the chemistry of acids and bases chapter 17. 2 strong and weak acids/bases acids and bases into...

The Chemistry of The Chemistry of Acids and BasesAcids and Bases

Chapter 17Chapter 17

22Strong and Weak Acids/BasesStrong and Weak Acids/Bases• Acids and bases into STRONG or WEAK ones.Acids and bases into STRONG or WEAK ones.

STRONG ACIDSTRONG ACID

OrOr

HNOHNO33(aq) ---> H(aq) ---> H++(aq) + NO(aq) + NO33-- (aq) (aq)

HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.

33

H2O

H—Cl

Cl-

H3O+

H2O

H—Cl

Cl-

H3O+

HNOHNO33, HCl, H, HCl, H22SOSO44 and HClO and HClO44 are are

among the strong acids.among the strong acids.

Strong and Weak Acids/BasesStrong and Weak Acids/Bases

44

• Weak acids are Weak acids are much lessmuch less than 100% than 100% ionized in water.ionized in water.

• One of the best known is acetic acid = One of the best known is acetic acid =

HCHC22HH33OO22 = CH = CH33COCO22H = HOAc = CHH = HOAc = CH33COOHCOOH

HCHC22HH33OO22 + H + H22OO <--> C<--> C22HH33OO22-- + H + H33OO

++

oror

HCHC22HH33OO22 <--> C<--> C22HH33OO22- - + H + H++


55

Strong BasesStrong Bases

100% dissociated in water100% dissociated in water

NaOH(aq) ---> NaNaOH(aq) ---> Na++(aq) + OH(aq) + OH--(aq)(aq)


Other common strong bases Other common strong bases

include KOH and include KOH and Ca(OH)Ca(OH)22..

CaOCaO

CaO (lime) + HCaO (lime) + H22O --> O -->

Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)

66

Weak BasesWeak Bases

Less than 100% ionized in waterLess than 100% ionized in water

One of the best known weak bases isOne of the best known weak bases is

ammoniaammonia..


NHNH3 3 (aq)(aq) + H + H22O O ((ll)) <--> <--> NHNH44++

(aq)(aq) + OH + OH-- (aq)(aq)

77

ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES

• The most general theory for common The most general theory for common

aqueous acids and bases is the aqueous acids and bases is the

BRØNSTED - LOWRYBRØNSTED - LOWRY theorytheory

• ACIDS DONATE HACIDS DONATE H++ IONS IONS

• BASES ACCEPT HBASES ACCEPT H++ IONS IONS

88


The Brønsted definition means NHThe Brønsted definition means NH33

is a is a BASEBASE in water — and water is in water — and water is

itself an itself an ACIDACID

BaseAcidAcidBaseNH 4

+ + OH -NH 3 + H 2 O

99


NHNH3 3 / NH/ NH44++ is a is a conjugate pairconjugate pair — related — related

by the gain or loss of Hby the gain or loss of H++

Every acid has a conjugate base - Every acid has a conjugate base - and vice-versa.and vice-versa.

NHNH33 is a is a BASEBASE in water — and water is in water — and water is itself an itself an ACIDACID..

1010

ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESA strong acid is 100% dissociated.A strong acid is 100% dissociated.

Therefore, a Therefore, a STRONG ACID STRONG ACID —— a good H a good H++ donor donor ——

must have a must have a WEAK CONJUGATE BASE WEAK CONJUGATE BASE —— a a

poor Hpoor H++ acceptor. acceptor.

HNOHNO33(aq)(aq) + H + H22OO((ll)) <--> <--> HH33OO++

(aq)(aq) + NO + NO33- -

(aq)(aq)

STRONG ASTRONG A basebase acid acid weak Bweak B

Notice that every A-B reaction has two Notice that every A-B reaction has two acids and two bases!acids and two bases!

1111ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES

We know that HNOWe know that HNO33 is a strong acid. is a strong acid.

1. It is a stronger acid than H1. It is a stronger acid than H33OO++

2. H2. H22O is a stronger base than NOO is a stronger base than NO33--

WEAK WEAK BASEBASE

STRONG STRONG ACIDACID

1212


Acetic acid is only 0.42% ionized when [HCAcetic acid is only 0.42% ionized when [HC22HH33OO22] = 1.0 ] = 1.0

M. It is a M. It is a WEAK ACIDWEAK ACID

Because [HBecause [H33OO++] is small, this must mean] is small, this must mean

1.1. H H33OO++ is a stronger acid than HC is a stronger acid than HC22HH33OO22

2.2. C C22HH33OO22-- is a stronger base than H is a stronger base than H22OO

1313

ACIDS AND BASESACIDS AND BASESACIDS AND BASESACIDS AND BASES

• MonoproticMonoprotic acidsacids can donate only one proton, can donate only one proton,

while while polyproticpolyprotic acidsacids can donate two or more can donate two or more

protons.protons.

• Monoprotic basesMonoprotic bases can accept only one proton, can accept only one proton,

while while polyprotic basespolyprotic bases can accept two or more can accept two or more

protons.protons.

• AmphiproticAmphiprotic substances can behave as either substances can behave as either

acids or bases.acids or bases.

1414

Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions

• Now we can describe Now we can describe reactions of acids with reactions of acids with bases and the bases and the direction of such direction of such reaction. reaction.

• Consider the acid HF Consider the acid HF reacting with the base reacting with the base NHNH33..

• HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--

• Now we can describe Now we can describe reactions of acids with reactions of acids with bases and the bases and the direction of such direction of such reaction. reaction.

• Consider the acid HF Consider the acid HF reacting with the base reacting with the base NHNH33..

• HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--

1515

Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions

Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions

Based on experiment, we can put acids and Based on experiment, we can put acids and bases on a chart.bases on a chart.

See Table 17.3 or Appendix H tables 16See Table 17.3 or Appendix H tables 16

ACIDSACIDS CONJUGATE BASESCONJUGATE BASES STRONGSTRONG weak weak

weak weak STRONG STRONG

This chart can be used to predict the This chart can be used to predict the direction of reactions between any A-B pair.direction of reactions between any A-B pair.

Reactions always go from the stronger A-B Reactions always go from the stronger A-B pair to the weaker A-B pair.pair to the weaker A-B pair.

1616


Predicting the direction of an acid-base reaction.Predicting the direction of an acid-base reaction.

ACID 1 BASE 2 ACID 2 BASE 1+ +

STRONG weak

Reactions always go from the Reactions always go from the

stronger A-B pairstronger A-B pair weaker A-B pair. weaker A-B pair.

1717

MORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATERHH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.

In pure water there can be In pure water there can be AUTOIONIZATIONAUTOIONIZATION

1818

KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

In a In a neutralneutral solution [Hsolution [H33OO++] = [OH] = [OH--]]

and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M

MORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATERAutoionizationAutoionization

2H2H22OO HH33OO+ + OH OH--2H2H22OO HH33OO+ + OH OH--

1919

Calculating [HCalculating [H++] & [OH] & [OH--]]Calculating [HCalculating [H++] & [OH] & [OH--]]

You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [Hpure water. Calculate [H33OO++] and [OH] and [OH--].].

