For Advanced
`High School
`Chemistry
`
`Brown LeMay Bursten
`
`CHEMISTRY
`
`THE CENTRAL SCIENCE
`Revised Eighth Edition
`
`DRL Ex. 1016, p. 001
`
`

`

`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
`
`DRL Ex. 1016, p. 002
`
`

`

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`
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`
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`
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`
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`
`DRL Ex. 1016, p. 003
`
`

`

`rotic Adds
`KaJ
`
`4.0 X:
`4.2 X:
`
`tply as '
`;tproton
`ht:
`zation)
`< w-z.
`
`37M. T.
`f carbor.
`
`~a mo"
`heequ;.
`
`x M
`
`.0 X 1
`matior
`
`16.7 I Weak Bases
`
`615
`
`x2 = (0.0037)(4.3 X 10- 7) = 1.6 X 10- 9
`lving for x, we have
`x = [H- ] = [HC03 - ] = ...J1.6 X 10- 9 = 4.0 X 10- 5 M
`The small value of x indicates that our simplifying assumption was justified. The
`'li is therefore
`
`pH= -log [H+] = -log (4.0 X 10- 5) = 4.40
`If we were asked to solve for [Col-], we would need to use l<,a,. Let's illustrate
`.tt calculation. Using the values of [HC03 -] and [H+] calculated above, and setting
`0 32-1 = y, we have the following initial and equilibrium concentration values:
`+
`
`0
`+yM
`yM
`
`4.0X 10-5M
`
`"<z2
`
`4.0x10- s M
`Initial
`+yM
`-yM
`Change
`(4.0 X 10- 5 + y) M
`(4.0x10- S-y)M
`Equilibrium
`-uming that y is small compared to 4.0 X 10- 5, we have
`v = [H. ][COl- 1 = (4.0 X 10-5)(y) = 56 X to- n
`4.0 X to-s
`.
`[HC03 -1
`y = 5.6 X 10-11 M = [COl- 1
`alue calculated for y is indeed very small in comparison to 4.0 X 10- 5, showing
`our assumption was justified. It also shows that the ionization of HC03- is neg-
`•• e in comparison to that of H2C03 as far as production of H .. is concerned. How-
`2-, which has a very low concentration in the solution.
`• it is the only source of C03
`Our calculations thus tell us that in a solution of carbon dioxide in water most of
`.::o2 is in the form of c~ or H2C03t a small fraction ionizes to form H+ and HC03- I
`an even smaller fraction ionizes to give col-.
`TICE EXERCISE
`.. date the pH and concentration of oxalate ion, [~ol-J, in a 0.020 M solution of ox-
`cid, H 2C20 4 (see Table 16.3). Answers: pH = 1.80; [C20l- l = 6.4 X 10- 5 M
`
`[16.30]
`
`-Weak Bases
`substances behave as weak bases in water. Such substances react with
`removing protons from H20, thereby forming the conjugate acid of the
`ndOH- ions:
`Weak base + H20 ~conjugate acid + OH -
`-ost commonly encountered weak base is ammonia:
`NH3(nq) + H 20(l) ~ Nfi.t -(nq) + OH- (aq)
`[16.31]
`~ "e equilibrium-constant expression for this reaction can be written as
`K = [NH4 +][oH-]
`[NH3][HzO]
`..e the concentration of water is essentially constant, the [H20] term is in-
`·ated into the equilibrium constant, giving
`K = K[H 0 ] = [NH4 +][oH- ]
`[NH3]
`2
`b
`--e constant Kb is called the base-dissociation constant, by analogy with the
`~sociation constant, K.v for weak acids. The constant K, always refers to the equi-
`
`[16.32]
`
`[16.33]
`
`DRL Ex. 1016, p. 004
`
`

