1 The Chemistry of Acids and Bases Chapter 17. 2 Acid and Bases.

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The Chemistry The Chemistry of Acids and of Acids and BasesBases

The Chemistry The Chemistry of Acids and of Acids and BasesBases

Chapter 17Chapter 17

22

Acid and BasesAcid and Bases

33


44


55

Strong and Weak Strong and Weak Acids/BasesAcids/Bases

• Generally divide acids and bases into Generally divide acids and bases into STRONG or WEAK ones.STRONG or WEAK ones.

STRONG ACID:STRONG ACID: HNO HNO33(aq) + H(aq) + H22O(liq) --->O(liq) --->

HH33OO++(aq) + NO(aq) + NO33--(aq)(aq)

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

66

HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.


77

• Weak acids are much less than 100% ionized in Weak acids are much less than 100% ionized in

water.water.

One of the best known is acetic acid = CHOne of the best known is acetic acid = CH33COCO22HH



88

• Strong Base:Strong Base: 100% dissociated in 100% dissociated in water.water.

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


Other common strong Other common strong bases include KOH and bases include KOH and Ca(OH)Ca(OH)22..

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

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

99

• Weak base:Weak base: less than 100% ionized less than 100% ionized in waterin water

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

NHNH33(aq) + H(aq) + H22O(liq) O(liq) ee NH NH44++(aq) + OH(aq) + OH--(aq)(aq)



1010

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 - LOWRY BRØNSTED - LOWRY theorytheory

• ACIDS DONATE HACIDS DONATE H++ IONS IONS

• BASES ACCEPT HBASES ACCEPT H++ IONS IONS

1111

The Brønsted definition means NHThe Brønsted definition means NH33 is a is a BASEBASE in water — and water is itself an in water — and water is itself an ACIDACID

BaseAcidAcidBaseNH4

+ + OH-NH3 + H2OBaseAcidAcidBase

NH4+ + OH-NH3 + H2O

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

1212

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

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

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

related by the gain or loss of Hrelated by the gain or loss of H++

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

1313

Conjugate PairsConjugate Pairs

1414

More About WaterMore About Water

HH22O 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

Equilibrium constant for autoion = KEquilibrium constant for autoion = Kww

KKww = [H = [H33OO++] [OH] [OH--] = ] = 1.00 x 101.00 x 10-14-14 at 25 at 25 ooCC

1515

More About WaterMore About Water

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--]]

so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22

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

OH-

H3O+

OH-

H3O+

AutoionizationAutoionization

1616Calculating [HCalculating [H33OO++] & ] & [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 [H3OO++] and [OH] and [OH--].].

SolutionSolution

2 H2 H22O(liq) O(liq) e e H H3OO++(aq) + OH(aq) + OH--(aq)(aq)

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

[H[H3OO++] < 10] < 10-7-7 at equilibrium. at equilibrium.

Set up a ICE table.Set up a ICE table.



SolutionSolution

2 H2 H22O(liq) O(liq) ee H H33OO++(aq) + OH(aq) + OH--(aq)(aq)

initialinitial 00 0.00100.0010 changechange +x+x +x+x equilibequilib xx 0.0010 + x0.0010 + xKKww = (x) (0.0010 + x) = (x) (0.0010 + x)

Because x << 0.0010 M, assume Because x << 0.0010 M, assume [OH[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



SolutionSolution

2 H2 H22O(liq) O(liq) e e H H3OO++(aq) + OH(aq) + OH--(aq)(aq)

[H[H3OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M

This solution is _________

because

[H3O+] < [OH-]

1919

[H[H33OO++], [OH], [OH--] and pH] and pH

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

pH = log (1/ [HpH = log (1/ [H33OO++]) ])

= - log [H= - log [H33OO++]]In a neutral solution, In a neutral solution,

[H[H3OO++] = [OH] = [OH--] = ] =

1.00 x 101.00 x 10-7-7 at 25 at 25 ooCC

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

2020

[H[H33OO++], [OH], [OH--] and pH] and pH

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

[H[H3OO++] = 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

2121

Figure 17.1Figure 17.1

2222

[H[H33OO++], [OH], [OH--] and pH] and pH[H[H33OO++], [OH], [OH--] and pH] and pH

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 [H3OO++] then] then

log [Hlog [H3OO++] = - pH] = - pH

Take antilog and getTake antilog and get

[H[H3OO++] = 10] = 10-pH-pH

[H[H3OO++] = 10] = 10-3.12-3.12 = =

7.6 x 107.6 x 10-4-4 M M

2323

pH of Common pH of Common SubstancesSubstances

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 = [H3OO++] [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 [H3OO++] + (-log [OH] + (-log [OH--])])

pKpKww = 14 = pH + pOH = 14 = pH + pOH

2525

Equilibria Involving Equilibria Involving Weak Acids and Weak Acids and

BasesBases

Equilibria Involving Equilibria Involving Weak Acids and Weak Acids and

BasesBases

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

Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and BasesEquilibria Involving Equilibria Involving