SolutionSolution

HOH <--> HOH <--> H H++ + OH + OH--

Le Chatelier predicts equilibrium shifts Le Chatelier predicts equilibrium shifts to the ____________. to the ____________.

[H[H++] < 10] < 10-7-7 at equilibrium. at equilibrium.

Set up a concentration table.Set up a concentration table.

leftleft

2020

You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of pure water. Calculate [Hpure water. Calculate [H3OO++] and [OH] and [OH--].].

SolutionSolution HH22OO((ll)) <--> H <--> H++ + OH + OH--

Calculating [HCalculating [H++] & [OH] & [OH--]]Calculating [HCalculating [H++] & [OH] & [OH--]]

initialinitial 0 0 0.0010 0.0010

changechange +x +x +x +x

equilibequilib x x 0.0010 + x 0.0010 + x

KKww = [H = [H++][OH][OH--] = (x) (0.0010 + x)] = (x) (0.0010 + x)Because x << 0.0010 M, assume [OHBecause x << 0.0010 M, assume [OH--] = 0.0010 M] = 0.0010 M

[H[H33OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M

This solution is This solution is BASICBASIC because [H because [H33OO++] < [OH] < [OH--]]

2121

[H[H++], [OH], [OH--] and pH] and pH[H[H++], [OH], [OH--] and pH] and pH

A common way to express acidity and A common way to express acidity and basicity is with pH.basicity is with pH.

pH = log (1/ [HpH = log (1/ [H++]) = - log [H]) = - log [H++]]

In a neutral solution,In a neutral solution,

[H[H++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 at 25 at 25 ooCC

pH = -log (1.00 x 10pH = -log (1.00 x 10-7-7) = - (-7.000) ) = - (-7.000) = 7.000 = 7.000

2222


What is the pH of the 0.0010 M NaOH What is the pH of the 0.0010 M NaOH solution? solution?

[H[H++] = 1.0 x 10] = 1.0 x 10-11-11 M M

pH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00

General conclusion —General conclusion —

Basic solution Basic solution pH > 7pH > 7

Neutral Neutral pH = 7pH = 7

Acidic solutionAcidic solution pH < 7pH < 7

2323


If the pH of Coke is 3.12, it is ____________.If the pH of Coke is 3.12, it is ____________.

Because pH = - log [HBecause pH = - log [H++] then] then

log [Hlog [H++] = - pH] = - pH

Take antilog and getTake antilog and get

[H[H++] = 10] = 10-pH-pH

[H[H++] = 10] = 10-3.12-3.12 = =

7.6 x 107.6 x 10-4-4 M M

AcidicAcidic

2424

Other pX ScalesOther pX ScalesOther pX ScalesOther pX Scales

In generalIn general pX = - log XpX = - log X

and so and so pOH = - log [OHpOH = - log [OH--]]

KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

Take the - log of both sidesTake the - log of both sides

- log (10- log (10-14-14) = - log [H) = - log [H33OO++] + (- log [OH] + (- log [OH--])])

pKpKww = 14.00 = pH + pOH = 14.00 = pH + pOH

2525

Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases


Aspirin is a good example Aspirin is a good example of a weak acid, of a weak acid, KKaa = 3.2 x 10 = 3.2 x 10-4-4

2626



AcidAcid Conjugate BaseConjugate Base

acetic, HCacetic, HC22HH33OO22 CC22HH33OO22--, acetate, acetate

ammonium, NHammonium, NH44++ NHNH33, ammonia, ammonia

bicarbonate, HCObicarbonate, HCO33-- COCO33

2-2-, carbonate, carbonate

A weak acid (or base) is one that A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).ionizes to a VERY small extent (< 5%).

2727

• For a weak acid, HA, and its For a weak acid, HA, and its conjugate base, Aconjugate base, A--, ,

KKaa .. K Kbb = K = Kww

• Thus, one can calculate KThus, one can calculate Kaa from K from Kbb and Kand Kbb from K from Kaa..

• Notice as KNotice as Kaa increases, K increases, Kbb decreases decreases and visa versa.and visa versa.

• The stronger the acid the weaker The stronger the acid the weaker the conjugate base.the conjugate base.



2828

Consider acetic acidConsider acetic acid

HCHC22HH33OO22 + H + H22O <--> HO <--> H33OO++ + C + C22HH33OO22--

AcidAcid Conj. Conj. basebase



(K is designated K(K is designated Kaa for an for an ACIDACID))

Because [HBecause [H33OO++] and [C] and [C22HH33OO22--] are SMALL, ] are SMALL,

KKaa << 1. << 1.

K a [H 3O+ ][CC22HH33OO22

-- ]

[HCC22HH33OO22] 1.8 x 10-5

2929



Values of KValues of Kaa for acid and K for acid and Kbb for bases for bases

are found in are found in TABLE 17.4TABLE 17.4 — page 799 — page 799

Notice the relation of TABLE 17.4 to Notice the relation of TABLE 17.4 to

the table of relative acid/base the table of relative acid/base

strengths (Table 17.3).strengths (Table 17.3).

3030

Incr

easi

ng

Aci

d S

tren

gth

Increasin

g B

ase Stren

gth

Partial table 17.3

3131

Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid

You have 1.00 M HOAc. You have 1.00 M HOAc.

Calculate the equilibrium concentrations of Calculate the equilibrium concentrations of HOAc, HHOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Step 1.Step 1. Define equilibrium concentrations. Define equilibrium concentrations.

[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]

initialinitial 1.001.00 0 0 0 0

changechange -x -x +x +x +x +x

equilibequilib 1.00-x1.00-x x x x x

Note that we neglect [HNote that we neglect [H33OO++] from H] from H22O.O.

3232Equilibria Involving A Weak AcidEquilibria Involving A Weak Acid

This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula or method of approximations formula or method of approximations

(see Appendix A.4).(see Appendix A.4).

You have 1.00 M HOAc. Calculate the equilibrium concentrations of HOAc, H3O+, OAc-, and the pH.

Step 2.Step 2. Write K Write Kaa expression expression

Ka 1.8 x 10 -5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - x

3333

And so x = [And so x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2

Ka 1.8 x 10 -5 = x2

1.00

First assume x is very small because KFirst assume x is very small because Kaa is is

so small.so small.

