Ch14 z5e acid base

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Z5e 14.1 The Nature of Acids and Bases Z5e 14.1 The Nature of Acids and Bases Arrhenius DefinitionArrhenius Definition

Acids produce hydrogen ions in Acids produce hydrogen ions in aqueous solution.aqueous solution.

Bases produce hydroxide ions Bases produce hydroxide ions when dissolved in water.when 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 an ammonia could not be an

Arrhenius base.Arrhenius base. http://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&searhttp://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&sear

chch= =

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Bronsted-Lowry DefinitionsBronsted-Lowry Definitions An An acidacid is an proton (H is an proton (H++) ) donordonor and a and a

basebase is a proton is a proton acceptoracceptor.. Acids and bases always come in Acids and bases always come in pairspairs.. HCl is an acid.HCl is an acid. When it dissolves in water it gives its When 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) H H33OO++ + Cl + Cl--

Water is a Water is a basebase: makes hydronium ion: makes hydronium ion

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Figure 14.1Figure 14.1The Reaction of HCI and HThe Reaction of HCI and H22OO

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Bronsted-Lowry Acids

Acids are hydrogen ion (H+) donors

Bases are hydrogen ion (H+) acceptors

HCl + H2O H3O+ + Cl-

donor acceptor + -

+ +

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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

+ +

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B. Definitions ppB. Definitions pp

Brønsted-Lowry

HCl + H2O Cl– + H3O+

•AcidsAcids are proton (H+) donors. •BasesBases are proton (H+) acceptors.

conjugate acidconjugate base

baseacid

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Conjugate Acid-Base Pairs

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B. DefinitionsB. Definitions

H2O + HNO3 H3O+ + NO3–

CBCAAB

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Amphoteric - can be an acid or a base. - can be an acid or a base.

NH3 + H2O NH4+ + OH-

CA CBB A

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F -

H2PO4-

H2O

HF

H3PO4

H3O+

Give the conjugate base for each of the following:

Polyprotic - an acid with more than one H - an acid with more than one H++

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Br -

HSO4-

CO32-

HBr

H2SO4

HCO3-

Give the conjugate acid for each of the following:

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Pairs Pairs pppp

General equation General equation HA(aq) + HHA(aq) + H22O(l) O(l) H H33OO++(aq) + A(aq) + A--(aq)(aq)

Acid + Base Acid + Base Conjugate acid +Conjugate acid + Conjugate Conjugate

basebase This is an equilibrium.This is an equilibrium.

CompetitionCompetition forfor H H++ betweenbetween H H22O 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.

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Acid dissociation constant KAcid dissociation constant Kaa The equilibrium constant for the The equilibrium constant for the

general equation.general equation. HA(aq) + HHA(aq) + H22O(l) O(l) H H33OO1+1+(aq) + A(aq) + A1- 1-

(aq)(aq)

KKaa = [H = [H33OO++][A][A--]]

[HA][HA] HH33OO++ is often written H is often written H++ ignoring ignoring

the water in equation (it is implied).the water in equation (it is implied).

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Acid Dissociation ReactionsAcid Dissociation Reactions Write the dissociation reaction for Write the dissociation reaction for

the following (omitting water).the following (omitting water). HCl . . .HCl . . . HClHCl(aq)(aq) HH1+1+

(aq)(aq) + Cl + Cl1-1-(aq)(aq)

NHNH441+1+ . . . . . .

NHNH441+1+

(aq)(aq) HH1+1+(aq)(aq) + NH + NH3(aq)3(aq)

[Al(H[Al(H22O)O)66]]3+3+ . . . . . .

[Al(H[Al(H22O)O)66]]3+3+ H H1+1+ + + [Al(H[Al(H22O)O)55]OH]OH2+2+

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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 We can write the expression for

any acid.any 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.

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Figure 14.5Figure 14.5Acid Strength Acid Strength Versus Conjugate Versus Conjugate Base Strength Base Strength z5e 661z5e 661

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Z5e 14.2 Acid StrengthZ5e 14.2 Acid Strength Strong acidsStrong acids

KKaa is largeis large

[H[H++] is about equal ] is about equal to to originaloriginal [HA] [HA]

AA-- is a weaker is a weaker base than waterbase than water

Weak acidsWeak acids

KKaa is smallis small

[H[H++] <<< [HA]] <<< [HA]

AA-- is a stronger is a stronger base than waterbase than water

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Z5e Figure 14.4 p. 661Z5e Figure 14.4 p. 661 pppp

Graphic Graphic Representation of Representation of the Behavior of the Behavior of Acids of Different Acids of Different Strengths in Strengths in Aqueous SolutionAqueous Solution

((a) = strong acid, a) = strong acid, complete complete dissociation.dissociation.

