2. 2
Z5e 14.1 The Nature of Acids and BasesZ5e 14.1 The Nature of Acids and Bases
Arrhenius DefinitionArrhenius Definition
Acids produce hydrogen ions inAcids produce hydrogen ions in
aqueous solution.aqueous solution.
Bases produce hydroxide ions whenBases produce hydroxide ions when
dissolved in water.dissolved in water.
Limited to aqueous solutions.Limited to aqueous solutions.
Only one kind of base (hydroxide).Only one kind of base (hydroxide).
NHNH33 ammonia could not be anammonia could not be an
Arrhenius base.Arrhenius base.
http://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&searchhttp://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&search==
3. 3
Bronsted-Lowry DefinitionsBronsted-Lowry Definitions
AnAn acidacid is an proton (His an proton (H++
)) donordonor and aand a
basebase is a protonis a proton acceptoracceptor..
Acids and bases always come inAcids and bases always come in pairspairs..
HCl is an acid.HCl is an acid.
When it dissolves in water it gives itsWhen it dissolves in water it gives its
proton to water (so Hproton to water (so H22O is a base).O is a base).
HClHCl(g)(g) + H+ H22OO(l)(l) →→ HH33OO++
+ Cl+ Cl--
Water is aWater is a basebase: makes hydronium ion: makes hydronium ion
5. 5
Bronsted-Lowry Acids
Acids are hydrogen ion (H+)
donors
Bases are hydrogen ion (H+
) acceptors
HCl + H2O H3O+
+ Cl-
donor acceptor + -
+ +
6. 6
NH3, A Bronsted-Lowry Base
When NH3 reacts with water, most of the
reactants remain dissolved as molecules, but a
few NH3 reacts with water to form NH4
+
and
hydroxide ion.
NH3 + H2O NH4
+
(aq) + OH-
(aq)
acceptor donor
+ +
7. 7
B. Definitions pp
Brønsted-Lowry
HCl + H2O → Cl–
+ H3O+
• AcidsAcids are proton (H+
) donors.
• BasesBases are proton (H+
) acceptors.
conjugate acid
conjugate base
baseacid
13. 13
PairsPairs pppp
General equationGeneral equation
HA(aq) + HHA(aq) + H22O(l)O(l) HH33OO++
(aq) + A(aq) + A--
(aq)(aq)
Acid + BaseAcid + Base Conjugate acid +Conjugate acid +
Conjugate baseConjugate base
This is an equilibrium.This is an equilibrium.
CompetitionCompetition forfor HH++
betweenbetween HH22O and AO and A--
The stronger base controls direction.The stronger base controls direction.
If HIf H22O is a stronger base it takes the HO is a stronger base it takes the H++
Equilibrium moves to right.Equilibrium moves to right.
14. 14
Acid dissociation constant KAcid dissociation constant Kaa
The equilibrium constant for theThe equilibrium constant for the
general equation.general equation.
HA(aq) + HHA(aq) + H22O(l)O(l) HH33OO1+1+
(aq) + A(aq) + A1-1-
(aq)(aq)
KKaa = [H= [H33OO++
][A][A--
]]
[HA][HA]
HH33OO++
is often written His often written H++
ignoring theignoring the
water in equation (it is implied).water in equation (it is implied).
16. 16
Acid dissociation constant KAcid dissociation constant Kaa
HA(aq)HA(aq) HH++
(aq) + A(aq) + A--
(aq)(aq)
KKaa = [H= [H++
][A][A--
]]
[HA][HA]
We can write the expression for anyWe can write the expression for any
acid.acid.
Strong acids dissociate completely.Strong acids dissociate completely.
Equilibrium far to right.Equilibrium far to right.
Conjugate base must be weak.Conjugate base must be weak.
17. 17
Figure 14.5Figure 14.5
Acid StrengthAcid Strength
Versus ConjugateVersus Conjugate
Base StrengthBase Strength z5e 661z5e 661
18. 18
Z5e 14.2 Acid StrengthZ5e 14.2 Acid Strength
Strong acidsStrong acids
KKaa is largeis large
[H[H++
] is about equal] is about equal
toto originaloriginal [HA][HA]
AA--
is a weaker baseis a weaker base
than waterthan water
Weak acidsWeak acids
KKaa is smallis small
[H[H++
] <<< [HA]] <<< [HA]
AA--
is a stronger baseis a stronger base
than waterthan water
19. 19
Z5e Figure 14.4 p. 661Z5e Figure 14.4 p. 661 pppp
GraphicGraphic
Representation ofRepresentation of
the Behavior ofthe Behavior of
Acids of DifferentAcids of Different
Strengths inStrengths in
Aqueous SolutionAqueous Solution
((a) = strong acid,a) = strong acid,
completecomplete
dissociation.dissociation.
((b) = weak acid,b) = weak acid,
slight dissociationslight dissociation..
20. 20
Which of the following
"molecular" pictures, “A”
or “B”, best represents
a concentrated solution
of the weak acid HA
with Ka = 10-5
? . . .
“B”, because . . .
There are more
molecules of
undissociated form.
“A” represents a strong
acid.
21. 21
Types of AcidsTypes of Acids
Polyprotic AcidsPolyprotic Acids - more than 1 acidic- more than 1 acidic
hydrogen (diprotic, triprotic).hydrogen (diprotic, triprotic).
OxyacidsOxyacids - Proton is attached to the- Proton is attached to the
oxygen of an ion.oxygen of an ion.