Step 3.Step 3. Solve K Solve Kaa expression expression



Ka 1.8 x 10 -5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - x

3434

Ka 1.8 x 10 -5 = x2

1.00

Step 3.Step 3. Solve K Solve Kaa approximateapproximate expression expression



x = [Hx = [H33OO++] = [OAc] = [OAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2

x = x = [H[H33OO++] = [OAc] = [OAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M

pH = - log [HpH = - log [H33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) = ) = 2.372.37


For many weak acidsFor many weak acids

[H[H33OO++] = [conj. base] = [K] = [conj. base] = [Kaa • C • Coo]]1/21/2

where Cwhere C00 = initial concentration of acid = initial concentration of acid

Useful Rule of Thumb:Useful Rule of Thumb:

If 1000 • KIf 1000 • Kaa < C < Coo, then [H, then [H33OO++] = [K] = [Kaa • C • Coo]]1/21/2

Consider the approximate expressionConsider the approximate expression

x [H3O+

] = [K a • 1.00]1/2

Ka 1.8 x 10 -5 = x2

1.00

3636

Approximation RulesApproximation Rules

10 times K for .1M, so 1010 times K for .1M, so 10K < [X]K < [X]

100 times K for .10M, so 100100 times K for .10M, so 100K < [X]K < [X]

1000 times K for .100M , so 10001000 times K for .100M , so 1000K < [X]K < [X]

Look at the number of decimal places to Look at the number of decimal places to determine factor to multiply K bydetermine factor to multiply K by


Calculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of formic acid, HCOformic acid, HCO22H.H.

HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++

KKaa = 1.8 x 10 = 1.8 x 10-4-4

Approximate solutionApproximate solution

[H[H33OO++] ] = = [K[Kaa • C • Coo]]1/21/2 = 4.2 x 10 = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37

Exact SolutionExact Solution

[H[H33OO++] = [] = [HCOHCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M

[[HCOHCO22HH] = 0.0010 - 3.4 x 10] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M

pH = 3.47 pH = 3.47 NOT ValidNOT Valid

3838Equilibria Involving A Weak BaseEquilibria Involving A Weak Base

You have 0.010 M NHYou have 0.010 M NH33. Calculate the pH.. Calculate the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 1.Step 1. Define equilibrium concentrations. Define equilibrium concentrations.

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial 0.0100.010 0 0 0 0


equilibequilib 0.010 - x0.010 - x x x x x

3939

Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so ), so

x = [OHx = [OH--] = [NH] = [NH44++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 = 0.010 M = 0.010 M

The approximation is validThe approximation is valid !!

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - x



KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 2.Step 2. Solve the equilibrium expression Solve the equilibrium expression

Equilibria Involving A Weak BaseEquilibria Involving A Weak Base

4040



KKbb = 1.8 x 10 = 1.8 x 10-5-5


Step 3.Step 3. Calculate pH Calculate pH

[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M

so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37

Because pH + pOH = 14,Because pH + pOH = 14,

pH = 10.63pH = 10.63

4141

1. The pH of a 0.10 M nicotinic acid solution is 1. The pH of a 0.10 M nicotinic acid solution is 2.92. Calculate K2.92. Calculate Kaa and the % ionization. and the % ionization.

1.4x101.4x10-5 -5 1.2%1.2%

2. A solution of propionic acid, 2. A solution of propionic acid, CHCH33CHCH22COOH, is 0.20 M. KCOOH, is 0.20 M. Kaa = 1.3x10 = 1.3x10-5-5. . Calculate the pH and % ionization.Calculate the pH and % ionization.

2.79 0.80%2.79 0.80%

3. A solution of hydrazine, N3. A solution of hydrazine, N22HH44, is 0.025 M. , is 0.025 M. K Kbb = 8.5x10 = 8.5x10-5-5. . Calculate the pH and % Calculate the pH and % ionization.ionization.

11.15 5.6%11.15 5.6% (4.) (4.)



4242

4. Calculate the pH and 4. Calculate the pH and

concentrations of the arsenate concentrations of the arsenate

containing species in a 0.15 M containing species in a 0.15 M

solution of Hsolution of H33AsOAsO44. . KK11= 2.5x10= 2.5x10-4-4, ,

KK22 = 5.6x10 = 5.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13

2.21, 0.14, 6.1x102.21, 0.14, 6.1x10-3-3,, 5.8x105.8x10-8-8, 2.8x10, 2.8x10-18-18

4343

MX + HMX + H22O ----> acidic or basic solution?O ----> acidic or basic solution?

Consider NHConsider NH44ClCl

NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)

(a)(a) Reaction of ClReaction of Cl-- with H with H22OO

ClCl-- + + HH22O <---->O <----> HCl + HCl + OHOH--

basebase acidacid acidacid basebase

ClCl-- ion is a VERY weak base because its conjugate ion is a VERY weak base because its conjugate acid is strong. acid is strong.

Acid-Base Properties of SaltsAcid-Base Properties of Salts

Therefore, ClTherefore, Cl-- ----> neutral solution ----> neutral solution

XX

4444

NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)

(b)(b) Reaction of NHReaction of NH44++ with H with H22OO

NHNH44++ + H + H22O <---->O <----> NH NH33 + H + H33OO++

acidacidbasebase basebase acidacid

NHNH44++ ion is a moderate acid because its ion is a moderate acid because its

conjugate base is weak. conjugate base is weak.

Therefore, NHTherefore, NH44++ ----> acidic solution ----> acidic solution

See TABLE 17.5 for a summary of acid-See TABLE 17.5 for a summary of acid-base properties of ions.base properties of ions.


4545

Determine if the following solutions are acidic, basic, or neutral. KBr,CrCl3, NaNO2, KHCO3, Na2CO3, NH4C2H3O2

4646

Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .

NaNa++ + H + H22O ---> neutralO ---> neutral

COCO332-2- ++ HH22OO HCOHCO33

-- ++ OHOH--basebase

acidacid acidacid basebase

KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 1.Step 1. Set up concentration tableSet up concentration table

[CO[CO332-2-]] [HCO[HCO33

--]] [OH[OH--]]

initialinitial 0.10 0.10 0 0 0 0


equilibequilib 0.10 - x 0.10 - x x x x x

Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts

4747




-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression


Assume 0.10 - x 0.10, because 100•KAssume 0.10 - x 0.10, because 100•Kbb < C < Coo

x = [HCOx = [HCO33--] = [OH] = [OH--] = 0.0046 M ] = 0.0046 M

Kb = 2.1 x 10-4 = [HCO3

-][OH- ]

[CO32 ]

x2

0.10 - x

4848




-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4


Step 3.Step 3. Calculate the pHCalculate the pH[OH[OH--] = 0.0046 M] = 0.0046 MpOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34pH + pOH = 14, pH + pOH = 14,

so so pH = 11.66pH = 11.66, and the solution is ________., and the solution is ________.Basic

4949

Review ProblemsReview ProblemsReview ProblemsReview Problems

• Calculate the pH and concentrations of Calculate the pH and concentrations of

the arsenate containing species in a 0.15 the arsenate containing species in a 0.15

M solution of HM solution of H33AsOAsO44. . KK11= 2.5x10= 2.5x10-4-4, K, K22 = =

5.6x105.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13

2.21, 0.14, 6.1x102.21, 0.14, 6.1x10-3-3,, 5.8x105.8x10-8-8, 2.8x10, 2.8x10-18-18

• Calculate the pH of a 0.050 M KCalculate the pH of a 0.050 M K22COCO33 solution. solution.