((b) = weak acid, b) = weak acid, slight dissociationslight dissociation..

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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.

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Types of AcidsTypes of Acids Polyprotic AcidsPolyprotic Acids - more than 1 - more than 1

acidic hydrogen (diprotic, triprotic).acidic hydrogen (diprotic, triprotic). OxyacidsOxyacids - Proton is attached to - Proton is attached to

the oxygen of an ion.the oxygen of an ion. Organic acidsOrganic acids contain the Carboxyl contain the Carboxyl

group -COOH with the H attached group -COOH with the H attached to Oto O

Polyprotic acids are generally very Polyprotic acids are generally very weak (but, Hweak (but, H22SOSO44 is strong with 1 is strong with 1stst HH++ and weak with 2 and weak with 2ndnd).).

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Relative Base Strength of AcidsRelative Base Strength of Acids pppp

Using Table 14.2 p. 628, arrange Using Table 14.2 p. 628, arrange the following in increasing the following 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 Hint: Strongest acid will have weakest base. Which conjugate of weakest base. Which conjugate of the above bases would be the the above bases would be the strongest acid? . . .strongest acid? . . .

That’s right, HCl. So, ClThat’s right, HCl. So, Cl1-1- must be must be the the weakestweakest base. base.

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Arrange HArrange H22OO, , FF1-1-, Cl, Cl1-1-, NO, NO221-1-, CN, CN1-1- in in

increasing increasing basebase strength. strength. Considering HConsidering H22OO,, the others are the others are

conjugate bases of conjugate bases of weakweak acids, so acids, so the above bases must be the above bases must be strongerstronger than water acting as a base. than water acting as a base. So . . .So . . .

ClCl-- < water < conjugate bases of < water < conjugate bases of weak acids . . .weak acids . . .

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Arrange HArrange H22OO, , FF1-1-, Cl, Cl1-1-, NO, NO221-1-, CN, CN1-1- in in

increasing increasing basebase strength. strength.ClCl-- < water < conj. bases of < water < conj. bases of

weak acids.weak acids. Now, use Table 14.2 p. 628 to Now, use Table 14.2 p. 628 to

“order” the rest. Your answer is . . .“order” the rest. Your answer is . . . ClCl-- < H < H22O < FO < F1-1- < NO < NO22

1-1- < CN < CN1-1-

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AmphotericAmphoteric Behaves as acid & base -- autoionizesBehaves as acid & base -- autoionizes 2H2H22O(l) O(l) H H33OO++(aq) + OH(aq) + OH--(aq)(aq) KKWW= [H= [H33OO++][OH][OH--]=[H]=[H++][OH][OH--]] HH22OO(l)(l) drops out of =m expression since drops out of =m expression since

it is a pure liquidit is a pure liquid At 25ºC KAt 25ºC KWW = 1.0 x10 = 1.0 x10-14-14 in in EVERYEVERY

aqueous solution.aqueous solution. KKww = ion product constant (or = ion product constant (or

dissociation constant)dissociation constant) So, we can use it to find [HSo, we can use it to find [H++] and [OH] and [OH--]]

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SolutionsSolutions In In EVERYEVERY aqueous solution - 3 aqueous solution - 3

possibilities.possibilities. NeutralNeutral solution [H solution [H++] = [OH] = [OH--]= 1.0 ]= 1.0

x10x10-7-7 AcidicAcidic solution [H solution [H++] > [OH] > [OH--]] BasicBasic solution [H solution [H++] < [OH] < [OH--]]

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Autoionization of Water problemAutoionization of Water problem

KKww = 1 x 10 = 1 x 10-13-13 at 60ºC. Using Le at 60ºC. Using Le Chatelier, predict if exo- or Chatelier, predict if exo- or endothermic endothermic 2H2H22OO(l)(l) HH33OO1+1+

(aq)(aq) + OH+ OH1-1-

(aq)(aq) Hint: compare KHint: compare Kww at 25ºC.at 25ºC.