Weak Acids and BasesWeak Acids and Bases

AcidAcid Conjugate BaseConjugate Base

acetic, CHacetic, CH33COCO22HH CHCH33COCO22--, acetate, acetate

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

bicarbonate, HCObicarbonate, HCO33-- COCO33

2-2-, carbonate, carbonate

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

2727

Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases

Consider acetic acid, CHConsider acetic acid, CH33COCO22H (HOAc)H (HOAc)

HOAc + HHOAc + H22O O ee H H33OO++ + OAc + OAc--

AcidAcid Conj. base Conj. base

Ka [H3O+][OAc- ]

[HOAc] 1.8 x 10-5Ka

[H3O+][OAc- ][HOAc]

1.8 x 10-5

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

Because [HBecause [H33OO++] and [OAc] and [OAc--] are SMALL, K] are SMALL, Kaa << 1. << 1.

2828

Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids

Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids

Weak acid has KWeak acid has Kaa < 1 < 1

Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7 ] and a pH of 2 - 7

2929

Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases

Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases

Weak base has KWeak base has Kbb < 1 < 1

Leads to small [OHLeads to small [OH--] and a pH of 12 - 7 ] and a pH of 12 - 7

3030Ionization Constants for Ionization Constants for Acids/Bases Acids/Bases

AcidsAcids ConjugateConjugateBasesBases

Increase strength

Increase strength

3131

Relation Relation

of Kof Kaa, K, Kbb, ,

[H[H33OO++] ]

and pHand pH

3232

K and Acid-Base Reactions K and Acid-Base Reactions

Reactions always go from the Reactions always go from the stronger A-B pair (larger K) to the stronger A-B pair (larger K) to the

weaker A-B pair (smaller K).weaker A-B pair (smaller K).

ACIDSACIDS CONJUGATE BASESCONJUGATE BASESACIDSACIDS CONJUGATE BASESCONJUGATE BASES

STRONGSTRONGSTRONGSTRONG

weakweakweakweak

weakweakweakweak

STRONGSTRONGSTRONGSTRONG

3333

K and Acid-Base ReactionsK and Acid-Base ReactionsK and Acid-Base ReactionsK and Acid-Base Reactions

A strong acid is 100% dissociated.A strong acid is 100% dissociated.

Therefore, a Therefore, a STRONG ACIDSTRONG ACID—a good H—a good H++ donor— donor—must have a must have a WEAK CONJUGATE BASEWEAK CONJUGATE BASE—a —a poor Hpoor H++ acceptor. acceptor.

HNOHNO33(aq) + H(aq) + H22O(liq) O(liq) ee H H33OO++(aq) + NO(aq) + NO33--(aq)(aq)

STRONG ASTRONG A basebase acid acid weak Bweak B

•Every A-B reaction has two acids and two Every A-B reaction has two acids and two bases.bases.•Equilibrium always lies toward the weaker pair.Equilibrium always lies toward the weaker pair.•Here K is very large.Here K is very large.

•Every A-B reaction has two acids and two Every A-B reaction has two acids and two bases.bases.•Equilibrium always lies toward the weaker pair.Equilibrium always lies toward the weaker pair.•Here K is very large.Here K is very large.

3434


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

1.1. It is a stronger acid than HIt is a stronger acid than H33OO++

2.2. HH22O is a stronger base than NOO is a stronger base than NO33--

3.3. K for this reaction is largeK for this reaction is large

WEAK BASE

ACID

STRONG ACID

BASEH3O+ + NO3

-HNO3 + H2OWEAK BASE

ACID

STRONG ACID

BASEH3O+ + NO3

-HNO3 + H2O

3535


Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It is a It is a WEAK ACIDWEAK ACID

HOAc + HHOAc + H22O O ee H H33OO++ + OAc + OAc--

WEAK AWEAK A basebase acid acid STRONG BSTRONG B

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

1.1. HH33OO++ is a stronger acid than HOAc is a stronger acid than HOAc

2.2. OAcOAc-- is a stronger base than H is a stronger base than H22OO

3.3. K for this reaction is smallK for this reaction is small

3636

Types of Acid/Base ReactionsTypes of Acid/Base Reactions

Strong acid + Strong baseStrong acid + Strong base

HH++ + Cl + Cl-- + Na + Na++ + OH + OH-- ee H H22O + NaO + Na++ + Cl + Cl--

Net ionic equationNet ionic equation

HH++(aq) + OH(aq) + OH--(aq) (aq) ee H H22O(liq)O(liq)

K = 1/KK = 1/Kww = 1 x 10 = 1 x 101414

Mixing equal molar quantities of a strong acid and strong

base produces a neutral solution.

3737


Weak acid + Strong baseWeak acid + Strong base

CHCH33COCO22H + OHH + OH- - ee H H22O + CHO + CH33COCO22--

This is the reverse of the reaction of CHThis is the reverse of the reaction of CH33COCO22--

(conjugate base) with H(conjugate base) with H22O.O.