11.5011.50

5050

• Lewis base = Lewis base = electron pair donor electron pair donor (NH(NH33))

Lewis Acids & BasesLewis Acids & Bases

• Lewis acid = Lewis acid = electron pair electron pair acceptor (BFacceptor (BF33))

5151

A Lewis acid and base can interact A Lewis acid and base can interact by sharing an electron pair.by sharing an electron pair.


5252

HH

H

BASEACID

O—H••••

••O—H

H+

++

A Lewis acid and base can interact A Lewis acid and base can interact by sharing an electron pair.by sharing an electron pair.

Formation of Formation of hydronium ion hydronium ion is an is an excellent example.excellent example.


5353

Other good examples involve metal ions.Other good examples involve metal ions.

Lewis Acids & Bases

Colorless MColorless Mxx(NO(NO33))yy form colorful form colorful

hydrated [Mhydrated [Mxx(H(H22O)O)yy]]z+z+ compounds compounds

5454

Such bonds as the HSuch bonds as the H22O ---> Co bond are often O ---> Co bond are often called called COORDINATE COVALENT BONDSCOORDINATE COVALENT BONDS because both electrons are supplied by one of because both electrons are supplied by one of the atoms of the bond.the atoms of the bond.


Other good examples involve metal ions.Other good examples involve metal ions.

HH

Co2+ ••

BASEACID

O—H••••

••O—H

Co2+

5555

The combination of metal The combination of metal ions (Lewis acids) with ions (Lewis acids) with Lewis bases such as HLewis bases such as H22O O

and NHand NH33 ------> ------>

COMPLEX IONSCOMPLEX IONS

All metal ions form All metal ions form complex ions with water complex ions with water — and are of the type — and are of the type [M(H[M(H22O)O)xx]]n+n+ where x = 4 where x = 4

and 6.and 6.


[Cu(NH3)4]2+

5656

Add NHAdd NH33 to light blue [Cu(H to light blue [Cu(H22O)O)44]]2+2+ ------> ------>

light blue Cu(OH)light blue Cu(OH)22 and then deep blue and then deep blue

[Cu(NH[Cu(NH33))44]]2+2+


5757

[Ni(H[Ni(H22O)O)66]]2+2+ + 6 NH + 6 NH33 ---> [Ni(NH ---> [Ni(NH33))66]]2+2+

+ DMG+ DMG


5858

The FeThe Fe2+2+ in heme can interact with O in heme can interact with O22 or or

CO in a Lewis acid-base reaction.CO in a Lewis acid-base reaction.


5959

Many complex ions containing water undergo Many complex ions containing water undergo HYDROLYSISHYDROLYSIS to give acidic solutions. to give acidic solutions.

[Cu(H[Cu(H22O)O)44]]2+2+ + H + H22O ---> [Cu(HO ---> [Cu(H22O)O)33(OH)](OH)]++ + H + H33OO++


6060

Many complex ions containing water undergo Many complex ions containing water undergo HYDROLYSISHYDROLYSIS to give acidic solutions. to give acidic solutions.

This explains why water solutions of FeThis explains why water solutions of Fe3+3+, Al, Al3+3+, , CuCu2+2+, Pb, Pb2+2+, etc. are acidic., etc. are acidic.


6161

This explains This explains AMPHOTERICAMPHOTERIC nature of some nature of some

metal hydroxides.metal hydroxides.

Al(OH)Al(OH)33(s) + 3 H(s) + 3 H++ --> Al --> Al3+3+ + 3 H + 3 H22OO

Here Al(OH)Here Al(OH)33 is a Brønsted base. is a Brønsted base.

Al(OH)Al(OH)33(s) + OH(s) + OH-- --> Al(OH) --> Al(OH)44--

Here Al(OH)Here Al(OH)33 is a Lewis acid. is a Lewis acid. Al 3+ O—H -

••••

••


6262

Amphoterism of Al(OH)Amphoterism of Al(OH)33

Al(OH)Al(OH)33 on right on right Add NaOHAdd NaOH Add HClAdd HCl

See Kotz / Treichel 5See Kotz / Treichel 5thth, page 722, page 722

6363

1. The electronegativity of 1. The electronegativity of the O atoms causes the the O atoms causes the H attached to O to be H attached to O to be highly positive.highly positive.

2. The O—H bond is 2. The O—H bond is highly polar.highly polar.

3. The H atom of O—H is 3. The H atom of O—H is readily attracted to polar readily attracted to polar HH22O. O.

6464

Neutral Lewis AcidNeutral Lewis AcidNeutral Lewis AcidNeutral Lewis AcidCarbon dioxide is a neutral (molecular) Lewis acid.Carbon dioxide is a neutral (molecular) Lewis acid.

+1.5+1.5 -0.75-0.75-0.75-0.75

6565


Many complex ions are very stable.Many complex ions are very stable.

CuCu2+2+ + 4 NH + 4 NH33 <--><--> [Cu(NH[Cu(NH33))44]]2+2+

K for the reaction is calledK for the reaction is called

KKformationformation

or a “formation constant”.or a “formation constant”.

Here K = 6.8 x 10Here K = 6.8 x 101212. .

Rxn. is strongly product-favored.Rxn. is strongly product-favored.

6666

Formation of complex ions explains why you Formation of complex ions explains why you can dissolve a precipitate by forming a can dissolve a precipitate by forming a complex ion. complex ion.

AgCl(s) + 2 NHAgCl(s) + 2 NH33 Ag(NHAg(NH33))22+ + + Cl+ Cl--


AgCl(s)

6767

AgCl(s) AgCl(s) <--> <--> AgAg++ + Cl + Cl-- K Kspsp = 1.8 x 10 = 1.8 x 10-10-10

AgAg++ + 2 NH + 2 NH33 <--> Ag(NH <--> Ag(NH33))22++ K Kformform = 1.6 x 10 = 1.6 x 1077

--------------------------------------------------------------------------

AgCl(s) + 2 NHAgCl(s) + 2 NH33 <--><--> Ag(NHAg(NH33))22+ + + Cl+ Cl--

KKnetnet = K = Kspsp • K • Kformform = 2.9 x 10 = 2.9 x 10-3-3

Lewis Acids & BasesLewis Acids & Bases Formation of complex ions explains why Formation of complex ions explains why

you can dissolve a precipitate by forming a you can dissolve a precipitate by forming a complex ion.complex ion.

Why?Why?Why?Why?• Why are some Why are some

compounds acids?compounds acids?

• Why are some Why are some compounds bases?compounds bases?

• Why do acids and bases Why do acids and bases vary in strength?vary in strength?

• Can we predict Can we predict variations in acidity or variations in acidity or basicity?basicity?

• Why are some Why are some compounds acids?compounds acids?

• Why are some Why are some compounds bases?compounds bases?

• Why do acids and bases Why do acids and bases vary in strength?vary in strength?

• Can we predict Can we predict variations in acidity or variations in acidity or basicity?basicity?

6969

Trichloroacetic acid is much stronger acid Trichloroacetic acid is much stronger acid owing to the high electronegativity of Cl, owing to the high electronegativity of Cl, which withdraws electrons from the rest of which withdraws electrons from the rest of the molecule. This makes the O—H bond the molecule. This makes the O—H bond highly polar. The H of O—H is very positive.highly polar. The H of O—H is very positive.