Organic acidsOrganic acids contain the Carboxylcontain the Carboxyl
group -COOH with the H attached to Ogroup -COOH with the H attached to O
Polyprotic acids are generally veryPolyprotic acids are generally very
weak (but, Hweak (but, H22SOSO44 is strong with 1is strong with 1stst
HH++
and weak with 2and weak with 2ndnd
).).
22. 22
Relative Base Strength of AcidsRelative Base Strength of Acids pppp
Using Table 14.2 p. 628, arrange theUsing Table 14.2 p. 628, arrange the
following in increasingfollowing in increasing basebase strength.strength.
HH22OO,, FF1-1-
, Cl, Cl1-1-
, NO, NO22
1-1-
, CN, CN1-1-
Steps . . .Steps . . .
Hint: Strongest acid will have weakestHint: Strongest acid will have weakest
base. Which conjugate of the abovebase. Which conjugate of the above
bases would be the strongest acid? . . .bases would be the strongest acid? . . .
That’s right, HCl. So, ClThat’s right, HCl. So, Cl1-1-
must be themust be the
weakestweakest base.base.
23. 23
Relative Base Strength of AcidsRelative Base Strength of Acids pppp
Arrange HArrange H22OO,, FF1-1-
, Cl, Cl1-1-
, NO, NO22
1-1-
, CN, CN1-1-
inin
increasingincreasing basebase strength.strength.
Considering HConsidering H22OO,, the others arethe others are
conjugate bases ofconjugate bases of weakweak acids, so theacids, so the
above bases must beabove bases must be strongerstronger thanthan
water acting as a base. So . . .water acting as a base. So . . .
ClCl--
< water < conjugate bases of weak< water < conjugate bases of weak
acids . . .acids . . .
24. 24
Relative Base Strength of AcidsRelative Base Strength of Acids pppp
Arrange HArrange H22OO,, FF1-1-
, Cl, Cl1-1-
, NO, NO22
1-1-
, CN, CN1-1-
inin
increasingincreasing basebase strength.strength.
ClCl--
< water < conj. bases of weak acids.< water < conj. bases of weak acids.
Now, use Table 14.2 p. 628 to “order”Now, use Table 14.2 p. 628 to “order”
the rest. Your answer is . . .the rest. Your answer is . . .
ClCl--
< H< H22O < FO < F1-1-
< NO< NO22
1-1-
< CN< CN1-1-
25. 25
AmphotericAmphoteric
Behaves as acid & base -- autoionizesBehaves as acid & base -- autoionizes
2H2H22O(l)O(l) HH33OO++
(aq) + OH(aq) + OH--
(aq)(aq)
KKWW= [H= [H33OO++
][OH][OH--
]=[H]=[H++
][OH][OH--
]]
HH22OO(l)(l) drops out of =m expression since itdrops out of =m expression since it
is a pure liquidis a pure liquid
At 25ºC KAt 25ºC KWW = 1.0 x10= 1.0 x10-14-14
inin EVERYEVERY aqueousaqueous
solution.solution.
KKww = ion product constant (or dissociation= ion product constant (or dissociation
constant)constant)
So, we can use it to find [HSo, we can use it to find [H++
] and [OH] and [OH--
]]
27. 27
Autoionization of Water problemAutoionization of Water problem
KKww = 1 x 10= 1 x 10-13-13
at 60ºC. Using Le Chatelier,at 60ºC. Using Le Chatelier,
predict if exo- or endothermicpredict if exo- or endothermic 2H2H22OO(l)(l)
HH33OO1+1+
(aq)(aq) + OH+ OH1-1-
(aq)(aq) Hint:Hint:
compare Kcompare Kww at 25ºC.at 25ºC.
Endothermic since KEndothermic since Kww increases withincreases with
temperature so energy is a reactant.temperature so energy is a reactant.
28. 28
Autoionization of Water problemAutoionization of Water problem
KKww = 1 x 10= 1 x 10-13-13
at 60ºC.at 60ºC.
2H2H22OO(l)(l) HH33OO1+1+
(aq)(aq) + OH+ OH1-1-
(aq)(aq)
Calculate [HCalculate [H++
] & [OH] & [OH--
] in] in neutralneutral soln.soln.
Hint: since neutral, they must be equal.Hint: since neutral, they must be equal.
Answer . . .Answer . . .
3 x 103 x 10-7-7
MM since [Hsince [H++
] [OH] [OH--
] = 1 x 10] = 1 x 10-13-13
29. 29
Figure 14.7Figure 14.7
Two Water Molecules React to FormTwo Water Molecules React to Form
HH33OO++
and OHand OH--
30. 30
Z5e 14.3 The pH ScaleZ5e 14.3 The pH Scale
pH= -log[HpH= -log[H++
]]
Used because [HUsed because [H++
] is usually very small] is usually very small
As pHAs pH decreasesdecreases, [H, [H++
]] increasesincreases
exponentiallyexponentially
Sig figs: Only the digitsSig figs: Only the digits afterafter thethe
decimal place of a pH are significant!decimal place of a pH are significant!
[H[H++
] = 1.0 x 10] = 1.0 x 10-8-8
pH= 8.00pH= 8.00 22 sig figssig figs
pOH= -log[OHpOH= -log[OH--
]]
pKa = -log KpKa = -log Kaa
31. 31
A solution with pH = 5 is 100 times more
acidic than a solution with a pH = ?
A) 7 B) 3 C) 0.05
• 7 because . . .
