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Multimedia Chemistry I & II (1996-9-11) [English].img
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chapt11.4c
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à 11.4cèBuffers
äèPlease determïe which systems satisfy ê requirements ë act as a buffer.
âèWill a mixture ç 0.1 mol ç acetic acid å 0.2 mol ç sodium
acetate ï 1.0 L ç solution function as a buffer?èThe requirements for
a buffer are met by this system.èWe need a weak acid å its conjugate
base for an acidic buffer.èAcetic acid is a weak acid.èSodium acetate
supplies ê conjugate base, ê acetate ion.
éSèBuffers are chemical systems that allow only small changes ï pH.
Moderatïg a change ï ê pH ç a solution is achieved by havïg rela-
tively large amounts ç a weak acid(base) å its conjugate base(acid) ï
comparison with ê Hó concentration.èTo see how an acidic buffer works,
we need ë examïe ê equilibrium constant expression for a weak acid.
HA(aq) + H½O = H╕Oó + Aú(aq).
èè èè [H╕Oó][Aú]
K╬ = ──────────.
[HA]
Sïce we are ïterested ï keepïg ê [H╕Oó] constant, we rearrange ê
equation ë solve for [H╕Oó].
èè [HA]
èèèèèèèèèèè [H╕Oó] = K╬ ∙ ────
èè [Aú]
From this equation, we can see that ê [H╕Oó] å thus ê pH will
remaï constant as long as ê ratio ç HA ë Aú does not change signifi-
cantly.èThis ratio will remaï approximately constant if ê amounts ç
HA å Aú are much greater than any added acids or bases.
èè Perhaps, a specific example will make ê operation ç ê buffer
clearer.èAcetic acid with sodium acetate can act as a buffer.èWhat is
ê pH ç a buffer that is 0.10 M acetic acid å 0.12 M sodium acetate?
We start with ê equilibrium, because it ïterrelates ê acetic acid,
acetate ion, å hydronium ion.
HC╖H╕O╖ + H╖O = H╕Oó + C╖H╕O╖ú, K╬ = 1.75x10úÉ.
Sïce acetic acid is a weak acid, its dissociation is slight å we can
use ê above molarities ï ê equilibrium expression.
èè [H╕Oó][C╖H╕O╖ú]èèèè [H╕Oó](0.12 M)
K╬ = ───────────────.èèèè────────────── = 1.75x10úÉ.
èèèè[HC╖H╕O╖]èèèèèèè (0.10 M)
[H╕Oó] = (1.75x10úÉ)(0.10)/(0.12).èè[H╕Oó] = 1.46x10úÉ Mè
The pH ç ê buffer is -log(1.46x10úÉ) = 4.84.è
èè What happens when we add HCl ë this buffer?èApplyïg LeChatelier's
prïciple, we know that ê equilibrium will shift ë ê left, some
acetate ion will be converted ïë acetic acid.èWhat would happen if we
added 1.0 mL ç 6.0 M HCl ë 500. mL ç ê 0.1 M HC╖H╕O╖-0.12 M NaC╖H╕O╖
buffer?èBefore addïg ê HCl, ê buffer contaïs:
èè (0.500 L)(0.100 M HC╖H╕O╖) = 0.0500 mol HC╖H╕O╖, å
èè (0.500 L)(0.120 M C╖H╕O╖ú) = 0.0600 mol C╖H╕O╖ú.
Addïg 1.0 mL ç 6.0 M HCl adds (0.0010 L)(6.0 M) = 0.0060 mol ç Hó ë
ê buffer.èThe added Hó converts 0.0060 mol ç ê acetate ion ïë
0.0060 mol ç acetic acid.èAfter ê addition, êre are:
èè 0.0500 + 0.0060 = 0.0560 mol HC╖H╕O╖, å
èè 0.0600 - 0.0060 = 0.0540 mol C╖H╕O╖ú ï ê system.
Puttïg êse values ïë ê equilibrium expression, we fïd ê new H╕Oó
concentration ë be:
èè [H╕Oó][C╖H╕O╖ú]èèèè [H╕Oó](0.0540 mol)
K╬ = ───────────────.èèèè────────────────── = 1.75x10úÉ.
èèèè[HC╖H╕O╖]èèèèèèè (0.0560 mol)
[H╕Oó] = (1.75x10úÉ)(0.0560)/(0.540).èè [H╕Oó] = 1.81x10úÉ M
We did not need ë use ê volume ç ê solution, because it cancels
when we take ê ratio.èThe new pH ç ê buffer is
èèèèèèèè pH = -log(1.81x10úÉ).èpH = 4.74.