Acetic acidAcetic acid Trichloroacetic acidTrichloroacetic acidKKaa = 1.8 x 10 = 1.8 x 10-5-5 KKaa = 0.3 = 0.3

7070

Oxyacid StrengthOxyacid StrengthOxyacid StrengthOxyacid Strength

• For oxyacids, XOFor oxyacids, XOnn(OH)(OH)mm, acid , acid

strength increases as n increases strength increases as n increases and is independent of m.and is independent of m.

• Rank in order of increasing Rank in order of increasing acidity: HBrOacidity: HBrO33, H, H33BOBO33, HIO, HIO44

BrO2(OH), B(OH)3, IO3(OH)

B(OH)B(OH)33 BrO BrO22(OH) IO(OH) IO33(OH)(OH)

7171

These ions are These ions are BASES.BASES.They become more and more basic as the They become more and more basic as the

negative charge increases.negative charge increases.

As the charge goes up, they interact more As the charge goes up, they interact more strongly with polar water molecules.strongly with polar water molecules.

NONO33--

COCO332-2-

Basicity of OxoanionsBasicity of OxoanionsBasicity of OxoanionsBasicity of Oxoanions

POPO443-3-

7272

Practice ProblemsPractice Problems1. Write the equation for the reaction of 1. Write the equation for the reaction of

hypochlorous acid and ammonia. Label the hypochlorous acid and ammonia. Label the acids and bases. Indicate the conjugate pairs.acids and bases. Indicate the conjugate pairs.

2. A 0.15 M weak acid solution is determined to 2. A 0.15 M weak acid solution is determined to be 2.0% dissociated. Calculate the Kbe 2.0% dissociated. Calculate the Kaa..

3. Calculate the pH of the following solutions:3. Calculate the pH of the following solutions:

a) 0.50 M nitric acida) 0.50 M nitric acid

b) 0.25 M potassium hydroxideb) 0.25 M potassium hydroxide

c) 0.15 M phosphoric acidc) 0.15 M phosphoric acid

d) 0.22 M sodium sulfited) 0.22 M sodium sulfite

e) 0.010 potassium cyanidee) 0.010 potassium cyanide

7373

Practice ProblemsPractice Problems

4. Calculate the carbonate ion concentration 4. Calculate the carbonate ion concentration in a 0.10 M carbonic acid solution.in a 0.10 M carbonic acid solution.

5. Determine whether a solution of Cs5. Determine whether a solution of Cs22SOSO33 is is acidic, neutral or basic.acidic, neutral or basic.

6. Write the Lewis acid/base reaction 6. Write the Lewis acid/base reaction between PHbetween PH33 and BF and BF33 using dot structures. using dot structures. Indicate the acid and the base.Indicate the acid and the base.

7474

Practice Problems AnswersPractice Problems Answers

1. HClO + NH1. HClO + NH33 <--> NH <--> NH44++ + ClO + ClO--

acidacid basebase acidacid basebase

2. 6.0 x 102. 6.0 x 10-5-5

3. a) .303. a) .30 b) 13.40b) 13.40 c) 1.52c) 1.52

d) 10.28d) 10.28 e) 10.70e) 10.70

4. 4.8 x 104. 4.8 x 10-11-11 5. basic 5. basic

7575

Practice Problems AnswersPractice Problems Answers

6. 6.

. H

H

P ... H . F

F

B ... F+ -->

. H

H P.

.. H

. F

F B.

. . F

The end.

7676

ammoniaammoniaNHNH33 + H + H22O <--> O <--> NHNH44

++ + OH + OH--

oror

NHNH44OH <--> OH <--> NHNH44+ + + OH + OH--

ReturnReturn

7777

Sample QuestionsSample QuestionsSample QuestionsSample Questions

Give the conjugate base of HCN. Give the conjugate base of HCN.

CNCN--

Give the conjugate acid of NOGive the conjugate acid of NO22--. .

HNOHNO22

Give the conjugate acid of NHGive the conjugate acid of NH33..

NHNH44++

The ion HCOThe ion HCO33-- has both a conjugate acid and has both a conjugate acid and

conjugate base. Give the formula of each.conjugate base. Give the formula of each.

HH22COCO33 CO CO332-2-

7878


Label the acid-base pairs: Label the acid-base pairs:

CNCN-- + HOH <--> HCN + OH + HOH <--> HCN + OH--


HCOHCO331-1- + HOH <--> OH + HOH <--> OH-- + H + H22COCO33

basebase acidacid base base acidacid

HCOHCO331-1- + HOH <--> CO + HOH <--> CO33

2-2- + H + H33OO++

acidacid base base basebase acidacid

7979


Consider the acid HF reacting with Consider the acid HF reacting with the base NHthe base NH33..

HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--

acid acid basebase acid acid basebase

Acids: HF > NHAcids: HF > NH44++

Bases: NHBases: NH33 > F > F--

Reaction favors productsReaction favors products

Consider the acid HF reacting with Consider the acid HF reacting with the base NHthe base NH33..

HF + NHHF + NH33 <--> NH <--> NH44++ + F + F--


Acids: HF > NHAcids: HF > NH44++

Bases: NHBases: NH33 > F > F--


8080

Incr

easi

ng

Aci

d S

tren

gth

Increasin

g B

ase Stren

gth

8181


Does the reaction between nitrous acid Does the reaction between nitrous acid and ammonia favor reactants or and ammonia favor reactants or products?products?

HNOHNO22 + NH + NH33 <--> NH <--> NH44++ + NO + NO22

--


Acids: HNOAcids: HNO22 > NH > NH44++

Bases: NHBases: NH33 > NO > NO22--


Does the reaction between nitrous acid Does the reaction between nitrous acid and ammonia favor reactants or and ammonia favor reactants or products?products?

HNOHNO22 + NH + NH33 <--> NH <--> NH44++ + NO + NO22

--


Acids: HNOAcids: HNO22 > NH > NH44++

Bases: NHBases: NH33 > NO > NO22--


8282


Does the reaction between hypochlorous Does the reaction between hypochlorous acid and fluoride ion favor reactants or acid and fluoride ion favor reactants or products? products?

HClO + FHClO + F-- <--> HF + ClO <--> HF + ClO--


Acids: HF > HClOAcids: HF > HClO

Bases: ClOBases: ClO-- > F > F--

Reaction favors reactantsReaction favors reactants

Does the reaction between hypochlorous Does the reaction between hypochlorous acid and fluoride ion favor reactants or acid and fluoride ion favor reactants or products? products?

HClO + FHClO + F-- <--> HF + ClO <--> HF + ClO--


Acids: HF > HClOAcids: HF > HClO

Bases: ClOBases: ClO-- > F > F--


8383

Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions Does the reaction between ammonium Does the reaction between ammonium

chloride and sodium sulfate favor chloride and sodium sulfate favor reactants or products? reactants or products? NHNH44Cl + NaCl + Na22SOSO44 <--> ??? <--> ???

