`
`Brown LeMay Bursten
`
`CHEMISTRY
`
`lSE‘d Eighth Ed.UOfl
`
`
`
`Mylan (IPR2020-00040) Ex. 1016 p. 001
`
`

`

`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
`
`
`
`
`
`Mylan (IPR2020-00040) Ex. 1016 p. 002
`
`

`

`1
`
`. C
`__ .1 _
`~1- 4-
`._. 'h'
`"‘
`13‘
`
`Editor: John Challice
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`
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`
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`
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`Printed in the United States of America
`10
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`
`
`
`Mylan (IPR2020-00040) Ex. 1016 p. 003
`
`

`

`
`
`
`
`x2 = (0.0037)(4.3 X 10‘?) = 1.6 X 10—9
`Solving for x, we have
`x = [Pr] = [HCO3‘] = «11.6 x 10—9 = 4.0 x 10‘5M
`
`16.7 r‘ Weak Bases
`
`615
`
`4'0 X 1‘."
`
`4.2 X 35'
`
`'
`
`
`
`The small value of 3: indicates that our simplifying assumption was justified. The
`pH is therefore
`
`pH = -log [H+] = —log (40 x 10-5} = 4.40
`
`If we were asked to solve for [COaz‘], we would need to use Kfl. Let’s illustrate
`first calculation. Using the values of [HCOf] and [H+] calculated above, and setting
`:COf‘] = y, we have the following initial and equilibrium concentration values:
`
`._
`
`HCOs‘iaq)
`
`v——‘ WW)
`
`+
`
`(3032—060
`
`Lply as “13.";
`:tprotorL;
`ht:
`
`
`
`
`(4.0 x 10-5 — y) M
`
`(4.0 x 10-5 + y) M
`
`zation}
`
`harming that y is small compared to 4.0 X 10—5, we have
`
`Kaz =
`
`IH-‘Hcoatl
`[HCOa‘]
`
`=
`
`-u
`
`
`
`
`
`
`
`(4.0 x 10—59(3)
`< 10“?
`
`4.0 x 10—5
`= 5.6 x 10
`.1 constants '
`_
`-
`as from the '
`
`y = 5'6 x 10 11 M = [C032 ]
`factor of 22‘
`nae value calculated for y is indeed very small in comparison to 4.0 X 10—5, showing
`[301},me
`
`2 our assumption was justified. It also shows that the ionization of HC03‘ is neg-
`hide in comparison to that of H2C03 as far as production of H“ is concerned. How-
`
`-
`-.
`it is the only source of C0321 which has a very low concentration in the solution.
`Our calculations thus tell us that in a solution of carbon dioxide in water most of
`
`C02 is in the form of C02 or H1C03, a small fraction ionizes to form H+ and HCOS',
`, ‘ an even smaller fraction ionizes to give C03 _.
`
`37M. The .
`f carbonic -
`CTICE EXERCISE
`
`'mlate the pH and concentration of oxalate ion, [©0421, in a 0.020 M solution of ox-
`
`" acid, Hzczo4 (see Table 16.3}. Answcrs: pH = 1.80; [(220421 = 6.4 x 10‘5M
`
`
`
`
`1‘2..ng _
`." Weak Bases
`
`0
`”M
`x M
`
`[16.30]
`
`[NHHIOH‘]
`K =W [16.32]
`
`
`he 8 '
`-
`--
`substances behave as weak bases in water. Such substances react with
`
`---*r. removing protons from H20, thereby forming the conjugate acid of the
`
`and OH‘ ions:
`
`Weak base + H20 = conjugate acid + OH‘
`most commonly encountered weak base is ammonia:
`
`[16.31]
`NH3(sq) + H200) = NI-I4+(sq) + 0H*(aq}
`
`The equilibrium-constant expression for this reaction can be written as
`
`
`__0 x 10-5
`
`mation ..- ‘
`
`
`
`- sse the concentration of water is essentially constant, the [H20] term is in-
`. stated into the equilibrium constant, giving
`[NH4+][OH-I
`{NHal
`
`K, = K[H20] =
`
`[16.33]
`
`Toe constant Kb is called the base-dissociation constant, by analogy with the
`association constant, K“, for weak acids. The Wawfmmjotfimm EX. 101 6 p_ 004
`
`
`
`Mylan (IPR2020-00040) Ex. 1016 p. 004
`
`