• 7 is 2 pH units more basic so 100 times
less [H3O1+
] (or 100 x more [OH1-
]
32. 32
RelationshipsRelationships pppp
KKWW = [H= [H++
][OH][OH--
]]
-log K-log KWW = -log([H= -log([H++
][OH][OH--
])])
-log K-log KWW = -log[H= -log[H++
]+ -log[OH]+ -log[OH--
]]
pKpKWW = pH + pOH= pH + pOH
KKWW = 1.0 x10= 1.0 x10-14-14
14.00 = pH + pOH14.00 = pH + pOH
[H[H++
] = 10] = 10-pH-pH
and [OHand [OH--
] = 10] = 10-pOH-pOH
[H[H++
],[OH],[OH--
],pH and pOH: Given any one],pH and pOH: Given any one
33. 33
Figure 14.8Figure 14.8
The pH Scale and pHThe pH Scale and pH
Values of SomeValues of Some
Common SubstancesCommon Substances
35. 35
Calculating pH of SolutionsCalculating pH of Solutions
AlwaysAlways write down the major specieswrite down the major species
in solution. If you forget you’llin solution. If you forget you’ll
probably get a wrong solution!probably get a wrong solution!
Remember, these are equilibria.Remember, these are equilibria.
Remember the chemistry.Remember the chemistry.
Don’t try to memorize, there is no oneDon’t try to memorize, there is no one
way to do this.way to do this.
36. 36
Calculating pH of SolutionsCalculating pH of Solutions
The pH of a blood sample was 7.41 atThe pH of a blood sample was 7.41 at
25ºC. Calculate pOH, [H25ºC. Calculate pOH, [H1+1+
] and [OH] and [OH1-1-
].].
Answers? . . .Answers? . . .
pOH = 6.59pOH = 6.59 (14 - 7.41 = 6.59) [H(14 - 7.41 = 6.59) [H1+1+
] = ?] = ?
[H[H1+1+
] = 3.9 x 10] = 3.9 x 10-8-8
(10(10-7.41-7.41
)) [OH[OH1-1-
] = ?] = ?
[OH[OH1-1-
] = 2.6 x 10] = 2.6 x 10-7-7
Either 10Either 10-6.59-6.59
or since Kor since Kww ==
[H[H1+1+
][OH][OH1-1-
],], [OH[OH1-1-
] = 10] = 10-14-14
/3.9 x 10/3.9 x 10-8-8
37. 37
z5e 14.4 Calculating the pH of Strong Acidsz5e 14.4 Calculating the pH of Strong Acids pppp
HBr, HI, HCl, HNOHBr, HI, HCl, HNO33, H, H22SOSO44, HClO, HClO44
ALWAYS WRITE THE MAJOR SPECIESALWAYS WRITE THE MAJOR SPECIES
Above are completely dissociated (strong acids)Above are completely dissociated (strong acids)
So, [HSo, [H++
] =] = originaloriginal [HA][HA]
[OH[OH--
] is going to be small because of] is going to be small because of
equilibrium (essentially none since shifted so farequilibrium (essentially none since shifted so far
to the right)to the right)
1010-14-14
= [H= [H++
][OH][OH--
]]
If [HA] < 10If [HA] < 10-7-7
thenthen waterwater contributes Hcontributes H++
andand
must be taken into account!!!must be taken into account!!!
38. 38
Strong AcidsStrong Acids pppp
Calculate the pH of 1.0 x 10Calculate the pH of 1.0 x 10-10-10
MM HCl. Ans.?HCl. Ans.?
pH = 7.00. Why? . . .pH = 7.00. Why? . . .
Always write the major species, which are . . .Always write the major species, which are . . .
HH22O, HO, H1+1+
and Cland Cl1-1-
Both contribute HBoth contribute H1+1+
But, the [HCl] is so small thatBut, the [HCl] is so small that it has no effectit has no effect!!
Essentially all the HEssentially all the H1+1+
is coming from water.is coming from water.
So, pH = 7.00.So, pH = 7.00.
39. 39
Z5e 14.5 Calculating the pH of Weak AcidsZ5e 14.5 Calculating the pH of Weak Acids pppp
Ka will be small.Ka will be small.
ALWAYS WRITE THE MAJOR SPECIESALWAYS WRITE THE MAJOR SPECIES..
It will be an equilibrium problemIt will be an equilibrium problem fromfrom
the startthe start (and shifted to the(and shifted to the leftleft).).
Determine whether most of the HDetermine whether most of the H++
willwill
come from the acid or the water!come from the acid or the water!
Compare to Ka or Kw per aboveCompare to Ka or Kw per above
Rest is just like last chapter (see,Rest is just like last chapter (see,
summary next slide)summary next slide)
41. 41
ExampleExample pppp
Calculate the pH of 2.0Calculate the pH of 2.0 MM acetic acid HCacetic acid HC22HH33OO22
with a Ka 1.8 x10with a Ka 1.8 x10-5-5
Steps. . .Steps. . .
Major species are . . .Major species are . . .
HCHC22HH33OO22 & H& H22O. pH answer is . . . (use RICE)O. pH answer is . . . (use RICE)
pH = 2.22pH = 2.22
Calculate pOH, [OHCalculate pOH, [OH--
], [H], [H++
] Ans. are . . .] Ans. are . . .
pOH = 11.78pOH = 11.78
[OH[OH--
] = 1.7 x 10] = 1.7 x 10-12-12
[H[H++
] = 6.0 x 10] = 6.0 x 10-3-3
42. 42
A mixture of Weak AcidsA mixture of Weak Acids pppp
The process is the same.The process is the same.
Determine the major species.Determine the major species.