NHNH44++ + SO + SO44

2-2- <--> NH <--> NH33 + HSO + HSO441-1-

acid acid basebase basebase acid acid

Acids: HSOAcids: HSO441-1- > NH > NH44

++

Bases: NHBases: NH33 > SO > SO442-2-


Does the reaction between ammonium Does the reaction between ammonium chloride and sodium sulfate favor chloride and sodium sulfate favor reactants or products? reactants or products? NHNH44Cl + NaCl + Na22SOSO44 <--> ??? <--> ???

NHNH44++ + SO + SO44

2-2- <--> NH <--> NH33 + HSO + HSO441-1-

acid acid basebase basebase acid acid

Acids: HSOAcids: HSO441-1- > NH > NH44

++

Bases: NHBases: NH33 > SO > SO442-2-


8484


1. Determine the pH, pOH, and [OH1. Determine the pH, pOH, and [OH--] ]

[H[H++] = 2.5 x 10] = 2.5 x 10-6-6

pH = - log [HpH = - log [H++] = 5.60] = 5.60

pOH = 14 - 5.60 = 8.40pOH = 14 - 5.60 = 8.40

pOH = - log [OH-]pOH = - log [OH-] or 1.00 x 10or 1.00 x 10-14-14 = [H = [H++][OH][OH--]]

8.40 = - log [OH-] 8.40 = - log [OH-] 1.00 x 101.00 x 10-14-14 = (2.5 x 10 = (2.5 x 10-6-6)[OH)[OH--]]

[OH[OH--] = 10] = 10-8.40-8.40

[OH[OH--] = 4.0 x 10] = 4.0 x 10-9 -9 MM

8585


2. Determine the pOH, [H2. Determine the pOH, [H++], and [OH], and [OH--] ]

pH = 6.52pH = 6.52

pOH = 7.48pOH = 7.48

[H[H++] = 3.0 x 10] = 3.0 x 10-7 -7 MM

[OH[OH--] = 3.3 x 10] = 3.3 x 10-8 -8 MM

How can we determine the pH of a solution?Acid/Base indicators is one method.

8686Name of Indicator pH Interval Color ChangeMethyl violet 0.2 - 3.0 yellow to blue-violetThymol blue 1.2 - 2.8 red to yellowOrange IV 1.3 - 3.0 red to yellowMethyl orange 3.1 - 4.4 red to orange to yellowBromophenol blue 3.0 - 4.6 yellow to blue-violetCongo red 3.0 - 5.0 blue to redBromocresol green 3.8 - 5.4 yellow to blueMethyl red 4.4 - 6.2 red to yellowBromocresol purple 5.2 - 6.8 yellow to purpleBromothymol blue 6.0 - 7.6 yellow to bluePhenol red 6.8 - 8.2 yellow to redThymol blue 8.0 - 9.6 yellow to bluePhenolphthalein 8.3 - 10.0 colorless to redThymolphthalein 9.3 - 10.5 colorless to blueAlizarin yellow 10.0 - 12.0 yellow to redIndigo carmine 11.4 - 13.0 blue to yellow

8787


3. Determine the pH, pOH, [H3. Determine the pH, pOH, [H++], and [OH], and [OH--] ]

0.035 M HCl.0.035 M HCl.

Strong acid --> [HStrong acid --> [H++] = 0.035] = 0.035

pH = 1.46pH = 1.46

pOH = 12.54pOH = 12.54

[OH[OH--] = 2.9 x 10] = 2.9 x 10-13-13 MM

8888


4. Determine the pH, pOH, [H4. Determine the pH, pOH, [H++], and [OH], and [OH--] ] 0.15 M NaOH.0.15 M NaOH.

Strong base --> [OHStrong base --> [OH--] = 0.15 M] = 0.15 M

pOH = 0.82pOH = 0.82

pH = 13.18pH = 13.18

[H[H++] = 6.7 x 10] = 6.7 x 10-14-14 M M

8989


You have 0.10 M HCYou have 0.10 M HC66HH44NONO22. .

Its pH is measured to be 2.92Its pH is measured to be 2.92

Calculate the equilibrium constant.Calculate the equilibrium constant.

Step 1.Step 1. Write the equation. Write the equation.

HA <--> HHA <--> H++ + A + A--

9090


Step 2.Step 2. Make a chart. Make a chart.

HA <--> HHA <--> H++ + A + A--

0.100.10 0 0 0 0

- 0.0012

0.00120.0012

0.00120.0012

0.10


Step 3.Step 3. Write K Write Kaa expression, without and expression, without and

with numbers.with numbers.

Ka = [H+][AA--]

[HAHA]

Ka =

(0.0012)(0.0012.0012)

(0.10 )


Step 4.Step 4. Solve. Solve.

Ka = 1.4 x 10-5

Step 5.Step 5. Answer question(s). Answer question(s).

Ka = 1.4 x 10-5

Ka =

(0.0012)(0.0012.0012)

(0.10 )

9393

Sample QuestionsSample Questions

1. You have 0.050 M HCN. Its pH is 1. You have 0.050 M HCN. Its pH is measured to be 5.4. Calculate the measured to be 5.4. Calculate the equilibrium constant.equilibrium constant.

HCN <--> HHCN <--> H++ + CN + CN--

0.0500.050 0 0 0 0

- 4 x10- 4 x10-6-6 + 4 x10+ 4 x10-6-6 + 4 x10+ 4 x10-6-6

4 x104 x10-6-6 4 x10-60.0500.050

Ka = [H+][CN-]

[HCN]= = 3 x 10-10

(4 x 10-6)2

(0.050)

9494


2. You have 0.1000 M HA which is 1.30% 2. You have 0.1000 M HA which is 1.30% dissociated. dissociated.


HA <--> HHA <--> H++ + A + A--

0.10000.1000 0 0 0 0

- 0.00130- 0.00130 + 0.00130+ 0.00130 + 0.00130+ 0.00130

0.001300.001300.001300.001300.09870.0987

Ka = [H+][A-]

[HA]= = 1.71 x 10-5

(0.00130)2

(0.0987)

9595


You have 1.00 M HCYou have 1.00 M HC22HH33OO22. .

Calculate the equilibrium concentrations Calculate the equilibrium concentrations of HCof HC22HH33OO22, H, H++, C, C22HH33OO22 --, and the pH., and the pH.


HCHC22HH33OO22 <--> H <--> H++ + C + C22HH33OO22--

9696



HCHC22HH33OO22 <--> H <--> H++ + C + C22HH33OO22--

1.001.00 0 0 0 0

-x

xx

+x+x

1.00-x


Step 3.Step 3. Write K Write Kaa expression, without and expression, without and


Ka = [H+][CC22HH33OO22

--]

[HCHC22HH33OO22]

1.8 x 10-5 = (x)(xx)

(1.00 - x)



1.8 x 10-5 = (x)(xx)

(1.00 - x)

1.8 x 10-5(1.00 - x) = x2

x2 + 1.8 x 10-5x - 1.8 x 10-5 = 0

x = 4.2 x 10-3 M

9999



x = [Hx = [H++] = [C] = [C22HH33OO22--] = 0.0042 M] = 0.0042 M

[HC[HC22HH33OO22] = 1.00 - 0.0042 = 1.00 M] = 1.00 - 0.0042 = 1.00 M

pH = 2.38pH = 2.38


Note: Note: Assume x is very small because KAssume x is very small because Kaa

is so small.is so small.

