50.
`Wefirst calculate the concentration of OH(aq) at equilibrium. We can calcules
`[OH™] by using either Equation 16.14 or Equation 16.17; we will use the lats=
`method here:
`
`
`
`
`
`pOH = 14.00 — pH = 14.00 — 10.50 = 3.50
`[OH] = 10739 = 3.2 x 10°*M
`This concentration is high enough that we can assumethat Equation 16.35 is the on™»
`source of OH”; that is, we can neglect any OH” produced by the autoionization —
`H,0. We now assumea valueofx for the initial concentration of CIO™ and solve ==
`_ equilibrium problem in the usual way:
`
`Oy
`ClO™(aq)
`+ HO) == HC1O(aqg)
`+ OH(aq)
`
`i)ae|
`
`+3.2 10-4
`
`
`
`
`
`
`3.2 107+
`
`
`
`
`
`We now use the expression for the base-dissociation constantto solve for x:
`_ [HCIO][OH7] _
`(32x 10-*?
`[Cclo7]
`x= 32x 10%
`
`= 3.3 x 1077
`
`K,
`
`Thus,
`
`_ G2 x 10-4
`3.3.x 1077
`
`+ (3.2 x 1074) = 031M
`
`Wesay that the solution is 0.31 M in NaClO, even though someof the ClO™ iow
`have reacted with water. Becausethe solution is 0.31 M in NaClO andthetotal vo
`umeof solution is 2.00 L, 0.62 mol of NaClO is the amountof thesalt that was ada=-
`to the water.
`
`PRACTICE EXERCISE
`A solution of NH; in water has a pH of 10.50. Whatis the molarity of the solutios”
`| Answer: 0.0058 M
`
`ech of these
`
`ow notice si
`wed 16.37 are
`wth just the z
`
`To determ
`ded reactio,
`venciples gov
`ordreaction, t
`es equilibrium
`
`16.8 Relationship Between K, and K,
`
`We've seenin a qualitative way that the stronger acids have the weaker com
`gate bases. Thefact thatthis qualitative relationship exists suggests that we nig
`be ableto find a quantitative relationship. Let’s explore this matter by consid
`ing the NH,” and NH;conjugate acid-base pair. Each of these species reac
`with water:
`
`NH,"(aq) == NH,(aq) + H*(aq)
`
`[16.3
`
`
`
`[1625
`NH;(aq) + H,O(1) —— NH," (aq) + OH™ (aq)
`
`
`7 of7
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`IPR2020-01045
`Teva Ex. 1016
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`7 of 7
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`IPR2020-01045
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