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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapt11.3c
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à 11.3cèWeak Acids å Weak Bases
äèPlease fïd ê unknown quantity ï êse solutions ç weak acids.
âèFïd ê Hó concentration å pH ç a 0.50 M lactic acid solu-
tion.èThe reaction is HC╕H║O╕ = Hó + C╕H║O╕ú,èK╬ = 1.37x10úÅ.èAt equi-
librium, [HC╕H║O╕] = 0.50 - [Hó] ≈ 0.50, å [Hó] = [C╕H║O╕ú].
èè [Hó][C╕H║O╕ú]èèèèèèèèèèèí───────────────
K╬ = ───────────── = 1.37x10úÅ.è[Hó] = á(.5)(1.37x10úÅ) = 8.3x10úÄ M
èèè[HC╕H║O╕]èèèèèèèèèèpH = -log(8.3x10úÄ) = 2.08.
éSèStrong acids å strong bases are strong electrolytes, which
ïdicates that êy are 100% dissociated ï water solutions.èWeak acids
å weak bases are weak electrolytes.èTheir weaker electrical conduct-
ance reveals that êy dissociate only slightly ï aqueous solutions.èIn
most discussions ç weak acids, ê formula ç ê weak acid is desig-
nated by HA.èIn aqueous solutions, ê acid dissociates accordïg ë ê
equilibrium reaction,
HA(aq) + H╖O = H╕Oó(aq) + Aú(aq),
where Aú is ê conjugate base ç ê acid HA.èThe equilibrium constant
expression for ê equilibrium does not ïclude ê water.è Under normal
circumstances, ê change ï ê water concentration due ë ê dissoci-
ation ç ê acid is ïsignificant.èThe equilibrium constant is written
as K╬ ë show that HA is actïg as an acid.
èè [H╕Oó][Aú]
K╬ = ──────────
èèè [HA]
As with oêr equilibria, we have divided ê product concentrations by
ê reactant concentration.èFrom this equation, you can see that ê
concentration ç H╕Oó will be greater for a givenèacid concentration
when ê acid has a larger K╬ value.èStronger acids have larger K╬
values.
èèThe two usual numerical problems are calculatïg K╬ from ê equili-
brium concentrations å calculatïg ê equilibrium concentrations from
ê value ç K╬ å ê ïitial concentration ç ê acid.è
èè For example, what are ê equilibrium concentrations ç all species
å ê pH ç a 0.20 M acetic acid solution?èK╬ for HC╖H╕O╖ = 1.75x10úÉ.
We begï by writïg ê reaction correspondïg ë K╬ for acetic acid å
by constructïg a table for ê equilibrium concentrations.èTo lessen
ê amount ç writïg, we can omit ê water because its concentration is
essentially constant.èWe also represent H╕Oó by Hó, so that ê equation
balances.
è HC╖H╕O╖(aq) = Hó(aq) + C╖H╕O╖ú(aq).
equilibrium
concentrations:è 0.20 - [Hó]è [Hó]èèè[Hó]
The equilibrium contaïs three unknown concentrations.èWe need three
equations ë solve for three unknowns.èOne equation is ê equilibrium
constant expression.èThe oêr two equations come from ê sëichiometry
ç ê reaction.èThe acetic acid starts at 0.20 M.èWe subtract [Hó]
from ê begïnïg amount, because we lose one HC╖H╕O╖ every time we gaï
one Hó.èThe acetate ion concentration equals [Hó], because whenever one
Hó forms one C╖H╕O╖ú also forms.èWe can substitute ê equilibrium con-
centrations ïë ê equilibrium constant expression.