`

`616
`
`Chapter 16 I Acid-Base Equilibria
`
`TABlE 16.4 Some Weak Bases and Their Aqueous Solution Equilibria
`Lewis
`Conjugate
`Structure
`Acid
`H-N-H
`
`Equilibrium Reaction
`
`NH4-
`
`1.8 X 10
`
`Base
`
`Ammonia
`(NH3)
`
`Pyridine
`(CsfisN)
`Hydroxylamine
`(H2NOH)
`
`Methylamine
`(NH2CH3)
`
`Hydrosulfide ion
`(HS-)
`
`Carbonate ion
`(C032- )
`
`Hypochlorite ion
`(CIO )
`
`I H @:
`
`CsHsNH-
`
`H3NOH-
`
`CsHsN + H20 ;=:: CsHsNH + + OH-
`
`1.7 X lC
`
`H2NOH + H20 ;=:: H~OH+ + OH-
`
`1.1 X l C
`
`NH3CH3+
`
`NH2CH3 + H20 ;=:: NH3CH3 + + OH-
`
`4.4 X lC
`
`..
`H-N-OH
`I
`H
`H-N-CH
`I
`3
`H
`[H-~J
`·o·
`:.(?:'~~9·
`..
`[c'l-o~-
`..
`
`H2S
`
`l ·· r
`
`HClO
`
`Hs-- + H20 ;=:: H2S + OH-
`
`1.8 X lC
`
`1.8 X lC
`
`3.3 X lC
`
`librium in which a base reacts with H20 to form the conjugate acid and OW. Tc0
`16.4 .&. lists the names, formulas, Lewis structures, equilibrium reactions, and ,.
`ues of~ for several weak bases in water. Appendix D includes a more extensi\·e :_
`Notice that these bases contain one or more lone pairs of electrons. A lone pair is oc
`essary to form the bond with H+. Notice also that in the neutral molecules the lc-
`pairs are on nitrogen atoms and that the other bases are anions derived from we::
`acids.
`
`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 probler.
`involving the ionization of weak acids. The first step is to write the ionization reactic-
`and the corresponding equilibrium-constant (Kh) expression:
`NHJ(aq) + H20( I);=:: NH/ (aq) + OH- (aq)
`K = [~ ... ][oH- ] = 1 8 x lo-s
`[NH3]
`.
`b
`We then tabulate the equilibrium concentrations involved in the equilibrium:
`
`Initial
`Change
`Equilibrium
`
`0.15M
`- xM
`(0.15 - x)M
`
`-
`-
`-
`
`0
`+xM
`xM
`
`0
`+xM
`xM
`
`'Jotice tha
`·brium-cor
`expression
`
`~auseKb
`~ompared ·
`15M. The
`
`... otice that
`15 M. The
`
`lACTICE
`ruch of tl
`"'0: pyrid:
`
`--pes of V
`w can\\
`-.e to beh
`,t catego
`.r of ele<
`d ing all
`ese subs
`-tines. In
`th a bon
`--1~ with
`'i-:--TH2).
`-ming an
`
`H-i
`
`e chemic
`1-! ,_.''-.JH 3 + .
`These·
`.ak acid~
`e N aCK
`is alwa
`~ oo- i·
`_ently, th
`
`CIO
`
`DRL Ex. 1016, p. 005
`
`

`

`1 6.7 I Weak Bases
`
`617
`
`:--Jotice that we ignore the concentration of H 20 because it is not involved in the equi-
`i brium-constant expression.) Inserting these quantities into the equilibrium-constant
`expression gives the following:
`
`Because~ is small, we can neglect the small amount of NH3 that reacts with water, as
`-ompared to the total NH3 concentration; that is, we can neglect x in comparison to
`15M. Then we have
`
`x2
`0.15 = 1.8 X 10-5
`r = (0.15)(1.8 X 10- 5) = 2.7 X 10-6
`x = (~ +] = (OH- ] = -./2.7 X 10- 6 = 1.6 X 10- 3 M
`otice that the value obtained for xis only about 1 percent of the NH3 concentration,
`15M. Therefore, our neglect of x in comparison with 0.15 is justified.
`
`'t4CTICE EXERCISE
`ruch of the following compounds should produce the highest pH as a 0.05 M solu-
`~= pyridine, methylamine, or nitrous acid? Allswer: methylamine
`
`- . pes of Weak Bases
`w can we recognize from a chemical formula whether a molecule or ion is
`• .e to behave as a weak base? Weak bases fall into two general categories. The
`"t category contains neutral substances that have an atom with a nonbonding
`.r 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.
`~ese substances include ammonia and a related class of compounds called
`:nines. In organic amines, one or more of the N-H bonds in NH3 is replaced
`:11 a bond between N and C. Thus, the replacement of one N- H bond in
`"11 with a N-CH3 bond gives methylamine, NH2CH3 (usually written
`"1.._'\IH~. Like NH3, amines can extract a proton from a water molecule by
`-ming an additional N-H bond, as shown here for methylamine:
`
`~e chemical formula for the conjugate acid of methylamine is usually written
`;,NH3+·
`The second general category of weak bases is composed of the anions of
`.. ak adds. Consider, for example, an aqueous solution of sodium hypochlo-
`e, NaClO. This salt dissolves in water to give Na+ and oo- ions. The Na+
`~is always a spectator ion in add-base reactions.
`However,
`e ClO- ion is the conjugate base of a weak add, hypochlorous acid. Conse-
`.ently, the ClO- ion acts as a weak base in water:
`
`Kb = 3.3 X 10- 7 [16.35]
`
`1.8 X 1 ..
`
`1.7 X 1
`
`1.1 X 1
`
`4.4 X 1
`
`1.8 X 1
`
`1.8 X 1
`
`3.3 X 1
`
`doH- . T.,
`ions, and
`extensive
`nepair isr
`:ulesthe ·
`!d from , ..
`
`ng probler-
`ttion reacb
`
`1ilibrium:
`
`+xM
`xM
`
`DRL Ex. 1016, p. 006
`
`