Endothermic since KEndothermic since Kww increases increases with temperature so energy is a with temperature so energy is a reactant.reactant.

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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 Hint: since neutral, they must be equal.equal.

Answer . . .Answer . . . 3 x 103 x 10-7-7 MM since [H since [H++] [OH] [OH--] = 1 x 10] = 1 x 10--

1313

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Figure 14.7Figure 14.7Two Water Molecules React to Form Two Water Molecules React to Form

HH33OO++ and OH and OH--

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Z5e 14.3 The pH ScaleZ5e 14.3 The pH Scale pH= -log[HpH= -log[H++]] Used because [HUsed because [H++] is usually very ] is usually very

smallsmall As pH As pH decreasesdecreases, [H, [H++] ] increasesincreases

exponentiallyexponentially Sig figs: Only the digits Sig figs: Only the digits afterafter the the

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= 8.00 pH= 8.00 22 sig figs sig figs pOH= -log[OHpOH= -log[OH--]] pKa = -log KpKa = -log Kaa

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A solution with pH = 5 is 100 times more acidic than a solution with a pH = ?

A) 7B) 3 C) 0.05

• 7 because . . .

• 7 is 2 pH units more basic so 100 times less [H3O1+] (or 100 x more [OH1-]

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Relationships Relationships 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 [OH and [OH--] = 10] = 10-pOH-pOH

[H[H++],[OH],[OH--],pH and pOH: Given any one ],pH and pOH: Given any one of these we can find the other three.of these we can find the other three.

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Figure 14.8Figure 14.8The pH Scale and pH The pH Scale and pH Values of Some Values of Some

Common SubstancesCommon Substances

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Basic

Acidic Neutral

100

10-1 10-3 10-

5

10-

7

10-

9

10-

11

10-

13

10-

14

[H+]

0 1 3 5 7 9 11

13

14

pH

Basic

100

10-110-310-

5

10-

7

10-

9

10-

11

10-

13

10-

14

[OH-]

01357911

13

14

pOH

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Calculating pH of SolutionsCalculating pH of Solutions AlwaysAlways write down the major write down the major

species in solution. If you forget species in solution. If you forget you’ll probably get a wrong you’ll probably get a wrong solution!solution!

Remember, these are equilibria.Remember, these are equilibria. Remember the chemistry.Remember the chemistry. Don’t try to memorize, there is no Don’t try to memorize, there is no

one way to do this.one way to do this.

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Calculating pH of SolutionsCalculating pH of Solutions The pH of a blood sample was 7.41 at The 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 10 Either 10-6.59-6.59 or or

since Ksince Kw w = [H= [H1+1+][OH][OH1-1-], ], [OH[OH1-1-] = 10] = 10--

1414/3.9 x 10/3.9 x 10-8-8

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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 Above are completely dissociated (strong

acids)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 equilibrium (essentially none since shifted so far to the right)so far to the right)

1010-14-14 = [H = [H++][OH][OH--]] If [HA] < 10If [HA] < 10-7-7 then then waterwater contributes H contributes H++

and must be taken into account!!!and must be taken into account!!!

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Strong Acids Strong Acids pppp

Calculate the pH of 1.0 x 10Calculate the pH of 1.0 x 10-10-10 MM HCl. HCl. Ans.?Ans.?

pH = 7.00. Why? . . .pH = 7.00. Why? . . . Always write the major species, which Always write the major species, which

are . . .are . . . HH22O, HO, H1+1+ and Cl and Cl1-1-

Both contribute HBoth contribute H1+1+ But, the [HCl] is so small that But, the [HCl] is so small that it has no it has no

effecteffect!! Essentially all the HEssentially all the H1+1+ is coming from is coming from

water.water. So, pH = 7.00.So, pH = 7.00.

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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 problem It will be an equilibrium problem

from the startfrom the start (and shifted to the (and shifted to the leftleft).).

Determine whether most of the HDetermine whether most of the H++ will come from the acid or the water!will 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)

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Z5e p637 Solving Weak Acid Equilibrium ProblemsZ5e p637 Solving Weak Acid Equilibrium Problems pppp

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ExampleExample pppp

Calculate the pH of 2.0 Calculate the pH of 2.0 MM acetic acid acetic acid HCHC22HH33OO22 with a Ka 1.8 x10 with a Ka 1.8 x10-5-5 Steps. . .Steps. . .