`

`616
`
`Chapter 16 {Acid-Base Equilibria
`
`
`
` ail-v. .
`
`Base
`
`Ammonia
`
`(NHs)
`
`Pyridine
`(CsHsN)
`
`Hydroxylamine
`(HzNOHJ
`
`Methylamine
`(NHZCHB)
`
`Hydrosulfide ion
`(H5?)
`
`Carbonate ion
`
`(C032?
`
`Hypochlorite ion
`(ClO‘)
`
`Lewis
`Structure
`
`H—I'il—H
`H
`
`Conjugate
`Acid
`
`NH;
`
`Equilibrium Reaction
`
`NH3 + H20 :NH; + OH‘
`
`Kg,
`
`1.3 x 10'
`
`‘
`
`< O N:
`
`C5H5NH'
`
`C5HSN + H20 :CSH5NH+ + OH“
`
`1.7 x 10--
`
`H—fTT—QH
`H
`
`H—I'xi—CH3
`H
`
`H3NOH‘
`
`HZNOH + H20 =H3NOH+ + on-
`
`1.1 x 10.
`
`'
`
`NH3CH3*
`
`NHZCHa + H20 = NH3CH3+ + 0H"
`
`4.4 x 10--
`
`[H—{sfl—
`
`H23
`
`HS" + H20: H25 + 0H-
`
`1.8 x 11‘"
`
`2—
`
`:'C|3:
`__ /C\\ -.
`-.Q.
`.9
`
`Hcog—
`
`cof- + H20 : HCOS" + OH"
`
`13 x 19'-
`
`|::¢'1—0::|_
`
`HCIO
`
`00* + H20 =Hc10 + OH‘
`
`3.3 x 1::-
`
`‘
`
`liimium in which a base reacts with H20 taform the conjugate acid and OH‘. Tab:
`16.4 A lists the names, formulas, Lewis structures, equilibrium reactions, and vs..-
`ues of Kb for several weak bases in water. Appendix D includes a more extensive 13-.
`Notice that these bases contain one or more lone pairs of electrons. A lone pair is rte:-
`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.
`
`SAMPLE EXERCISE 16.14
`
`Calculate the concentration of OH" in a 0.15 M solution of LII-I3.
`
`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 (Kb) expression:
`
`NH3(aa) + H200) = NHJQaq} + OH'(aa)
`
`Kb
`
`2 [NHFHOH‘I
`[NH3]
`
`= 1.8 x 10—5
`
`We then tabulate the equilibrium concentrations involved in the equilibrium:
`
`NH3{aa)
`
`+
`
`H200) : NH4+[aq) + OH‘Iaa}
`
`
`
`[Notice tha
`libriurn—cor
`
`expression
`
`Because Kb
`compared -
`0.15 M. Th:
`
`Notice that
`0.15 M. Th:
`
`PRACTICE
`Which of ti
`
`Son: pyrid:
`
`Types of 11
`How can in
`role to beh.
`
`i-st catego
`:air of elet
`finding all
`These subs
`amines. In
`3 3th a bon
`
`RH; with
`CH3NH2).
`smiling an
`
`the chemi:
`
`CHQIHJ.
`The so
`weak acids
`
`ate, NaClC
`Dr} is alwa
`fee 00‘ it
`
`raently, thl
`
`C10"
`
`
`
`Mylan (IPR2020-00040) Ex. 1016 p. 005
`
`

`

`
`
`16.7 ,e’ Weak Bases
`
`617
`
`(Notice that we ignore the concentration of H20 because it is not involved in the equi-
`librium-constant expression.) Inserting these quantities into the equilibrium-constant
`expression gives the following:
`
`Kb:
`
`(WI)
`[NHIIIOH—I
`—— = —— = 1. x
`[NH3]
`0.15 — x
`8
`
`10
`
`-5
`
`Because K, 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
`
`.12
`——= . x
`0.15
`18
`
`‘5
`
`10
`
`x2 = (0.15)(1.8 x 10“) = 2.7 x 10-6
`
`x = [NI-14+] = [OH’] = m x 10'6 = 1.6 x 10‘3M
`
`Notice that the value obtained for x is only about 1 percent of the NH; concentration,
`3.15 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-
`s'on: pyridine, methylamine, or nitrous acid? Answer: methylamine
`
`1.7X1C '
`
`1.1 X 1C '
`
`4.4 X If"
`
`1.8X1C -
`
`1.8><1C '
`
`3.3 X 1'."-
`
`Types of Weak Bases
`
`d OH‘. Table
`
`ions, and vs.-
`extensive is.
`
`ne pair is he;-
`cules the lore
`ad from wees
`
`ng problems
`ition reacticr.
`
`Jilibriurn:
`
`01-1—03?)
`
`How can we recognize from a chemical formula whether a molecule or ion is
`xle to behave as a weak base? Weak bases fall into two general categories. The
`test 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-
`;.-_:ding all the uncharged bases listed in Table 16.4, contain a nitrogen atom.
`.‘rtese substances include ammonia and a related class of compounds called
`mines. In organic amines, one or more of the N—H bonds in NH3 is replaced
`a 1th a bond between N and C. Thus, the replacement of one N-H bond in
`‘ii—I; with a N—CH3 bond gives methylamine, NHZCHs (usually written
`:E-I;NH2). Like NH3, amines can extract a proton from a water molecule by
`:rming an additional N—H bond, as shown here for methylamine:
`
`‘ Acids and Bases
`simulation
`
`H
`..
`|
`H—hll—CHafizq) + H200) = H—I\Ii-CH3 (sq) + OH‘Qflq)
`
`H
`
`H
`
`[16.34]
`
`The chemical formula for the conjugate acid of methylamine is usually written
`:HNHJ.
`The second general category of weak bases is composed of the anions of
`weak acids. Consider, for example, an aqueous solution of sodium hypochlo—
`:te. NaClO. This salt dissolves in water to give Na+ and C10" ions. The Na+
`no is always a spectator ion in acid—base reactions.
`:Sectinn 4.11 However,
`:e C10“ ion is the conjugate base of a weak acid, hypochlorous acid. Conse-
`.jsently, the C10‘ ion acts as a weak base in water:
`
`Clo—(fig) + H20(1)=HC10(“q) + OH‘WIanfiFfiiOfifiEOBES?EX. 1016 p. 006
`
`-
`
`A
`
`Mylan (IPR2020-00040) Ex. 1016 p. 006
`
`