The pH ç ê buffer changed from 4.84 ë 4.74 or only one tenth ç a pH
unit.è500 mL ç an unbuffered solution at pH 4.84 would have changed
from 4.84 ë 2.52 or a change ç 2.32 pH units.
Notice also that ê pH ç ê buffer dropped when we added ê acid.èWe
should expect a lower pH because ê solution is more acidic.
Of course, if we contïue ë add more å more acid ë ê buffer, event-
ually all ç ê acetate ion would be converted ïë acetic acid.èThe
buffer is exhausted, å it no longer acts as a buffer.èAddïg an excess
ç base has ê opposite effect.èThe acetic acid is converted ïë ê
acetate ion.èEventually êre would be only acetate ion present å ê
buffer is ruïed.
1èWhich one ç ê followïg pairs could you use ë make a
èèbuffer?
èèA) H╕PO╣ + HCl B) HCl + KCl
èèC) NaHCO╕ + Na╖CO╕ D) KC╖H╕O╖ + KOH
üè We need a weak acid å its conjugate base or a weak base å
its conjugate acid ë prepare a buffer.èAmong ê choices ï this
problem, CO╕ìú is a weak base å HCO╕ú is its conjugate acid.èThe oêr
combïations contaï only acids, (A) å (B), or only bases, (D).
Ç C
2èWhich one ç ê followïg combïations would form a buffer ï
èèèèè 1.0 L ç solution?
èèA) 0.1 mol HC╖H╕O╖ + 0.1 mol HCl
èèB) 0.2 mol HC╖H╕O╖ + 0.1 mol NaC╖H╕O╖
èèC) 0.1 mol HC╖H╕O╖ + 0.2 mol NaOH
èèD) 0.2 mol NaC╖H╕O╖ + 0.1 mol NaOH
üèIn order ë make a buffer, we need ë combïe or ë generate a
weak acid - weak conjugate base combïation.èChoice (B) satisfies those
conditions because ê 0.2 mol HC╖H╕O╖ is ê weak acid å ê 0.1 mol
NaC╖H╕O╖ supplies ê conjugate base ç acetic acid.èChoice (A) contaïs
only ê weak acid å a strong acid.èIn choice (C), êèNaOH would
convert ê acetic acid ïë sodium acetate. so no weak acid would exist.
Choice (D) contaïs only bases, which is what happens ï (C).
Ç B
3èWhich one ç ê followïg combïations would form a buffer ï
èèèèè 1.0 L ç solution?
èèA) 0.1 mol NH╣Cl + 0.2 mol NaOH
èèB) 0.2 mol NH╣Cl + 0.1 mol HCl
èèC) 0.2 mol NH╕è + 0.1 mol HCl
èèD) 0.1 mol NH╕è + 0.1 mol NaOH
üèIn order ë make a buffer, we need ë combïe or ë generate a
weak acid - weak conjugate base combïation.èChoice (C) satisfies ê
conditions for a buffer.èNH╕ is a weak base å HCl is a strong acid.
The 0.1 mol ç HCl will react with 0.1 mol ç ê origïal 0.2 mol ç NH╕
ë form 0.1 mol NH╣Cl. The reaction generates a buffer contaïïg
0.1 mole NH╣ó å 0.1 mole NH╕.èNH╣ó is a weak acid å NH╕ is its con-
jugate base.
Ç C
4èWhich compound would you add ë an aqueous solution ç mono-
chloroacetic acid, a weak acid, ClCH╖CO╖H,ï order ë prepare a buffer?
èèA) HCl B) KCl
èèC) KOH D) None ç êse.
üèWe need ë form ê conjugate base ç monochloroacetic acid ï
this solution ï order ë make ê buffer.èKOH will react with ê
ClCH╖CO╖H ë form potassium chloroacetate.èAddïg enough KOH ë convert
some, but not all, ç ê ClCH╖CO╖H ïë ê chloroacetate ion will leave
ê system contaïïg both ê acid å its conjugate base.
Ç C
5èWhat net reaction occurs when NaOHaq) is added ë a
èèèèè HPO╣ìú - PO╣Äú buffer?
èèA)èOHú + HPO╣ìú ──¥ HPO║Äú.
èèB)èOHú + PO╣Äúè──¥ O╖ + HPO╕Åú.
èèC)èOHú + HPO╣ìú ──¥ H╖ + PO║Äú.
èèD)èOHú + HPO╣ìú ──¥ H╖O + PO╣Äú.