The stronger will predominate.The stronger will predominate.
So, use bigger Ka if concentrations areSo, use bigger Ka if concentrations are
comparable (comparable (i.e.,i.e., can ignore smaller Ka)can ignore smaller Ka)
43. 43
A mixture of Weak AcidsA mixture of Weak Acids pppp
Calculate the pH of aCalculate the pH of a mixturemixture of 1.20of 1.20 MM
HF (Ka = 7.2 x 10HF (Ka = 7.2 x 10-4-4
) and 3.4) and 3.4 MM HOCHOC66HH55
(Ka = 1.6 x 10(Ka = 1.6 x 10-10-10
))
Major species: HF, HOCMajor species: HF, HOC66HH55, H, H22OO
Since HF is the dominant producer ofSince HF is the dominant producer of
HH++
, use its Ka, use its Ka
Set up “RICE” table, useSet up “RICE” table, use
approximations and 5% rule to get . . .approximations and 5% rule to get . . .
[H[H++
] = .030 & pH = 1.53] = .030 & pH = 1.53
44. 44
A mixture of Weak AcidsA mixture of Weak Acids pppp
Calculate theCalculate the [OC[OC66HH55
1-1-
]] of a mixture ofof a mixture of
1.201.20 MM HF (Ka = 7.2 x 10HF (Ka = 7.2 x 10-4-4
) and 3.4) and 3.4 MM
HOCHOC66HH55 (Ka = 1.6 x 10(Ka = 1.6 x 10-10-10
))
Use Ka and “RICE” for HOCUse Ka and “RICE” for HOC66HH55,, butbut
remember that [Hremember that [H++
] concentration came] concentration came
fromfrom bothboth acidsacids
So, 1.6 x 10So, 1.6 x 10-10-10
= (= (0.0300.030)(x)/(3.4))(x)/(3.4)
[OC[OC66HH55
1-1-
] = . . .] = . . .
45. 45
Percent dissociationPercent dissociation pppp
== amount dissociatedamount dissociated x 100%x 100%
initialinitial concentrationconcentration
For aFor a weakweak acid percent dissociationacid percent dissociation increasesincreases
as acid becomesas acid becomes more dilutemore dilute..
Calculate the % dissociation of 1.00Calculate the % dissociation of 1.00 MM andand
0.1000.100 MM Acetic acid (Ka = 1.8 x 10Acetic acid (Ka = 1.8 x 10-5-5
). Ans. . .). Ans. . .
% diss 1.00% diss 1.00 MM = 0.42%, 0.00100= 0.42%, 0.00100 MM = 1.3%= 1.3%
As [HA]As [HA]00 decreases [Hdecreases [H++
]] decreasesdecreases but %but %
dissociationdissociation increasesincreases..
Le Chatelier’s principle operates hereLe Chatelier’s principle operates here
46. 46
The other wayThe other way pppp
What is the Ka of a weak acid that isWhat is the Ka of a weak acid that is
8.1 % dissociated as 0.1008.1 % dissociated as 0.100 MM solution?solution?
Ans. . .Ans. . .
6.6 x 106.6 x 10-4-4
Solving method . . .Solving method . . .
Since (X/0.100)(100%) = 8.1%,Since (X/0.100)(100%) = 8.1%,
x = 8.1 x 10x = 8.1 x 10-3-3
= [H= [H++
] = [A] = [A--
]]
Then plug into =m to solve for Ka.Then plug into =m to solve for Ka.
(8.1 x 10(8.1 x 10-3-3
))22
÷ 0.100 = 6.6 x 10÷ 0.100 = 6.6 x 10-4-4
(remember, divide by initial [ ])(remember, divide by initial [ ])
47. 47
Figure 14.10Figure 14.10
p. 643.p. 643.
The Effect ofThe Effect of
Dilution onDilution on
the Percentthe Percent
DissociationDissociation
and (H+) of aand (H+) of a
Weak AcidWeak Acid
SolutionSolution
48. 48
Z5e 14.6 BasesZ5e 14.6 Bases
The OHThe OH--
is a strong base.is a strong base.
Hydroxides of theHydroxides of the alkalialkali metals aremetals are
strong bases because they dissociatestrong bases because they dissociate
completely when dissolved.completely when dissolved.
The hydroxides ofThe hydroxides of alkaline earthsalkaline earths
Ca(OH)Ca(OH)22 etc. are strong dibasic bases,etc. are strong dibasic bases,
butbut they don’t dissolve well in water.they don’t dissolve well in water.
So, used as antacids because [OHSo, used as antacids because [OH--
] can’t] can’t
build up with them, which would hurtbuild up with them, which would hurt
the stomach lining..the stomach lining..
49. 49
BasesBases withoutwithout OHOH--
B-L bases are protonB-L bases are proton acceptorsacceptors..
NHNH33 + H+ H22OO NHNH44
++
+ OH+ OH--
It is the lone pair on nitrogen (N:) thatIt is the lone pair on nitrogen (N:) that
accepts the proton.accepts the proton.
Many weak bases contain NMany weak bases contain N
BB(aq)(aq) + H+ H22O(l)O(l) BHBH++
(aq)(aq) + OH+ OH--
(aq)(aq)
KKbb = [= [BHBH++
][][OHOH--
]]
[[BB]]
50. 50
Strength of BasesStrength of Bases
N:
Hydroxides are strong.Hydroxides are strong.
Others are weak.Others are weak.
SmallerSmaller KKbb == weaker base.weaker base.