`

`618
`
`Chapter 16 f Acid-Base Equilibria
`
`
`
`
`
`
`
`SAMPLE EXERCISE 16.15
`
`A solution is made by adding solid sodium hypochlorite, NaClO, to enough water r.
`make 2.00 L of solution. If the solution has a pH of 10.50, how many moles of NaCJC!
`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 1', which is a spec-
`tator ion, and CIO‘ ion, which is a weak base with Kb = 3.3 X 10‘7 (Equation 16.33.
`We wish to determine the concentration of ClO‘ in solution that would gene-rm
`enough OH‘ ion to raise the pH to 10.50.
`We first calculate the concentration of OH‘{aq) at equilibrium. We can calculz:
`[OH'] by using either Equation 16.14 or Equation 16.17; we will use the lam:
`method here:
`
`pOH = 14.00 — pH = 14.00 — 10.50 = 3.50
`
`[OI-1‘] = 10‘350 = 3.2 x 10-4M
`
`This concentration is high enough that we can assume that Equation 16.35 is the «is
`source of OH'; that is, we can neglect any OH‘ produced by the autoionization :1'
`H20. We now assume a value of x for the initial concentration of C10' and solve £1
`E equilibrium problem in the usual way:
`
`C10'(aq)
`+
`H200) = HClO(aq)
`
`——
`
`s... "—
`-—— -
`
`+ OH_(aq)
`
`
`
`
`
`We now use the expression for the base-dissociation constant to solve for x:
`
`K.
`
`_ [HClOJIOH‘] m (3.2 x 10*4)2
`_
`[Clo-1
`' 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 0.31 M in NaClO and the total \‘ce-
`. tune of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was adder.
`' to the water.
`
`I PRACTICE EXERClSE
`
`A solution of NH3 in water has a pH of 10.50. What is the molarity of the soluticr.‘
`' Answer: 0.0058 M
`
`16.8 Relationship Between Kn and Kl,
`
`Ad
`
`Many amine
`fishy” odor
`.i-ic (abseno
`matter. Two
`
`H:N(CH2}41\
`known as re
`
`Many d
`mphetamil
`guinea, thes
`gas is readily
`asulting pn
`fitneviation
`
`nth hydrocl
`sometimes i
`:‘rochloride.
`the acid salt .
`
`@.
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`
`Each of these
`
`New notice 54
`mi 16.37 are
`
`rah just the e
`
`To deterrr.
`aided reactio:
`
`;t-_nciples gov
`:icvj reaction, i
`n: equilibrium
`
`We’ve seen in a qualitative way that the stronger acids have the weaker con".—
`gate bases. The fact that this qualitative relationship exists suggests that we mar.
`be able to find a quantitative relationship. Let’s explore this matter by conside—
`ing the NH: and N113 conjugate acid—base pair. Each of these species rear:
`with water:
`
`
`N'H3{aq) + H200) =NH4+(aq) + OH‘(aq)
`
`NI-I4+(aq)=NI-13(aq) + H""(aq)
`
`[16.3w
`
`
`
`
`[16?"
`
`Mylan (IPR2020-00040) Ex. 1016 p. 007
`
`