üèNaOH is a strong base ï water, å ê aqueous solution contaïs
Naó å OHú ions.èThe hydroxide ion will react with ê acid ç ê
buffer, which is HPO╣ìú.
èèèèThe net reaction is OHú + HPO╣ìú ──¥ H╖O + PO╣Äú.
Ç D
6èWhat net reaction occurs when HBr(aq) is added ë a formic
èèacid - sodium formate buffer, HCO½H - NaHCO½?
èèA)èNaó + Brú ──¥ NaBr(s).
èèB)èHó + HCO½ú ──¥ HCO½H.
èèC)èHó + HCO╖H ──¥ HC(OH)╖ó.
èèD)èHó + HCO½H ──¥ H╖ + HCO╖ú.
üèHBr is a strong acid ï water å dissociates completely ï
aqueous solutions formïg Hó(aq) å Brú(aq) ions.èThe Hó will react
with ê base ç ê buffer, which is HCO½ú.èThe buffer reacts ë added
strong acid by convertïg ê base ïë its conjugate acid.èThe conjugate
acid ç HCO½ú is HCO½H.èThe net reaction is: Hó + HCO½ú ──¥ HCO½H.
Ç B
7èWhich compound would you add ë an aqueous solution ç
èèdiethylamïe, (CH╕CH╖)╖NH, ï order ë prepare a buffer?
èèA) H╖SO╣ B) NaOH
èèC) NaNO╕ D) NH╕
üèAmïes are weak bases.èTo prepare a buffer usïg diethylamïe,
(CH╕CH╖)╖NH, we need ë make ê conjugate acid ç diethylamïe.èThe
conjugate acid is (CH╕CH╖)╖NH╖ó.èWe must add enough strong acid ë con-
vert some, but not all, ç ê amïe ïë its conjugate acid. Sulfuric
acid will react with ê amïe ë form diethylammonium sulfate:
H╖SO╣ + 2(CH╕CH╖)╖NH ─¥ 2(CH╕CH╖)╖NH╖ó + SO╣ìú.èNone ç ê oêr
choices are acids that would react with ê amïe.
Ç A
äèPlease select ê best buffer for a given pH or calculate ê pH ç a given buffer.
âèWhat is ê pH at which ê acetic acid - sodium acetate buffer
is most effective?èK╬ ç acetic acid is 1.75x10úÉ.èA buffer is most
effective when it can håle added acid or base equally well.èThis occurs
when ê acid å conjugate base concentrations are equal.
èè [Hó][C╖H╕O╖ú]èè When [HC╖H╕O╖] = [C╖H╕O╖ú], ên [Hó] = K╬.èThe
K╬ = ─────────────.èèmost effective buffer has [Hó] = 1.75x10úÉ or a
èèè [HC╖HO╖]èèèèpH equal ë -log(1.75x10úÉ) = 4.76.
éSèCalculatïg ê pH ç a buffer is somewhat simpler than calcu-
latïg ê pH ç a solution ç a weak acid or weak base.èSïce ê
buffer contaïs large amounts ç ê acid(base) å its conjugate base
(acid), ê extent ç dissociation is reduced below that ï a solution
ç ê acid or base alone.èThis is anoêr application ç LeChatelier's
prïciple.
What is ê pH ç a buffer that is 0.20 M NaH╖PO╣ å 0.30 M Na╖HPO╣?
We begï with ê equilibrium reaction:
H╖PO╣ú = Hó + HPO╣ìú,èK╬ = 6.3x10úô.
èè [Hó][HPO╣ìú]èèèèè[Hó](0.30 M)
K╬ = ────────────èè K╬ = ──────────── = 6.3x10úô
èèè[H╖PO╣ú]èèèèèèè (0.20 M)
èèèèèèèèèèèèèèèèèèèèèèèèèè┌───────────┐
[Hó] = 6.3x10úô(0.20/0.30) = 4.2x10úô Mèè | pH = 7.38 |
èèèèèèèèèèèèèèèèèèèèèèèèèè└───────────┘
A 0.20 M NaH╖PO╣ solution is 0.20 M Naó å 0.20 M H╖PO╣ú.
A 0.30 M Na╖HPO╣ solution is 0.60 M Naó å 0.30 M HPO╣ú.èWe simply sub-
stitute ê concentrations ï ê equilibrium constant expression.èWe do
not need ë worry about ê changes ï ê molarities due ë ê estab-
lishment ç ê equilibrium, because êse changes are very small.
A buffer will be equally or most effective ëward eiêr added acid or
base when ê concentrations ç ê components ç ê buffer are equal.