Calculate the pH of a solution of 4.0Calculate the pH of a solution of 4.0 MM
pyridinepyridine (K(Kbb = 1.7 x 10= 1.7 x 10-9-9
). Answer . . .). Answer . . .
51. 51
Strength of BasesStrength of Bases
Calculate the pH of a solution of 4.0Calculate the pH of a solution of 4.0 MM
pyridine (Kpyridine (Kbb = 1.7 x 10= 1.7 x 10-9-9
)) DidDid
you get pH = 4.08?you get pH = 4.08?
Oops!!!Oops!!!
You got pYou got pOHOH = 4.08 (= 4.08 (from [OHfrom [OH1-1-
] = 1.2 x 10] = 1.2 x 10-10-10
-- you were given K-- you were given Kbb))
pH = 14 - pOH = 9.92pH = 14 - pOH = 9.92
This makes sense since pyridine is aThis makes sense since pyridine is a
weakweak basebase! (so, high pH)! (so, high pH)
52. 52
Z5e 14.7 Polyprotic acidsZ5e 14.7 Polyprotic acids pppp
These always dissociateThese always dissociate stepwisestepwise..
TheThe firstfirst HH++
comes offcomes off muchmuch easiereasier thanthan
the second.the second.
So, the KSo, the Kaa for the first step is muchfor the first step is much
bigger than Ka for the second.bigger than Ka for the second.
Denoted KaDenoted Ka11, Ka, Ka22, Ka, Ka33
Overall K = KaOverall K = Ka11•Ka•Ka22•Ka•Ka33 . . .. . . (AP quest)(AP quest)
For a typical polyprotic acid in HFor a typical polyprotic acid in H220,0,
only the 1st dissociation step isonly the 1st dissociation step is
important in determining the pHimportant in determining the pH
53. 53
Polyprotic acidPolyprotic acid
HH22COCO33 HH++
+ HCO+ HCO33
--
KaKa11= 4.3 x 10= 4.3 x 10-7-7
HCOHCO33
--
HH++
+ CO+ CO33
-2-2
KaKa22= 4.3 x 10= 4.3 x 10-10-10
The base in first step is acid in second.The base in first step is acid in second.
InIn pHpH calculations we cancalculations we can normallynormally
ignore the second dissociation.ignore the second dissociation.
But, you may have to calculate KaBut, you may have to calculate Ka11 andand
KaKa22 in Lab 14, depending on your acid.in Lab 14, depending on your acid.
54. 54
Calculate the ConcentrationCalculate the Concentration pppp
Of all the ions in a solution of 1.00Of all the ions in a solution of 1.00 MM
Arsenic acid HArsenic acid H33AsOAsO44
KaKa11 = 5.0 x 10= 5.0 x 10-3-3
KaKa22 = 8.0 x 10= 8.0 x 10-8-8
KaKa33 = 6.0 x 10= 6.0 x 10-10-10
Steps. . .Steps. . .
What’s the first step? . . .What’s the first step? . . .
Write the major species, which are . . .Write the major species, which are . . .
HH22O and HO and H33AsOAsO44
55. 55
Calculate the ConcentrationCalculate the Concentration pppp
Of all the ions in a solution of 1.00Of all the ions in a solution of 1.00 MM
Arsenic acid HArsenic acid H33AsOAsO44
Write the =m expression for HWrite the =m expression for H33AsOAsO44 . . .. . .
HH33AsOAsO44 HH1+1+
+ H+ H22AsOAsO44
1-1-
xx22
/ [H/ [H33AsOAsO44] = K] = Ka1a1 = 5.0 x 10= 5.0 x 10-3-3
x = 7.1 x 10x = 7.1 x 10-2-2
= [H= [H1+1+
]] andand [H[H22AsOAsO44
1-1-
]]
56. 56
Calculate the ConcentrationCalculate the Concentration pppp
Of all the ions in a solution of 1.00Of all the ions in a solution of 1.00 MM
Arsenic acid HArsenic acid H33AsOAsO44
Now, write =m expression for HNow, write =m expression for H22AsOAsO44
1-1-
HH22AsOAsO44
1-1-
HH1+1+
+ HAsO+ HAsO44
2-2-
We cannot useWe cannot use xx22
/ [H/ [H22AsOAsO44
1-1-
] = K] = Ka2a2
Why?Why?
[H[H1+1+
] ≠ [HAsO] ≠ [HAsO44
2-2-
] because we have [H] because we have [H1+1+
] of] of
7.1 x 107.1 x 10-2-2
from thefrom the firstfirst =m.=m.
57. 57
Calculate the ConcentrationCalculate the Concentration pppp
Of all the ions in a solution of 1.00Of all the ions in a solution of 1.00 MM
Arsenic acid HArsenic acid H33AsOAsO44
HH22AsOAsO44
1-1-
HH1+1+
+ HAsO+ HAsO44
2-2-
[H[H1+1+
][HAsO][HAsO44
2-2-
]/[H]/[H22AsOAsO44
1-1-
] = . . .] = . . .
(7.1 x 10(7.1 x 10-2-2
)(x) = k)(x) = ka2a2 = 8.0 x 10= 8.0 x 10-8-8
7.1 x 107.1 x 10-2-2
So, x = [HAsOSo, x = [HAsO44
2-2-
] = k] = ka2a2
This is important forThis is important for titrationstitrations..
58. 58
Calculate the ConcentrationCalculate the Concentration pppp
Of all the ions in a solution of 1.00 MOf all the ions in a solution of 1.00 M
Arsenic acid HArsenic acid H33AsOAsO44..