Major species are . . .Major species are . . . HCHC22HH33OO22 & H & H22O. pH answer is . . . (use O. pH answer is . . . (use

RICE)RICE) pH = 2.22 pH = 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

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A mixture of Weak Acids A 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 So, use bigger Ka if concentrations

are comparable (are comparable (i.e., i.e., can ignore can ignore smaller Ka)smaller Ka)

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Calculate the pH of a Calculate the pH of a mixturemixture of 1.20 of 1.20

MM HF (Ka = 7.2 x 10 HF (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 of Since HF is the dominant producer of

HH++, use its Ka, use its Ka Set up “RICE” table, use Set up “RICE” table, use

approximations and 5% rule to approximations and 5% rule to get . . .get . . .

[H[H++] = .030 & pH = 1.53] = .030 & pH = 1.53

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Calculate the Calculate the [OC[OC66HH551-1-]] of a mixture of a mixture

of 1.20 of 1.20 MM HF (Ka = 7.2 x 10 HF (Ka = 7.2 x 10-4-4) and 3.4 ) and 3.4

MM HOC HOC66HH55 (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 ] concentration came from came from bothboth acids acids

So, 1.6 x 10So, 1.6 x 10-10-10 = ( = (0.0300.030)(x)/(3.4))(x)/(3.4) [OC[OC66HH55

1-1-] = . . . ] = . . .

1.81 x 101.81 x 10-8-8

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Percent dissociation Percent dissociation pppp

= = amount dissociated amount dissociated x 100% x 100% initialinitial concentration concentration

For a For a weakweak acid percent dissociation acid percent dissociation increasesincreases as acid becomes as acid becomes more dilutemore dilute..

Calculate the % dissociation of 1.00 Calculate the % dissociation of 1.00 MM andand 0.100 0.100 MM Acetic acid (Ka = 1.8 x 10 Acetic 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 [H decreases [H++] ] decreasesdecreases but % but %

dissociation dissociation increasesincreases.. Le Chatelier’s principle operates hereLe Chatelier’s principle operates here

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The other wayThe other way pppp

What is the Ka of a weak acid that What is the Ka of a weak acid that is 8.1 % dissociated as 0.100 is 8.1 % dissociated as 0.100 MM solution? Ans. . .solution? 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 ÷ 0.100 = 6.6 x 1010-4-4 (remember, divide by initial (remember, divide by initial [ ])[ ])

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Figure 14.10 Figure 14.10 p. 643.p. 643.The Effect of The Effect of Dilution on Dilution on the Percent the Percent Dissociation Dissociation and (H+) of a and (H+) of a Weak Acid Weak Acid

SolutionSolution

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Z5e 14.6 BasesZ5e 14.6 Bases The OHThe OH--

is a strong base. is a strong base. Hydroxides of the Hydroxides of the alkalialkali metals are metals are

strong bases because they dissociate strong bases because they dissociate completely when dissolved.completely when dissolved.

The hydroxides of The 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 build up with them, which would can’t build up with them, which would hurt the stomach lining..hurt the stomach lining..

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Bases Bases withoutwithout OH OH--

B-L bases are proton B-L bases are proton acceptorsacceptors..

NHNH33 + H + H22O O NHNH44++ + OH+ OH--

It is the lone pair on nitrogen (N:) It is the lone pair on nitrogen (N:) that accepts the proton.that accepts the proton.

Many weak bases contain NMany weak bases contain N BB(aq) (aq) + H+ H22O(l)O(l) BH BH++(aq) (aq) + OH+ OH- -

(aq)(aq)

KKbb = [= [BHBH++][][OHOH- - ]] [ [BB] ]

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Strength of BasesStrength of Bases

N:

Hydroxides are strong.Hydroxides are strong. Others are weak.Others are weak.

Smaller Smaller KKb b == weaker base. weaker base.

Calculate the pH of a solution of Calculate the pH of a solution of 4.0 4.0 MM pyridine pyridine (K(Kbb = 1.7 x 10 = 1.7 x 10-9-9). ). Answer . . .Answer . . .

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Strength of BasesStrength of Bases Calculate the pH of a solution of 4.0 Calculate the pH of a solution of 4.0

MM pyridine (K pyridine (Kbb = 1.7 x 10 = 1.7 x 10-9-9))Did you get pH = 4.08?Did you get pH = 4.08?