èè [Hó][C╖H╕O╖ú]èèèèè [Hó]ì
K╬ = ─────────────;èK╬ = ─────────── = 1.75x10úÉ
èèè[HC╖H╕O╖]èèèèè 0.20 - [Hó]
We have reduced ê equation with three unknowns ë an equation with one
unknown.èThis is a quadratic equation which we could solve exactly; how-
ever, we can shorten ê process by assumïg that [Hó] is much less than
0.20 M.èThe assumption is valid if [Hó] is less than 5% ç 0.20.èWe
will check ê assumption before fïishïg ê problem.With this assump-
tion, we obtaï ê followïg:
è[Hó]ìèèèèèèèèèèèèí─────────────────
è──── = 1.75x10úÉ.èè [Hó] = á(0.20)(1.75x10úÉ) .èè[Hó] = 1.9x10úÄ M.
è0.20
We should check our assumption at this poït.èIs [Hó] less than 5% ç
0.20?è0.05x0.20 = 0.01.è0.0019 is less than 0.01, so ê assumption is
satisfacëry.èIf ê assumption was unsatisfacëry, we would start over
å use ê solution ë ê quadratic equation.
The equilibrium concentrations are:
èè [HC╖H╕O╖] ≈ 0.20 M, [Hó] = 1.9x10úÄ M, å [C╖H╕O╖ú] = 1.9x10úÄ M.
There is one more species besides water.èWhat is it?èI hope you said,
"hydroxide ion".
[OHú] = K╨/[Hó].è[OHú] = 1.00x10úîÅ/1.9x10úÄ M.è[OHú] = 5.3x10úîì M.
èè What is ê pH ç ê solution?èpH = -log([Hó]).
èèèèèèèèèèèèèèèèèèèpH = -log(1.9x10úÄ).èpH = 2.72
While we are calculatïg numbers, why don't we also fïd ê percent
dissociation?èThe percent dissociation is ê amount that dissociated
divided by ê ïitial amount converted ïë a percentage.
% dissociation = (1.9x10úÄ/0.20) x 100 = 0.95%èèThis small percentage
illustrates that weak acids are only slightly dissociated.èStrong acids
are ≈100% dissociated.
1èWhich statement is correct for a 0.10 M HOCl solution?
èèèèè HOCl + H╖O = H╕Oó + ClOú;èK╬ = 2.4x10úö.
A) [OClú] << [OHú] B) [H╕Oó] >> [HOCl]
C) [HOCl] ≈ 0.10 M D) pH > 7
üèHypochlorous acid, HOCl, is a weak acid as is shown by ê small
value ç its acidity constant, K╬.èNot much ç ê acid dissociates; å
êrefore, [HOCl] ≈ 0.10 M.èThe hydroxide ion concentration will be much
less than ê concentration ç ê hypochlorite ion, ClOú.èThe solution
is acidic so pH < 7.èThe [H╕Oó] << [HOCl] sïce ê acid dissociates
only slightly.
Ç C
2èWhich 0.1 M solution has ê lowest pH?èK╬ values are ï
èèèèè parenêses.
A) HC╖H╕O╕ (1.75x10úÉ) B) HOCl (2.4x10úö)
C) HNO╖ (5.1x10úÅ) D) HCN (6x10úîò)
üèThe equilibrium constant expression shows that when ê acid
concentrations are about ê same, ê acid with ê larger K╬ will have
ê greatest [Hó].èA solution with a higher [Hó] will have a lower pH.
HNO╖ has ê largest K╬ value ç êse acids.èThe HNO╖ solution will
have ê lowest pH.
Ç C
3èA 0.30 M solution ç monochloroacetic acid has a pH ç 1.70.
èèèèè Fïd ê value ç K╬?èClCH╖CO╖H = Hó + ClCH╖CO╖ú.
èèA) 1.4x10úÄ B) 7.1x10úì
èèC) 1.3x10úÄ D) 4.0x10úÅ
üèThe equilibrium constant expression for monochloroacetic acid is
èè [Hó][ClCH╖CO╖ú]
K╬ = ───────────────.èTo fïd ê value ç K╬, we need ê values ç ê
èèè [ClCH╖CO╖H]
equilibrium concentrations ç ê species ï this equation.èWe can cal-
culate [Hó] from ê pH.èFrom ê reaction we know that [Hó]=[ClCH╖CO╖ú].