You do it for the last KYou do it for the last Ka3a3 = 6.0 x 10= 6.0 x 10-10-10
[AsO[AsO44
3-3-
] = ???] = ???
6.8 x 106.8 x 10-16-16
from (x)(7.1 x 10from (x)(7.1 x 10-2-2
) = 6.0 x 10) = 6.0 x 10-10-10
8.0 x 108.0 x 10-8-8
59. 59
Sulfuric acid is specialSulfuric acid is special
InIn firstfirst step it is astep it is a strongstrong acid.acid.
In the second step, KaIn the second step, Ka22 = 1.2 x 10= 1.2 x 10-2-2
Calculate the [ ]s of theCalculate the [ ]s of the major speciesmajor species in ain a
3.03.0 MM solution of Hsolution of H22SOSO44. Answer . . .. Answer . . .
Both [HBoth [H++
] and [HSO4] and [HSO4--
] are 3.0] are 3.0 MM, since, since
completely dissociated.completely dissociated.
We can neglect the effect of the 2nd stepWe can neglect the effect of the 2nd step
because it is so small (<5%) compared tobecause it is so small (<5%) compared to
the 1st step complete disassociation.the 1st step complete disassociation.
60. 60
Sulfuric acid is specialSulfuric acid is special
In first step it is a strong acid.In first step it is a strong acid.
In the second step, KaIn the second step, Ka22 = 1.2 x 10= 1.2 x 10-2-2
Calculate the concentrations in a 2.0Calculate the concentrations in a 2.0 x 10x 10-3-3
MM solution of Hsolution of H22SOSO44 (tricky: why?)(tricky: why?)
SinceSince dilutedilute, the 2nd step, the 2nd step doesdoes contributecontribute
HH++
, so quadratic expression must be used, so quadratic expression must be used
because Kabecause Ka22 is not small enough tois not small enough to
simplify the expression.simplify the expression.
A pre-test question requires you to use theA pre-test question requires you to use the
quadratic for an Hquadratic for an H22SOSO44 problem.problem.
61. 61
Z5e 14.8 Salts as acids and basesZ5e 14.8 Salts as acids and bases
Salts areSalts are ionicionic compounds.compounds.
Salts of theSalts of the cationcation ofof strongstrong basesbases andand
thethe anionanion ofof strongstrong acidsacids are neutral.are neutral.
for example, NaCl, KNOfor example, NaCl, KNO33
There isThere is no equilibriumno equilibrium forfor strongstrong acidsacids
and bases.and bases.
So, we can ignore the reverse reaction.So, we can ignore the reverse reaction.
62. 62
BasicBasic SaltsSalts
If theIf the anionanion of a salt is the conjugate base of aof a salt is the conjugate base of a
weakweak acidacid →→ basicbasic solution.solution.
In an aqueous solution of NaFIn an aqueous solution of NaF
thethe major speciesmajor species are Naare Na++
, F, F--
, and H, and H22OO
FF--
+ H+ H22OO HF + OHHF + OH--
KKbb =[HF][=[HF][OHOH--
]]
[F[F--
]]
but Kbut Kaa = [= [HH++
][F][F--
]]
[HF][HF]
66. 66
Basic SaltsBasic Salts
KKaa x Kx Kbb = [HF][OH= [HF][OH--
]] x [Hx [H++
][F][F--
]]
[F[F--
]] [HF][HF]
KKaa x Kx Kbb =[OH=[OH--
] [H] [H++
]]
67. 67
Basic SaltsBasic Salts
KKaa x Kx Kbb = [HF][OH= [HF][OH--
]] x [Hx [H++
][F][F--
]]
[F[F--
]] [HF][HF]
KKaa x Kx Kbb = [OH= [OH--
] [H] [H++
]]
Since [OHSince [OH--
] [H] [H++
] = K] = KWW
KKaa x Kx Kbb = K= KWW
68. 68
KKaa tells us Ktells us Kbb pppp
Calculate the pH of 1.00Calculate the pH of 1.00 MM NaCN.NaCN.
KKaa of HCN is 6.2 x 10of HCN is 6.2 x 10-10-10
Answer . . .Answer . . .
Answer: pH = 11.60. Here’s why . . .Answer: pH = 11.60. Here’s why . . .
The anion of a weak acid is a weakThe anion of a weak acid is a weak
base. Why? Consider the above:base. Why? Consider the above:
2 reactions:2 reactions: (1) dissociation of HCN(1) dissociation of HCN
(2) CN(2) CN--
reacting with Hreacting with H22O.O.
The CNThe CN--
ion competes with OHion competes with OH--
for Hfor H++
Base strength = OHBase strength = OH--
> CN> CN--
> H> H22O.O.
69. 69
KKaa tells us Ktells us Kbb pppp
Calculate the pH of a solution of 1.00Calculate the pH of a solution of 1.00 MM NaCN.NaCN.
KKaa of HCN is 6.2 x 10of HCN is 6.2 x 10-10-10
Ans. pH = 11.60Ans. pH = 11.60
Write major species: NaWrite major species: Na1+1+
, CN, CN1-1-
and Hand H22OO
Set up =m betweenSet up =m between anionanion and waterand water
CNCN1-1-
+ HOH+ HOH HCN + OHHCN + OH1-1-
Since product includes hydroxide ion, mustSince product includes hydroxide ion, must
use Kuse Kbb so need to get from Kso need to get from Kaa (using K(using Kww))
Calculate hydroxide concentration, convert toCalculate hydroxide concentration, convert to
pOH,pOH, subtract from 14 to get pHsubtract from 14 to get pH
70. 70
AcidicAcidic saltssalts
A salt with the cation of aA salt with the cation of a weakweak base and thebase and the
anion of aanion of a strongstrong acid will beacid will be acidicacidic..