Oops!!!Oops!!! You got pYou got pOHOH = 4.08 ( = 4.08 (from [OHfrom [OH1-1-] = 1.2 ] = 1.2

x 10x 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 This makes sense since pyridine is

a weak a weak basebase! (so, high pH)! (so, high pH)

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Z5e 14.7 Polyprotic acids Z5e 14.7 Polyprotic acids pppp

These always dissociate These always dissociate stepwisestepwise.. The The firstfirst H H++ comes off comes off muchmuch easier easier

than the second.than the second. So, the KSo, the Kaa for the first step is much for 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 is only the 1st dissociation step is important in determining the pHimportant in determining the pH

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Polyprotic acidPolyprotic acid HH22COCO33 HH++ + HCO + HCO33

-- KaKa11= 4.3 x = 4.3 x 1010-7-7

HCOHCO33-- HH++ + CO + CO33

-2-2 KaKa22= 4.3 x = 4.3 x 1010-10-10

The base in first step is acid in second.The base in first step is acid in second. In In pHpH calculations we can calculations we can normallynormally

ignore the second dissociation.ignore the second dissociation. But, you may have to calculate KaBut, you may have to calculate Ka11 and and

KaKa22 in Lab 14, depending on your acid. in Lab 14, depending on your acid.

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Calculate the Concentration Calculate the Concentration pppp

Of all the ions in a solution of 1.00 Of all the ions in a solution of 1.00 MM Arsenic acid HArsenic acid H33AsOAsO44 KaKa11 = =

5.0 x 105.0 x 10-3-3 KaKa22 = = 8.0 x 108.0 x 10-8-8 KaKa33 = = 6.0 x 106.0 x 10-10-10 Steps. . . Steps. . .

What’s the first step? . . .What’s the first step? . . . Write the major species, which Write the major species, which

are . . .are . . . HH22O and HO and H33AsOAsO44

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Of all the ions in a solution of 1.00 Of all the ions in a solution of 1.00 MM Arsenic acid H Arsenic acid H33AsOAsO44

Write the =m expression for Write the =m expression for HH33AsOAsO44 . . . . . .

HH33AsOAsO44 H H1+1+ + H + H22AsOAsO441-1-

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-]]

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Of all the ions in a solution of 1.00 Of all the ions in a solution of 1.00 MM Arsenic acid HArsenic acid H33AsOAsO44

Now, write =m expression for Now, write =m expression for HH22AsOAsO44

1-1-

HH22AsOAsO441-1- H H1+1+ + HAsO + HAsO44

2-2-

We cannot use We cannot use xx22 / [H / [H22AsOAsO441-1-] = K] = Ka2a2

Why?Why? [H[H1+1+] ≠ [HAsO] ≠ [HAsO44

2-2-] because we have ] because we have [H[H1+1+] of 7.1 x 10] of 7.1 x 10-2-2 from the from the firstfirst =m. =m.

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Of all the ions in a solution of 1.00 Of all the ions in a solution of 1.00 MM Arsenic acid H Arsenic acid H33AsOAsO44

HH22AsOAsO441-1- H H1+1+ + HAsO + HAsO44

2-2-

[H[H1+1+][HAsO][HAsO442-2-]/[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 for This is important for titrationstitrations..

58


Of all the ions in a solution of 1.00 Of all the ions in a solution of 1.00 M Arsenic acid HM Arsenic acid H33AsOAsO44. .

You do it for the last KYou do it for the last Ka3a3 = 6.0 x 10 = 6.0 x 10--

1010

[AsO[AsO443-3-] = ???] = ???

6.8 x 106.8 x 10-16-16 from (x)(7.1 x 10 from (x)(7.1 x 10-2-2) = ) = 6.0 x 106.0 x 10-10-10 8.0 x 108.0 x 10-8-8

59

Sulfuric acid is specialSulfuric acid is special In In firstfirst step it is a step 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 the Calculate the [ ]s of the major speciesmajor species in a 3.0 in a 3.0 MM solution of H solution 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 We can neglect the effect of the 2nd step because it is so small (<5%) step because it is so small (<5%) compared to the 1st step complete compared to the 1st step complete disassociation.disassociation.

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 = 1.2 x

1010-2-2

Calculate the concentrations in a 2.0 Calculate the concentrations in a 2.0 x x 1010-3-3 MM solution of H solution of H22SOSO4 4 (tricky: why?)(tricky: why?)