OHOH-- stronger base than CH stronger base than CH33COCO22--

K = 1/KK = 1/Kbb = 5.6 x 10 = 5.6 x 1044

Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate

base. The solution is basic.

Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate

base. The solution is basic.

3838


Strong acid + Weak baseStrong acid + Weak base

HH33OO++ + NH + NH33 ee H H22O + NHO + NH44++

This is the reverse of the reaction of NHThis is the reverse of the reaction of NH44++

(conjugate acid of NH(conjugate acid of NH33) with H) with H22O.O.

HH33OO++ stronger acid than NH stronger acid than NH44++

K = 1/KK = 1/Kaa = 5.6 x 10 = 5.6 x 1044

Mixing equal molar quantities of a strong acid and weak base produces the bases’s conjugate

acid. The solution is acid.

Mixing equal molar quantities of a strong acid and weak base produces the bases’s conjugate

acid. The solution is acid.

3939


Weak acid + Weak baseWeak acid + Weak base

•Product cation = conjugate acid of weak base.Product cation = conjugate acid of weak base.•Product anion = conjugate base of weak acid.Product anion = conjugate base of weak acid.•pH of solution depends on relative strengths pH of solution depends on relative strengths

of cation and anion.of cation and anion.

4040

Types of Acid/Base Reactions: Types of Acid/Base Reactions: SummarySummary

4141

Calculations Calculations with with

Equilibrium Equilibrium ConstantsConstants

pH of an acetic pH of an acetic

acid solution.acid solution.

What are your What are your

observations?observations?

pH of an acetic pH of an acetic

acid solution.acid solution.

What are your What are your

observations?observations?

0.0001 M0.0001 M

0.003 M0.003 M

0.06 M0.06 M

2.0 M2.0 M

a pH meter, Screen 17.9a pH meter, Screen 17.9

4242Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid

Equilibria Involving A Weak Equilibria Involving A Weak AcidAcid

You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,

and the pH.and the pH.

Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE

table.table.

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

initialinitial

changechange

equilibequilib



You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,

and the pH.and the pH.

Step 1.Step 1. Define equilibrium concs. in ICE Define equilibrium concs. in ICE

table.table.

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

initialinitial 1.001.00 00 00

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

equilibequilib 1.00-x1.00-x xx xx

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



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

You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

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

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

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

Appendix A).Appendix A).



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


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

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

First assume x is very small because First assume x is very small because KKaa is so small. is so small.

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.

4646

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

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


Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

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

x = x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M

pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) = ) = 2.372.37



Consider the approximate expression Consider the approximate expression

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00 x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2

For many weak acidsFor many weak acids

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

where Cwhere C00 = initial conc. of acid = initial conc. of acid

Useful Rule of Thumb:Useful Rule of Thumb:

If 100•KIf 100•Ka a < C< Coo, then [H, then [H33OO++] = [K] = [Kaa•C•Coo]]1/21/2



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 ee 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

4949

Weak BasesWeak Bases

5050Equilibria Involving A Weak Equilibria Involving A Weak BaseBase

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

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

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

Step 1.Step 1. Define equilibrium concs. in ICE table Define equilibrium concs. in ICE table

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

initialinitial

changechange

equilibequilib




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

Step 1.Step 1. Define equilibrium concs. in ICE table Define equilibrium concs. in ICE table

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

initialinitial 0.0100.010 00 00


equilibequilib 0.010 - x0.010 - x x x xx



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

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

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

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

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 !!




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

5454

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

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

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

5555

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 ----> NHNH33 + + HH33OO++

acidacid basebase 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 17.4 for a summary of See TABLE 17.4 for a summary of acid-base properties of ions.acid-base properties of ions.

Acid-Base Properties of SaltsAcid-Base Properties of Salts

5656

Acid-Base Properties of Acid-Base Properties of SaltsSalts

5757

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- + + HH22O O ee HCOHCO33

-- + OH + OH--


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

changechange equilibequilib


5858



COCO332-2- + + HH22O O ee HCOHCO33

-- + OH + OH--


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

Step 1.Step 1. Set up ICE tableSet up ICE table

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

--]] [OH[OH--] ] initialinitial 0.100.10

00 00


equilibequilib 0.10 - x0.10 - x xx xx


5959



COCO332-2- ++ HH22O O ee HCOHCO33

-- ++ OHOH--


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


Kb = 2.1 x 10-4 = [HCO3

- ][OH- ]

[CO32 ]

x2

0.10 - xKb = 2.1 x 10-4 =

[HCO3- ][OH- ]

[CO32 ]

x2

0.10 - x

Assume 0.10 - x ≈ 0.10, because 100•Kb < Co

x = [HCO3-] = [OH-] = 0.0046 M

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

6060



COCO332-2- + H+ H22O O ee HCOHCO33

-- + OH + OH--


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 M

pOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34

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

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

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