1.8 x 10-5 = (x)(xx)

(1.00 - x)

1.8 x 10-5(1.00) = x2

x = 4.2 x 10-3 M

1.8 x 10-5 = (x)(xx)

(1.00)

General Rule: Simplfy if [HA] / Ka > 1000

101101


1. Calculate the pH of a 0.100 M HF solution.1. Calculate the pH of a 0.100 M HF solution.

HF <--> HHF <--> H++ + F + F--

0.1000.100 0 0 0 0 - x + x + x

x x0.100 - x

Ka = [H+][F-]

[HF]= = 7.2 x 10

- 4 x2

(0.100 - x)

X = 0.0081 M = {H+] pH = 2.09

102102


2. Calculate the pH of a 0.20 M HClO solution.2. Calculate the pH of a 0.20 M HClO solution.

HClO <--> HHClO <--> H++ + ClO + ClO--

0.200.20 0 0 0 0 - x + x + x

x x0.20

Ka = [H+][ClO-]

[HClO]= = 3.5 x 10

- 8 x2

(0.20)

X = 8.4 x 10 - 5 M = {H+] pH = 4.08

103103


3. Calculate the pH of a 0.20 M HCOOH solution.3. Calculate the pH of a 0.20 M HCOOH solution.

HA <--> HHA <--> H++ + A + A--

0.200.20 0 0 0 0 - x + x + x

x x0.20 - x

Ka = [H+][A-]

[HA]= = 1.8 x 10-4

x2

(0.20 - x)

X = 0.0059 M = {H+] pH = 2.23

104104


4. Calculate the pH of a 0.15 M H4. Calculate the pH of a 0.15 M H22COCO33 solution. solution.

HH22COCO33 <--> H <--> H++ + HCO + HCO33--

0.150.15 0 0 0 0- x + x + x

x x0.15

Ka = [H+][HCO3

-]

[H2CO3]= = 4.2 x 10-7

x2

0.15

X = 2.5 x 10-4 M

105105Sample QuestionsSample Questions Does the second equilibrium have a Does the second equilibrium have a

significant effect on the pH? significant effect on the pH?

HCOHCO331-1- <--> H <--> H++ + CO + CO33

2-2-

2.5 x 102.5 x 10-4 -4 2.5 x 102.5 x 10-4 -4 00

- y + y + y

y2.5 x 10-42.5 x 10-4

Ka = [H+][CO3

2-]

[HCO31-]

= = 4.8 x 10-11(2.5 x 10-4)y

2.5 x 10-4

y = 4.8 x 10-11 M pH = 3.60

NO!NO!

106106


You have 0.10 M BOH. Its pH is measured You have 0.10 M BOH. Its pH is measured to be 10.11 to be 10.11



BOH <--> BBOH <--> B++ + OH + OH--

[OH][OH]-- = 10= 10-pOH-pOH = 10= 10-3.89-3.89 = 1.3 x 10= 1.3 x 10-4 -4 MM

107107



BOH <--> BBOH <--> B++ + OH + OH--

0.100.10 0 0 0 0

- 0.00013

0.000130.000130.00013

0.000130.00013

00.10


Step 3.Step 3. Write K Write Kbb expression, without and expression, without and


Kb = [B+][OHOH--]

[BOHBOH]

Kb =

(0.00013)(0.00013.00013)

(0.10)



Kb = 1.7 x 10-7

Kb =

(0.00013)(0.00013.00013)

(0.10)


Kb = 1.7 x 10-7

110110

Sample QuestionSample Question You have 0.0200 M B. You have 0.0200 M B.

Its is 3.0% dissociated. Its is 3.0% dissociated.


B + HOH <--> HBB + HOH <--> HB++ + OH + OH--

0.02000.0200 0 0 0 0- 6.0 x 10- 6.0 x 10-4-4 + 6.0 x 10+ 6.0 x 10-4-4 + 6.0 x 10+ 6.0 x 10-4-4

6.0 x 106.0 x 10-4-4 6.0 x 106.0 x 10-4-40.01940.0194

Kb = [HB+ ][OH-]

[B]= = 1.9x10-5

(6.0 x 10-4)2

(0.0194)

111111


You have 0.100 M NHYou have 0.100 M NH33

Calculate all equilibrium concentrations Calculate all equilibrium concentrations and the pH.and the pH.


NHNH33 + HOH <--> NH + HOH <--> NH44++ + OH + OH--

112112



NHNH33 + HOH <--> NH + HOH <--> NH44++ + OH + OH--

0.1000.100 0 0 0 0

-x

xx

+x+x

0.100




Kb = [NH4

+][OHOH--]

[NHNH33]

1.8 x 10-5 = (x)(xx)

(0.100)



1.8 x 10-5 = (x)(xx)

(0.100)

x = 1.3 x 10-3 M

115115



[NH[NH44++] = [OH] = [OH--] = 0.0013 M] = 0.0013 M

[NH[NH33] = 0.099 M] = 0.099 M

pOH = 2.89pOH = 2.89

pH = 11.11pH = 11.11

116116

Sample QuestionsSample Questions Calculate the pH of a 0.015 M B solution.Calculate the pH of a 0.015 M B solution.

KKbb = 2.5 x 10 = 2.5 x 10-7-7

B + HOH <--> HBB + HOH <--> HB++ + OH + OH--

0.015 0 00.015 0 0- x + x + x

x x0.015

Kb = [HB+ ][OH-]

[B]= = 2.5x10-7

x2

(0.015)

X = 6.5 x 10-5 M = [OH-] pH = 9.79

117117

1. The pH of a 0.10 M nicotinic acid solution is 1. The pH of a 0.10 M nicotinic acid solution is 2.92. Calculate K2.92. Calculate Kaa and the % ionization. and the % ionization.

Sample Problems: Weak Acids and BasesSample Problems: Weak Acids and Bases

HA <--> HHA <--> H++ + A + A--

0.100.10 0 0 0 0

- 0.0012 + 0.0012 + 0.0012

0.0012 0.00120.10

Ka = [H+][A-]

[HA]= = 1.4 x 10-5

(0.0012)2

(0.10)

% ionization = 0.0012/0.10 x 100 = 1.2%

118118

2. A solution of propionic acid, 2. A solution of propionic acid, CHCH33CHCH22COOH, is 0.20 M. KCOOH, is 0.20 M. Kaa = 1.3x10 = 1.3x10-5-5. .

Calculate the pH and % ionization.Calculate the pH and % ionization.