When this condition is met, ê [Hó] equals ê K╬.èSïce we usually
specify ê pH ç ê buffer, we can also view this condition as
pH = -log(K╬)èorèK╬ = ╢╡-pH.èThese relationships provide a means for
selectïg a buffer.
The value ç K╬ for acetic acid is 1.75x10úÉ.èFïdïg -log(1.75x10úÉ)
gives 4.76.èThe result shows that an acetic acid - sodium acetate buffer
will be most effective around pH 4 ë pH 5.èThe NaH╖PO╣ - Na╖HPO╣ buffer
with a K╬ ç 6.3x10úô is most effective around a pH ç 7.
8èWhich system would be most effective for a pH 9 buffer?
èèK╬ values ï parenêses.
A) NaHSO╣ - Na╖SO╣ (1.2x10úì) B) HC╖H╕O╖ - NaC╖H╕O╖ (1.75x10úÉ)
C) NaH╖PO╣ - Na╖HPO╣ (6.3x10úô) D) NH╣Cl - NH╕ (5.6x10úîò)
üèWe want a system whose -log(K╬) equals ê pH ç ê desired
buffer.èIn ê NH╣Cl - NH╕ system, K╬ = 5.6x10úîò.èTakïg ê -log ç
K╬ yields 9.25.èThe NH╣Cl - NH╕ system is ê most effective buffer at
pH 9.
Ç D
9èWhich system would be most effective for a pH 2 buffer?
èèK╬ values ï parenêses.
A) H╖C╖O╣ - KHC╖O╣ (5.9x10úì) B) KHC╗H╣O╣ - K╖C╗H╣O╣ (3.9x10úæ)
C) C║H╗NCl - C║H║N (1.8x10úæ) D) NaH╖PO╣ - Na╖HPO╣ (6.3x10úô)
üèWe want a system whose -log(K╬) equals ê pH ç ê desired
buffer.èIn ê H╖C╖O╣ - KHC╖O╣ system, K╬ equals 5.9x10úì.èTakïg ê
-log ç K╬ yields 1.23.èThe -log(K╬) for ê H╖C╖O╣ - KHC╖O╣ system is
closest ë ê desired pH ç ê buffer, so it would be ê most effec-
tive buffer at pH 2 ç ê choices ï this problem.
Ç A
10èWhat is ê pH ç a buffer that is 0.15 M H╖C╖O╣ å
èèèèèè0.30 M KHC╖O╣?èH╖C╖O╣ = Hó + HC╖O╣ú, K╬ = 5.9x10úì.
èè A) 1.23 B) 1.53
èè C) 0.93 D) 2.58
üèWe need ë fïd ê concentration ç Hó before we can calculate
ê pH.èSïce ê extent ç dissociation ç ê acid is slight ï a
buffer usïg a weak acid,èwe just need ë fïd ê concentration ç Hó
ï equilibrium with ê H╖C╖O╣ å ê HC╖O╣ú.èThe equilibrium expression
isèèè [Hó][HC╖O╣ú]
èèK╬ = ──────────── = 5.9x10úì.èRearrangïg ë fïd [Hó] givesè
èèèèè [H╖C╖O╣]
èèèèèèèèè [H╖C╖O╣]èèèèè 0.15
èè[Hó] = 0.059 x ──────── = 0.059 x ────è= 0.029║ M
èèèèèèèèè [HC╖O╣ú]èèèèè 0.30
The pH ç ê buffer is -log(0.0295).èpH = 1.53
Ç B
11èWhat is ê pH ç a buffer that is 0.50 M NH╣Cl å
èèèèèè1.50 M NH╕?èNH╣ó = Hó + NH╕, K╬ = 5.6x10úîò
èè A) 9.73 B) 9.38
èè C) 8.77 D) 9.25
üèWe need ë fïd ê concentration ç Hó ï order ë fïd ê pH.
The extent ç dissociation ç ê weak acid or weak base is slight
buffers.èConsequently, we just need ë fïd ê concentration ç Hó ï
equilibrium with ê NH╣ó å ê NH╕.èThe equilibrium expression
isèèè [Hó][NH╕]
èèK╬ = ───────── = 5.6x10úîò.èRearrangïg ë fïd [Hó] givesè
èèèèè[NH╣ó]
èèèèèèèèèèè [NH╣ó]èèèèèèè 0.50
èè[Hó] = 5.6x10úîò x ────── = 5.6x10úîò x ────è= 1.87x10úîò M
èèèèèèèèèèè [NH╕]èèèèèèèè1.50
The pH ç ê buffer is -log(1.87x10úîò).èpH = 9.73.
Ç A