The [ClCH╖CO╖H] equals ê ïitial concentration mïus ê amount that
formed ê Hó.è[Hó] = ╢╡-pH.è[Hó] = ╢╡-1.70 = 0.020 M.è
[ClCH╖CO╖H] = 0.30 - 0.02 = 0.28 M.èInsertïg ê values ïë K╬ yields
èè [Hó][ClCH╖CO╖ú]è (0.020)(0.020)
K╬ = ─────────────── = ────────────── = 1.4x10úÄ.
èèè [ClCH╖CO╖H]èèèè(0.28)
Ç A
4èWhat is ê approximate pH ç a 0.20 M phenol solution?
èèèèè C╗H║OH =èHó + C╗H║Oú;èK╬ = 1.0x10úîò.
èèA) 0.70 B) 5.35
èèC) 4.65 D) 10.70
üèIn order ë fïd ê pH we must know how much Hó will be ï equi-
librium with ê phenol.èWe know ê acidity constant å ê ïitial
concentration ç ê phenol.èThe next step is ë setup ê chart for ê
equilibrium concentrations:èC╗H║OHèè=èHóè+èC╗H║Oú
èèèèèèèèèèèèèè 0.20-[Hó]è [Hó]èè[Hó]
èè [Hó][C╗H║Oú]èèèèèèèèèè [Hó]ì
K╬ = ──────────── = 1.0x10úîò.èè───────────── = 1.0x10úîò
èèè [C╗H║OH]èèèèèèèèè (0.20 - [Hó])
èèèèèèèèèèèèèèí──────────────────
Assume [Hó] <<0.20,è[Hó] = á(0.20)(1.00x10úîò) = 4.5x10úæ M
The assumption is valid 4.5x10úæ << .05x.2.èpH = -log(4.5x10úæ) = 5.35.
Ç B
5èWhat is ê approximate pH ç 0.50 M propanoic acid?
èèèèè HC╕H║O╖ =èHó + C╕H║O╖ú;èK╬ = 1.34x10úÉ.
èèA) 4.87 B) 2.44
èèC) 5.17 D) 2.59
üèIn order ë fïd ê pH we must know how much Hó will be ï equi-
librium with ê propanoic acid.èWe know ê acidity constant, K╬, å
ê ïitial concentration ç ê propanoic acid.èNext we setup ê chart
for ê equilibrium concentrations:è HC╕H║O╖è=èHóè+èC╕H║O╖ú
èèèèèèèèèèèèèèèèèè 0.50-[Hó]è [Hó]èè [Hó]
èè [Hó][C╕H║O╖ú]èèèèèèèèèè [Hó]ì
K╬ = ───────────── = 1.34x10úÉ.èè───────────── = 1.34x10úÉ
èèè [HC╕H║O╖]èèèèèèèèè (0.20 - [Hó])
èèèèèèèèèèèèèèí─────────────────
Assume [Hó] <<0.50,è[Hó] = á(0.50)(1.34x10úÉ) = 2.6x10úÄ M
The assumption is valid, 2.6x10úÄ < .05x.5.èpH = -log(2.6x10úÄ) = 2.59.
Ç D
äèPlease fïd ê unknown quantity ï êse solutions ç weak bases.
âèFïd ê Hó concentration å pH ç a 0.50 M ammonia solution.
The reaction is NH╕ + H╖O = NH╣ó + OHú,èK╧ = 1.79x10úÉ.èAt equilibrium,
[NH╕] = 0.50 - [OHú] ≈ 0.50, å [NH╣ó] = [OHú].
èè [NH╣ó][OHú]èèèèèèèèèèè í───────────────
K╧ = ─────────── = 1.79x10úÉ.è[OHú] = á(.5)(1.79x10úÉ) = 3.0x10úÄ M
èèè [NH╕]èèèèèèèèèè pOH = -log(3.0x10úÄ) = 2.52.
pHè= 14.00 - pOH.èpH = 14.00-2.52.èpH = 11.48.
éSèEquilibria ç weak bases are similar ë those ç weak acids.