`

`618
`
`Chapter 16 I Acid-Base Equilibria
`
`I SAMPLE EXERCISE 16.15
`
`I
`
`A solution is made by adding solid sodium hypochlorite, NaCIO, to enough wat~
`make 2.00 L of solution. If the solution has a pH of 10.50, how many moles of NaC
`were added to the water?
`Solution NaClO is an ionic compound consisting of Na + and CIO- ions. As sue::
`is a strong electrolyte that completely dissociates in solution into Na-, which is a ~
`tator ion, and ClO- ion, which is a weak base with Kb = 3.3 X 10-7 (Equation 16 --
`We wish to determine the concentration of CIO- in solution that would gener
`enough OH- ion to raise the pH to 10.50.
`We first calculate the concentration of OH- (aq) at equilibrium. We can calcu..
`[OH- ] by using either Equation 16.14 or Equation 16.17; we will use the Ia
`method here:
`
`pOH = 14.00 - pH = 14.00 - 10.50 = 3.50
`[OH- ] = 10- 3.so = 3.2 X 10- 4M
`This concentration is high enough that we can assume that Equation 16.35 is the c-
`source of OH-; that is, we can neglect any OH- produced by the autoionizatio:-
`H20. We now assume a value of x for the initial concentration of oo- and solw
`equilibrium problem in the usual way:
`
`Initial
`Change
`Final
`
`xM
`- 3.2 X10- 4M
`(x - 3.2 X 10- 4) M
`
`-
`-
`-
`
`0
`+3.2x1o- 4
`3.2 X10- 4
`
`0
`+ 3.2 x w-,
`3.2 x l0-,
`
`We now use the expression for the base-dissociation constant to solve for x:
`Kb = [HCIO][OH- ]
`(3.2 X 10- 4)2 = 3 3 X 10-7
`X - 3.2 X 10- 4
`[CIO- ]
`•
`
`Thus,
`
`4f + (3.2 X 10-~) = 0.31 M
`x = (3.2 X l0-
`3.3 x w-7
`We say that the solution is 0.31 M in NaClO, even though some of the CIO-.
`have reacted with water. Because the solution is 0.31 Min NaCIO and the total
`ume of solution is 2.00 L, 0.62 mol of NaCIO is the amount of the salt that was ad.::
`to the water.
`I
`PRACTICE EXERCISE
`A solution of NH3 in water has a pH of 10.50. What is the molarity of the solut
`Answer: 0.0058 M
`
`16.8 Relationship Betw·een K 7 and Kv
`We've seen in a qualitative way that the stronger acids have the weaker co:
`gate bases. The fact that this qualitative relationship exists suggests that we rr_,
`be able to find a quantitative relationship. Let's explore this matter by consic
`ing the NH4 + and NH3 conjugate acid-base pair. Each of these species re;:
`with water:
`
`NH/(aq) ;:::=:! NH3(aq) + H +(aq)
`NH3(aq) + H20(l);:::=:!NH4+(aq) + OH- (aq)
`
`Ct
`ny aminE
`-hy" odor
`..:(abseno
`:ter. Two
`'\(CH2)41'
`wnas ca
`Manyd
`phetamil
`~es, thes
`., is readil)
`o!lting pr<
`"reviation
`"lhydrocl
`'etimes'
`.chloride.
`acid salt
`
`]/-c
`
`~of these
`
`notice S•
`16.37 are
`just thee
`
`To determ
`ed reactio:
`-dples gov
`i reaction, t
`11lilibrium
`
`DRL Ex. 1016, p. 007
`
`