HA <--> HHA <--> H++ + A + A--

0.200.20 0 0 0 0

- x + x + x

x x0.20

Ka = [H+][A-]

[HA]= = 1.3 x 10-5

x2

(0.20)

% ionization = 0.0016/0.20 x 100 = 0.80%

x = 1.6 x 10-3 = [H+], pH = 2.80

119119

3. A solution of hydrazine, N3. A solution of hydrazine, N22HH44, is 0.025 M. , is 0.025 M.

KKbb = 8.5x10 = 8.5x10-5-5. Calculate the pH and % ionization.. Calculate the pH and % ionization.


NN22HH44 + HOH <--> N + HOH <--> N22HH55++ + OH + OH--

0.0250.025 0 00 0- x + x + x

x x0.025

Kb = [N2H5

+][OH-]

[N2H4]= = 8.5 x 10-5

x2

(0.025)

% ionization = 0.0015/0.025 x 100 = 6.0%

x = 1.5 x 10-3 = [OH-], pOH = 2.82 pH = 11.18

1201204. Calculate the pH and concentrations of the 4. Calculate the pH and concentrations of the arsenate containing species in a 0.15 M solution of arsenate containing species in a 0.15 M solution of HH33AsOAsO44. . KK11= 2.5x10= 2.5x10-4-4, K, K22 = =

5.6x105.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13

HH33AsOAsO44 <--> H <--> H++ + + HH22AsOAsO44

--

0.15 0 00.15 0 0- x + x + x

x x0.15 - x

Ka = [H+][H2AsO4

-]

[H3AsO4]= = 2.5 x 10-4

x2

(0.15 - x)

[H3AsO4 ] = 0.14 M [H2AsO4-] = 6.0 x 10-3 M

x = 6.0 x 10-3 M = [H+], pH = 2.22

1211214. Calculate the pH and concentrations of 4. Calculate the pH and concentrations of

the arsenate containing species in a the arsenate containing species in a 0.15 M solution of H0.15 M solution of H33AsOAsO44. .

KK11= 2.5x10= 2.5x10-4-4, K, K22 = 5.6x10 = 5.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13

HH22AsOAsO44-- <--> H <--> H++ + +

HAsOHAsO442-2-

6.0 x 106.0 x 10-3-3 6.0 x 10 6.0 x 10-3-3 0 0- y + y + y

y 6.0 x 10-36.0 x 10-3

Ka = [H+][HAsO4

2-]

[H2AsO4-]

= = 5.6 x 10-86.0 x 10-3 y

6.0 x 10-3

y = 5.6 x 10-8 M = [HAsO42-]

1221224. Calculate the pH and concentrations 4. Calculate the pH and concentrations

of the arsenate containing species in a of the arsenate containing species in a 0.15 M solution of H0.15 M solution of H33AsOAsO44. .

KK11= 2.5x10= 2.5x10-4-4, K, K22 = 5.6x10 = 5.6x10-8-8, K, K33 = 3.0x10 = 3.0x10-13-13

HAsOHAsO442-2- <--> H <--> H++ + AsO + AsO44

3-3-

5.6 x 105.6 x 10-8-8 6.0 x 106.0 x 10-3-3 0 0- z + z + z

z 6.0 x 10-35.6 x 10-8

Ka = [H+][AsO4

3-]

[HAsO42-]

= = 3.0 x 10-136.0 x 10-3 z

5.6 x 10-8

z = 2.8 x 10-18 M = [AsO43-]

123123

Determine if the following solutions are Determine if the following solutions are acidic, basic, or neutral.acidic, basic, or neutral.

KBrKBr

KK++ + HOH <--> KOH + H + HOH <--> KOH + H++

BrBr-- + HOH <--> HBr + OH + HOH <--> HBr + OH--


CrCl3

Cr3+ + HOH <--> CrOH2+ + H+

Cl- + HOH <--> HCl + OH-

XXXX

neutral

XXacidic

124124


NaNONaNO22

NaNa++ + HOH <--> NaOH + H + HOH <--> NaOH + H++

NONO22-- + HOH <--> HNO + HOH <--> HNO22 + OH + OH--


KHCO3

K+ + HOH <--> KOH + H+

HCO3- + HOH <--> H2CO3 + OH- Kb = 2.4x10-8

HCO3- + HOH <--> CO3

2- + H3O+ Ka = 4.8x10-11

XXbasic

XX

basic

125125


NaNa22COCO33

NaNa++ + HOH <--> NaOH + H + HOH <--> NaOH + H++

COCO332-2- + HOH <--> HCO + HOH <--> HCO33

-- + OH + OH--


NH4C2H3O2

NH4+ + HOH <--> NH3 + H3O+ Ka = 5.6x10-10

C2H3O2- + HOH<-->HC2H3O2 +OH- Kb =5.6x10-10

XXbasic

neutral

126126


You have 0.10 M NaYou have 0.10 M Na22COCO33

Calculate the pH.Calculate the pH.



-- + OH + OH--

127127




-- + OH + OH--

0.100.10 0 0 0 0

-x

xx

+x+x

0.10 - x

128128Acid-Base Properties of SaltsAcid-Base Properties of Salts



Kb = [HCO3

-][OHOH--]

[COCO332-2-]

2.1 x 10-4 = (x)(xx)

(0.10 - x)

129129Acid-Base Properties of SaltsAcid-Base Properties of Salts


2.1 x 10-4 = (x)(xx)

(0.10 - x)

x = 4.5 x 10-3 M

130130



[OH[OH--] = 0.0045 M] = 0.0045 M

pOH = 2.35pOH = 2.35

pH = 11.65pH = 11.65

131131


1. Calculate the pH of a 0.10 M K1. Calculate the pH of a 0.10 M K33POPO44 solution.solution.

POPO443-3- + HOH <--> HPO + HOH <--> HPO44

2-2- + OH + OH--

0.100.10 0 0 0 0- x + x + x

x x0.10 - x

Kb = [HPO4

2- ][OH-]

[PO43- ]

= = 0.028 x2

(0.10 - x)

X = 0.041 pH = 12.61

132132


2. Calculate the pH of a 0.10 M Na2. Calculate the pH of a 0.10 M Na22CC22OO44 solution.solution.

CC22OO442-2- + HOH <--> HC + HOH <--> HC22OO44

-- + OH + OH--

0.100.10 0 00 0- x + x + x

x x 0.10

Kb = [HCC22OO44

- ][OH-]

[CC22OO442- ]

= =1.6x10-10 x2

(0.10)

X = 4.0 x 10-6 pH = 8.60

133133

Sample QuestionsSample Questions3. Calculate the pH of a 0.045 M KCN 3. Calculate the pH of a 0.045 M KCN

solution. solution.

CNCN-- + HOH <--> HCN + OH + HOH <--> HCN + OH --

0.045 0 00.045 0 0

- x + x + x

x x 0.045

Kb = [HCNCN][OH-]

[CNCN- ]= = 2.5x10-5

x2

(0.045)

x = 1.1 x 10-3 = [OH-] pOH = 2.96, pH = 11.04

the chemistry of acids and bases chapter 17. 2 strong and weak acids/bases acids and bases into...

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