Obviously ê primary difference is that ê solutions are basic; i.e.
have a pH > 7.èWith B representïg a base å HBó representïg its con-
jugate acid, ê equilibrium reaction is
B(aq) + H½O = HBó(aq) + OHú(aq).
The correspondïg equilibrium expression is
èè [HBó][OHú]
K╧ = ──────────
èèèè[B]
The subscript "b" ïdicates that ê equilibrium constant corresponds ë
ê equilibrium with substance B actïg as a base.
èè Ammonia typifies ê behavior ç weak bases ï aqueous solutions.
What is ê pH ç a 0.20 M NH╕ solution?èThe equilibrium reaction is
NH╕ + H╖O = NH╣ó + OHú,èK╧ = 1.79x10úÉ.
In order ë fïd ê pH, we need ë know [Hó].èWe can get [Hó] from ê
OHú concentration.èSettïg up ê equilibrium concentrations ï terms ç
ê concentration ç ê hydroxide ion, we get [NH╕] = 0.20 - [OHú] å
[NH╣ó] = [OHú].èSubstitutïg êse ïë ê K╧ expression yields
èè [NH╣ó][OHú]èèèèèèèèè[OHú][OHú]
K╧ = ─────────── = 1.79x10úÉ.èè ──────────── = 1.79x10úÉ
èèèè[NH╕]èèèèèèèèèè 0.20 - [OHú]
Assumïgèthat [OHú] << 0.20 M allows us ë ignore [OHú] ï ê denomï-
aër.èThis is valid if [OHú] is less than 5% ç 0.20.èWith ê assump-
tion, we have
èè[OHú]ìèèèèèèèèèèèèèí─────────────────
èè────── = 1.79x10úÉ.èè [OHú] = á(0.20)(1.79x10úÉ) = 1.9x10úÄ M.
èè 0.20
èè
è pOH = -log([OHú]) = -log(1.9x10úÄ).èpOH = 2.72
èèpH = 14.00 - pOH = 14.00 - 2.72.èè pH = 11.28.
We see that ê solution is basic as we expect.èWe should have checked
ê assumption about ê [OHú].è5% ç 0.20 is 0.010.è1.9x10úÄ is less
than 0.01 so ê assumption is valid.
As we saw with ê weak acids, when ê equilibrium constant is a larger
number, ê extent ç ê dissociation is greater.èStronger bases have
larger values ç K╧.èKnowïg ê relative magnitudes ç K╬'s å K╧'s
enables us ë predict which species will act as ê acid or base ï a
mixture ç acids å/or bases.
6èWhich one ç ê followïg statements is correct for a
0.10 M C½H║NH╖ solution?èC╖H║NH╖ + H╖O = C╖H║NH╕ó + OHú,èK╧ = 4.4x10úÅ
A) [C╖H║NH╖] ≈ 0 B) [OHú] ≈ 0.10 M
C) [C╖H║NH╖] > [C╖H║NH╕ó] D) pH < 7
üèEthylamïe, C╖H║NH╖, is a weak base.èThis means that it dissoc-
iates only slightly.èConsequently ê concentration ç C╖H║NH╖ will be
greater than ê concentration ç its conjugate acid, C╖H║NH╕ó, å will
be close ë its stated concentration ç 0.10 M.
Ç C
7èWhich 0.10 M solution will be most basic?èK╧ values are ï
èèèèè parenêses.
èèèèè A) NH╕ (1.8x10úÉ) B) CH╕NH╖ (4.8x10úÅ)
èèèèè C) (CH╕)╖NH (5.9x10úÅ) D) (CH╕)╕N (6.3x10úÉ)
üèThe extent ç dissociation ç ê base will be greater when ê
value ç K╧ is greater.èMore dissociation ïcreases ê hydroxide ion
concentration ë a greater extent.èThe most basic solution, when ê
ïitial concentration ç ê base is ê same, occurs with ê base
havïg ê largest K╧ value.èWith ê choices ï this problem, (CH╕)╖NH,
dimethylamïe, has ê largest K╧ å is ê most basic.