Since Since dilutedilute, the 2nd step , the 2nd step doesdoes contribute Hcontribute H++, so quadratic expression , so quadratic expression must be used because Kamust be used because Ka22 is not small is not small enough to simplify the expression.enough to simplify the expression.

A pre-test question requires you to use A pre-test question requires you to use the quadratic for an Hthe quadratic for an H22SOSO4 4 problem. problem.

61

Z5e 14.8 Salts as acids and basesZ5e 14.8 Salts as acids and bases Salts are Salts are ionicionic compounds. compounds. Salts of the Salts of the cationcation of of strong strong basesbases

and the and the anionanion of of strong strong acidsacids are are neutral.neutral.

for example, NaCl, KNOfor example, NaCl, KNO33

There is There is no equilibriumno equilibrium for for strongstrong acids and bases.acids and bases.

So, we can ignore the reverse So, we can ignore the reverse reaction.reaction.

62

BasicBasic Salts Salts If the If the anionanion of a salt is the conjugate of a salt is the conjugate

base of a base of a weakweak acid acid basicbasic solution. solution. In an aqueous solution of NaFIn an aqueous solution of NaF

the the major speciesmajor species are Na are Na++, F, F--, and H, and H22OO

FF- - + H+ H22O O HF + OHHF + OH--

KKbb =[HF][=[HF][OHOH--]]

[F [F- - ]]

but Kbut Kaa = [ = [HH++][F][F--]] [HF] [HF]

63

Basic SaltsBasic Salts KKbb x K x Ka a = [HF][OH= [HF][OH--]] x [Hx [H++][F][F--]]

[F[F- - ] ] [HF][HF]

64

Basic SaltsBasic Salts KKaa x K x Kb b = [HF][OH= [HF][OH--]] x [Hx [H++][F][F--]]

[F[F- - ] ] [HF][HF]

65


[F[F- - ] ] [HF][HF]

66


[F[F- - ] ] [HF][HF]

KKaa x K x Kb b =[OH=[OH--] [H] [H++]]

67


[F[F- - ] ] [HF][HF]

KKaa x K x Kb b = [OH= [OH--] [H] [H++]]

Since [OHSince [OH--] [H] [H++] = K] = KWW

KKaa x K x Kb b = K= KWW

68

KKaa tells us K tells us Kbb pppp

Calculate the pH of 1.00 Calculate 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 weak The 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 reacting with HH22O.O.

The CNThe CN-- ion competes with OH ion competes with OH-- for H for H+ +

Base strength = OHBase strength = OH-- > CN > CN-- > H > H22O.O.

69

KKaa tells us K tells us Kbb pppp

Calculate the pH of a solution of 1.00 Calculate the pH of a solution of 1.00 MM NaCN. KNaCN. Kaa of HCN is 6.2 x 10 of HCN is 6.2 x 10-10-10 Ans. pH = Ans. pH = 11.6011.60

Write major species: NaWrite major species: Na1+1+, CN, CN1-1- and H and H22OO Set up =m between Set up =m between anionanion and water and water

CNCN1-1- + HOH + HOH HCN + OH HCN + OH1-1- Since product includes hydroxide ion, must Since product includes hydroxide ion, must

use Kuse Kbb so need to get from K so need to get from Kaa (using K (using Kww)) Calculate hydroxide concentration, convert Calculate hydroxide concentration, convert

to pOH, to pOH, subtract from 14 to get pHsubtract from 14 to get pH

70

AcidicAcidic salts salts A salt with the cation of a A salt with the cation of a weakweak base base

and the anion of a and the anion of a strongstrong acid will be acid will be acidicacidic..

The same development as bases leads The same development as bases leads to to

KKaa x K x Kb b = K= KWW

Other acidic salts are those of highly Other acidic salts are those of highly charged metal ions (e.g., aluminum ion charged metal ions (e.g., aluminum ion -- 3+).-- 3+).

More on this later.More on this later.

71

Acidic salts Acidic salts pppp

Calculate the pH of a solution of 0.40 Calculate the pH of a solution of 0.40 MM NH NH44Cl (the Cl (the

KKbb of NH of NH33 1.8 x 10 1.8 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-- is is notnot conj. base of WA, but of SA (no conj. base of WA, but of SA (no

equilibrium), so look at NHequilibrium), so look at NH44++ which is the CA of WB. which is the CA of WB.