The same development as bases leads toThe same development as bases leads to
KKaa x Kx Kbb = K= KWW
Other acidic salts are those of highly chargedOther acidic salts are those of highly charged
metal ions (e.g., aluminum ion -- 3+).metal ions (e.g., aluminum ion -- 3+).
More on this later.More on this later.
71. 71
Acidic saltsAcidic salts pppp
Calculate the pH of a solution of 0.40Calculate the pH of a solution of 0.40 MM NHNH44Cl (the KCl (the Kbb
of NHof NH33 1.8 x 101.8 x 10-5-5
). Steps). Steps
How do we know the correct =m rxn to use? . . .How do we know the correct =m rxn to use? . . .
Write the major species! NHWrite the major species! NH44
1+1+
, Cl, Cl1-1-
, H, H22OO
ClCl--
isis notnot conj. base of WA, but of SA (no equilibrium),conj. base of WA, but of SA (no equilibrium),
so look at NHso look at NH44
++
which is the CA of WB.which is the CA of WB.
Since NHSince NH44
++
isis a major species and NHa major species and NH33 isn’t, the =misn’t, the =m
must be: NHmust be: NH44
++
NHNH33 + H+ H++
and use Kand use Kaa
So,So, don’tdon’t use NHuse NH33 + OH+ OH1-1-
NHNH44
1+1+
or Kor Kbb !!!!
Use KUse Kaa x Kx Kbb = K= Kww to calculate Kto calculate Kaa
Answer for pH is . . .Answer for pH is . . .
72. 72
Anion of weak acid, cation of weak baseAnion of weak acid, cation of weak base
KKaa > K> Kbb acidicacidic
KKaa < K< Kbb basicbasic
KKaa = K= Kbb NeutralNeutral
See Table 14.6, p. 660 for summarySee Table 14.6, p. 660 for summary
74. 74
Z5e 14.9 Structure & Acid base PropertiesZ5e 14.9 Structure & Acid base Properties
A molecule with an H in it is aA molecule with an H in it is a potentialpotential acid.acid.
TheThe strongerstronger the X-Hthe X-H bondbond thethe lessless acidicacidic;; i.e.i.e.,,
more basic (compare bond dissociationmore basic (compare bond dissociation
energies).energies).
TheThe more polarmore polar the X-H bond thethe X-H bond the stronger thestronger the
acidacid (use electronegativities).(use electronegativities).
The more polar H-O-X bond - stronger acid.The more polar H-O-X bond - stronger acid.
75. 75
Strength of oxyacidsStrength of oxyacids
The more oxygen hooked to the centralThe more oxygen hooked to the central
atom, the more acidic the hydrogen.atom, the more acidic the hydrogen.
HClOHClO44 > HClO> HClO33 > HClO> HClO22 > HClO> HClO
Remember that the H is attached to anRemember that the H is attached to an
oxygen atom (oxygen atom (notnot the central atom).the central atom).
The oxygens are electronegative.The oxygens are electronegative.
So, they pull electrons away fromSo, they pull electrons away from
hydrogenhydrogen
80. 80
Hydrated metalsHydrated metals
Highly chargedHighly charged
metalmetal ionsions pull thepull the
electrons ofelectrons of
surrounding watersurrounding water
molecules towardmolecules toward
them.them.
Makes it easier forMakes it easier for
HH++
to come off.to come off.
Al+3 O
H
H
82. 82
Z5e 14.10 Acid-Base Properties of OxidesZ5e 14.10 Acid-Base Properties of Oxides
Non-metalNon-metal oxides dissolved in wateroxides dissolved in water
can make acids.can make acids.
SOSO33(g) + H(g) + H22O(l)O(l) HH22SOSO44(aq)(aq)
IonicIonic (e.g., metal) oxides dissolve in(e.g., metal) oxides dissolve in
water to produce bases.water to produce bases.
CaO(s) + HCaO(s) + H22O(l)O(l) Ca(OH)Ca(OH)22(aq)(aq)
83. 83
Z5e 14.11 Lewis Acids and BasesZ5e 14.11 Lewis Acids and Bases
Most general definition.Most general definition.
Acids are electron pair acceptors.Acids are electron pair acceptors.
Bases are electron pair donors.Bases are electron pair donors.
B F
F
F
:N
H
H
H
84. 84
Lewis Acids and BasesLewis Acids and Bases
Boron trifluoride wants more electrons.Boron trifluoride wants more electrons.
B F
F
F
:N
H
H
H
85. 85
Lewis Acids and BasesLewis Acids and Bases
Boron trifluoride wants more electrons.Boron trifluoride wants more electrons.
BFBF33 is Lewis base NHis Lewis base NH33 is a Lewis Acid.is a Lewis Acid.
BF
F
F
N
H
H
H
86. 86
B. Definitions
Lewis
• AcidsAcids are electron pair acceptors.
• BasesBases are electron pair donors.
Lewis
base
Lewis
acid
87. 87
Lewis Acids and BasesLewis Acids and Bases
Al+3
( )
H
H
O
Al
( )6
H
H
O
+ 6
+3
88. 88
Z5e 14.12 Strategy for solving Acid-Z5e 14.12 Strategy for solving Acid-
Base ProblemsBase Problems pppp
What major species are present?What major species are present?