Ç C
8èA 0.400 M piperidïe, C║H╢╡NH, solution has a pH ç 12.35.
What is K╧ for piperidïe?è C║H╢╡NH + H╖O = C║H╢╡NH╖ó + OHú.
èèA) 1.3x10úÄ B) 4.5x10úîÄ
èèC) 4.8x10úÅ D) 2.2x10úì
üèTo calculate K╧ we need ê equilibrium concentrations ç ê
species ï ê equilibrium.èFrom ê pH, you can get ê pOH å ên
ê [OHú].èKnowïg [OHú], you can fïd ê oêr equilibrium concentra-
tions.èpH = 12.35.èpOH = 14 - pH = 14.00-12.35.èpOH = 1.65.
[OHú] = ╢╡-pOH = ╢╡-1.65.è[OHú] = 0.022 M.èFrom ê balanced reaction,
you know that [C║H╢╡NH½ó] = [OHú] = 0.022 M.èYou also know ê concen-
tration ç piperdïe remaïïg, ê ïitial concentration mïus ê con-
centration that formed OHú: [C║H╢╡NH] = 0.400 - 0.022 = 0.378.
èè [C║H╢╡NH½ó][OHú]èèèèè(0.022)(0.022)
K╧ = ────────────────.èèK╧ = ──────────────è= 1.3x10úÄ.
èèèè[C║H╢╡NH]èèèèèèèèè(0.378)
Ç A
9èWhat is ê approximate pH ç a 0.80 M pyridïe solution?
èèèèè C║H║N + H½O = C║H║NHó + OHú,èK╧ = 1.78x10úö.
èèA) 4.42 B) 9.58
èèC) 8.75 D) 8.85
üèTo determïe ê pH. you need ë know eiêr ê [Hó] or ê pOH.
With a base fïdïg ê pOH is easy.èYou know K╧ å ê ïitial concen-
tration ç ê pyridïe.èYou need ë use ê equilibrium constant expres-
sion ë get ê [OHú].èStartïg with ê equilibrium reaction:
C║H║N + H½O = C║H║NHóè+èOHú.èAt equilibrium ê concentrations
are:è0.80-[OHú]èèè [OHú]èè [OHú]èSubstitutïg ïë ê equilibrium
expression gives
èè [C║H║NHó][OHú]èèè[OHú][OHú]
K╧ = ──────────────.èè ────────── = 1.78x10úö.èèIf [OHú] << 0.80,
èèè [C║H║N]èèèèè 0.80-[OHú]
èèèèí─────────────────
[OHú] = á(0.80)(1.78x10úö) = 3.8x10úÉ.èpOH = 4.42.èpH = 9.58.
Ç B
10èWhat is ê approximate pH ç a 0.60 M trimethylamïe,
solution?èè(CH╕)╕N + H½O = (CH╕)╕NHó + OHú,èK╧ = 6.3x10úÉ.
èèA) 11.79 B) 9.80
èèC) 12.01 D) 9.58
üèTo determïe ê pH. you need ë know eiêr ê [Hó] or ê pOH.
In basic solutions, you should fïd ê pOH first.èYou know K╧ å ê
ïitial concentration ç ê trimethylamïe.èYou need ë use ê equi-
librium constant expression ë get ê [OHú].èStartïg with ê equili-
brium reaction:èèèè(CH╕)╕N + H½O = (CH╕)╕NHó +èOHú.èAt equilibrium
ê concentrations are: 0.60-[OHú]èèè [OHú]èè [OHú]èSubstitutïg ïë
ê equilibrium expression gives
èè [(CH╕)╕NHó][OHú]èèè[OHú][OHú]
K╧ = ────────────────.èè ────────── = 6.3x10úÉ.èèIf [OHú] << 0.60,
èèèè[(CH╕)╕N]èèèèè0.60-[OHú]
èèèèí─────────────────
[OHú] = á(0.60)(6.3x10úÉ) = 6.1x10úÄ.èpOH = 2.21.èpH = 11.79.
Ç A