Since NHSince NH44++ isis a major species and NH a major species and NH33 isn’t, the isn’t, the

=m must be: NH=m must be: NH44++ NH NH33 + H + H++ and use K and use Kaa

So, So, don’tdon’t use NH use NH33 + OH + OH1-1- NH NH441+1+ or K or Kbb !! !!

Use KUse Kaa x K x Kbb = K = Kww to calculate K to calculate Kaa

Answer for pH is . . . Answer for pH is . . . [H[H++] = 1.5 x 10] = 1.5 x 10-5-5 and pH = 4.83 and pH = 4.83

72

Anion of weak acid, cation of weak baseAnion of weak acid, cation of weak base

KKaa > K > Kb b acidicacidic

KKaa < K < Kb b basicbasic

KKaa = K = Kb b NeutralNeutral

See Table 14.6, p. 660 for See Table 14.6, p. 660 for summarysummary

73

Acid-base Properties of Aqueous Ions

74

Z5e 14.9 Structure & Acid base PropertiesZ5e 14.9 Structure & Acid base Properties

A molecule with an H in it is a A molecule with an H in it is a potentialpotential acid.acid.

The The strongerstronger the X-H the X-H bondbond the the lessless acidic acidic; ; i.e.i.e., more basic (compare bond , more basic (compare bond dissociation energies).dissociation energies).

The The more polarmore polar the X-H bond the the X-H bond the stronger stronger the acidthe acid (use electronegativities). (use electronegativities).

The more polar H-O-X bond - stronger The more polar H-O-X bond - stronger acid.acid.

75

Strength of oxyacidsStrength of oxyacids The more oxygen hooked to the The more oxygen hooked to the

central atom, the more acidic the central atom, the more acidic the hydrogen.hydrogen.

HClOHClO44 > HClO > HClO33 > HClO > HClO22 > HClO > HClO

Remember that the H is attached to Remember that the H is attached to an oxygen atom (an oxygen atom (notnot the central the central atom).atom).

The oxygens are electronegative.The oxygens are electronegative. So, they pull electrons away from So, they pull electrons away from

hydrogenhydrogen

76

Strength of oxyacidsStrength of oxyacids

Electron Density

Cl O H

77


Electron Density

Cl O HO

78


Cl O H

O

O

Electron Density

79


Cl O H

O

O

O

Electron Density

80

Hydrated metalsHydrated metals Highly charged Highly charged

metalmetal ionsions pull pull the electrons of the electrons of surrounding surrounding water molecules water molecules toward them.toward them.

Makes it easier Makes it easier for Hfor H++ to come to come off.off.

Al+3 OH

H

81

Metal Ion Surrounded by Water Molecules

82

Z5e 14.10 Acid-Base Properties of OxidesZ5e 14.10 Acid-Base Properties of Oxides

Non-metalNon-metal oxides dissolved in oxides dissolved in water can make acids.water can make acids.

SOSO33(g) + H(g) + H22O(l) O(l) H H22SOSO44(aq)(aq)

IonicIonic (e.g., metal) oxides dissolve (e.g., metal) oxides dissolve in water to produce bases.in water to produce bases.

CaO(s) + HCaO(s) + H22O(l) O(l) Ca(OH) Ca(OH)22(aq)(aq)

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 FF

F

:NH

H

H

84

Lewis Acids and BasesLewis Acids and Bases Boron trifluoride wants more Boron trifluoride wants more

electrons.electrons.

B FF

F

:NH

H

H

85

Lewis Acids and BasesLewis Acids and Bases Boron trifluoride wants more Boron trifluoride wants more

electrons.electrons. BFBF33 is Lewis base NH is Lewis base NH33 is a Lewis is a Lewis

Acid. Acid.

BF

F

F

N

H

H

H

86


Lewis

•AcidsAcids are electron pair acceptors. •BasesBases are electron pair donors.

Lewis base

Lewis acid

87

Lewis Acids and BasesLewis Acids and Bases

Al+3 ( )H

HO

Al ( )6

H

HO

+ 6

+3

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 be Does a reaction occur that can be

assumed to go to completion?assumed to go to completion? What equilibrium What equilibrium dominatesdominates the the

solution?solution? See guide on next slide and page See guide on next slide and page

667.667.

89

Z5e Solving Acid-Base Problems p. 667Z5e Solving Acid-Base Problems p. 667 pppp

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Ch14 z5e acid base