Does a reaction occur that can beDoes a reaction occur that can be
assumed to go to completion?assumed to go to completion?
What equilibriumWhat equilibrium dominatesdominates thethe
solution?solution?
See guide on next slide and page 667.See guide on next slide and page 667.
89. 89
Z5e Solving Acid-Base Problems p. 667Z5e Solving Acid-Base Problems p. 667 pppp
QuickTime™ and a
None decompressor
are needed to see this picture.
Hinweis der Redaktion
Z5e 14.1 658 The Nature of Acids and Bases
Alternative link if can’t get through to YouTube:
http://dailycackle.com/2007/01/20/science-experiment-gone-wrong/
Z53 Fig. 14.1 659: reaction of HCl and H20
Z5e 660 SE 14.1 a, c, e
Z5e fig. 14.5 p. 661
Z5e 661 14.2 Acid Strength; see Table 14.1
For SE 14.2 -- you need to “remember” that HCl is a strong acid, so Cl- is a weak base (since id does not easily accept a proton (or a H+).
That is, acid strength is inversely proportional to its conjugate base strength.
Z5e 663 SE 14.2
Z5e 663 SE 14.2
Z5e 663 SE 14.2
Rf. Z5e 664
Note: in =m expression, H2O (l) is a pure liquid, so constant and drops out of the expression.
Kw called the ion-product constant (or the dissociation constant)
Rf. Z5e 664
Note: in =m expression, H2O (l) is a pure liquid, so constant and drops out of the expression.
Kw called the ion-product constant (or the dissociation constant)
Z5e 666 SE 14.4
Z5e 664 Fig. 14.7
Z5e 14.3 The pH Scale
Z5e Fig. 14.8 667
Z5e 668. SE 14.6
Z5e 669 Section 14.4: Calculating the pH of Strong Acid Solutions
Z5e 670 SE 14.7b
Z5e 671 Section 14.5: Calculating the pH of Weak Acid Solutions
[H+] = .0060, so pH = 2.22
pOH = 11.78; [OH-] = 1.7 x 10-12 & [H+] = 6.0 x 10-3
Z5e 675
Major species: HF, HOC6H5, H2O
Since HF is dominant producer of H+, use its Ka
Set up “RICE” table, use approximations and 5% rule
[H+] = .030 & pH = 1.53
Z5e 675
Major species: HF, HOC6H5, H2O
Since HF is dominant producer of H+, use its Ka
Set up “RICE” table, use approximations and 5% rule
[H+] = .030 & pH = 1.53
Rf. Z5e 675 SE 14.9
Z5e 677
[H] = 4.24 x 10-3 vs. 1.34 x 10-4
So % dissociation (aka % ionization) = 0.42% vs. 1.3%
Rf. Z5e 680 SE 14.11
Since (X/0.100)(100%) = 8.1%, x = 8.1 x 10-3 = [H+] = [A-], then plug into =m to solve for Ka.
Ans. = 6.6 x 10-4
Z5e 680 fig. 14.10
Z5e 681 Section 14.6 Bases
See next slide for explanation
[hydroxide] = 1.2 x 10-10
Z5e 688 Section 14.7 Polyprotic Acids
[H+] = 7.1 x 10-2
[H2AsO42-] = 7.1 x 10-2 also
[HAsO43-] = 8.0 x 10-8 = Ka2
[AsO43-] = 6.8 x 10-16
Remember: Must consider [H+] from all sources
[H+] = 7.1 x 10-2
[H2AsO42-] = 7.1 x 10-2 also
[HAsO43-] = 8.0 x 10-8 = Ka2
[AsO43-] = 6.8 x 10-16
Remember: Must consider [H+] from all sources
[H+] = 7.1 x 10-2
[H2AsO42-] = 7.1 x 10-2 also
[HAsO43-] = 8.0 x 10-8 = Ka2
[AsO43-] = 6.8 x 10-16
Remember: Must consider [H+] from all sources
[H+] = 7.1 x 10-2
[H2AsO42-] = 7.1 x 10-2 also
[HAsO43-] = 8.0 x 10-8 = Ka2
[AsO43-] = 6.8 x 10-16
Remember: Must consider [H+] from all sources
[H+] = 7.1 x 10-2
[H2AsO42-] = 7.1 x 10-2 also
[HAsO43-] = 8.0 x 10-8 = Ka2
[AsO43-] = 6.8 x 10-16
Remember: Must consider [H+] from all sources
Both [H+] and [HSO4-] are 2 M, since completely dissociated.
Both [H+] and [HSO4-] are 2 M, since completely dissociated.
Z5e 694 Section 14.8 Acid-Base properties of Salts
Z5e 695
Write major species
Set up =m between anion and water
Since product includes hydroxide ion, must use Kb so need to convert from Ka
Calculate hydroxide concentration, convert to pOH, subtract from 14 to get pH
Z5e 695
Write major species
Set up =m between anion and water
Since product includes hydroxide ion, must use Kb so need to convert from Ka
Calculate hydroxide concentration, convert to pOH, subtract from 14 to get pH
Z5e 697
Z5e 697
Z5e 699 Table 14.5
Z5e 701 Section 14.9 The Effect of Structure on Acid-Base Properties
Z5e 702 Figure 14.11
Z5e Section 14.10 Acid-Base Properties of Oxides
Z5e 704 Section 14.11 The Lewis Acid-Base Model
Al ion is Lewis acid and water is Lewis base
Z5e 707 Section 14.12 Strategy for Solving Acid-Base Problems: A Summary
