Notes Solutions Chapter 07

7/29/2019 Notes Solutions Chapter 07

1/15

Chapter 7 Acids and Bases

7.1 The Nature of Acids and Bases

7.2 Acid Strength

7.3 The pH Scale7.4 Calculating the pH of Strong Acid Solutions

7.5 Calculating the pH of weak Acid Solutions

7.6 Bases

7.7 Polyprotic Acids

7.8 Acid-Base Properties of Salts

7.9 Acid Solutions in Which Water Contributes to the

H+ Concentration

7.10 Strong Acid Solutions in Which Water Contributes

to the H+ Concentration

7.11 Strategy for solving Acid-Base Problems: A Summary

Arrhenius (or Classical) Acid-Base Definition

An acid is a substance that contains hydrogen and dissociates

in water to yield a hydronium ion : H3O+

A base is a substance that contains the hydroxyl group and

dissociates in water to yield OH -

Neutralization is the reaction of an H+ (H3O+) ion from the

acid and the OH - ion from the base to form water, H2O.

The neutralization reaction is exothermic:

H+(aq) + OH-(aq) H2O(l) + ENERGY

Hydrating water

molecules

H+ with water

More general definition of acids and bases

Brnsted-Lowry Acid - A substance that can donate a hydrogen

ion, e.g. when chloric acid is dissolved in water.

HClO3(aq) + H2O(l) = H3O+(aq) + ClO3

-(aq)

HClO3(aq) = H+(aq) + ClO3

-(aq)

Brnsted-Lowry Base - A substance that can accept a hydrogenion, e.g. when ammonia is dissolved in water:

H2O(l) + NH3(aq) = NH4+(aq) + OH-(aq)

note: The Brnsted-Lowry scheme is not limited to aqueous solutions:

HCl(in NH3) + NH3(l) = NH4+(in NH3) + Cl

-(in NH3)

Acid strength is determined by a competition of

H2O and the conjugate base of the acid for H+

HClO3(aq) + H2O(l) = H3O+(aq) + ClO3

-(aq)acid base acid base

HClO3 (acid) and ClO3- (conjugate base) form a conjugate pair.

H3O+ (acid) and H2O (conjugate base) form a conjugate pair.

H2O(l) + NH3(aq) = NH4+(aq) + OH-(aq)

acid base acid base

H2O (acid) and OH- (conjugate base) form a conjugate pair.

NH4+ (acid) and NH3 (conjugate base) form a conjugate pair.

Figure 7.2:

Acid and

conjugate base

strength

HCl (aq) Cl-

(aq) + H+

(aq)

Acid Conjugate

Base

Conjugate Base

Acid


2/15

EP 120: Which of the following can act as Brnsted-Lowry acids?

Give the formula of the conjugate Brnsted-Lowry base for each

(a) Cl-

not a Brnsted-Lowry acid because it has no hydrogen.

(b) HSO4-

An acid. Conjugate base: SO42-

(c) NH4+

An acid. Conjugate base: NH3

(d) NH3

An acid. Conjugate base: NH2-

(e) H2O

An acid. Conjugate base: OH-

Amphoteric Molecules - Can Function as either

acids or bases, depending on reaction conditions.

Water - (a) acts as an acid in donating a hydrogen ion to

NH3

, (b) acts as a base in accepting a hydrogen ion from

HClO3

Hydrogen phosphate ion can act as an acid:

HPO42-(aq) + H2O(l) = H3O

+(aq) + PO43-(aq)

or as a base:

HPO42-(aq) + H2O(l) = H2PO4

-(aq) + OH-(aq)

Water Dissociates Slightly:H2O(l) + H2O(l) = H3O

+(aq) + OH-(aq)acid base acid base

H2O(l) = H+(aq) + OH-(aq)

We can express this process of autoionization in terms of an

equilibrium constant:

Can assume that concentration of H2O remains constant andcan be absorbed into equilibrium constant, Kw.

[ ][ ]

+ -

23 + -

2 32

2

H O OHH O H O OH

H OwK K K

= = =

[ ]2 -11000g 1

H O 55.518.02g L 1.00

n

V L= = =

Measurements show that the concentration of H+ and

OH- are 1.0 x 10-7 M in pure water.

Kw =[H3O+][OH-]=(1.0 x 10-7)(1.0 x 10-7)= 1.0 x 10-14

at 25 oC.

This relation is always true, independent of adding

an acid or base to the water.

Remember that H2O(l) does not enter into the

equilibrium constant Kw.

Aqueous Acid Solution: contains an excess

of H3O+ ions over OH- ions.

Strong Acid: Ionizes completely in aqueous solution.

Example:

HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)

The single arrow indicates that the reaction goes to

completion. Other strong acids include HBr, HI, H2SO4,

HNO3 and HClO4

EP 121: What is the concentration of OH- ions in a

0.10 M solution of HCl?

[OH-] = Kw/[H3O+] = 1.0 x 10-14/0.10 = 1.0 x 10-13 M

Graphical representation

of the behavior of

acids of different

strengths in aqueous

solution.

(a) Strong acid.

(b) Weak AcidQuantify acid strength

by Ka

( ) ( ) ( )

[ ]

+

+

-HA aq H aq + A aq

H A

HAaK

=

R


3/15

Table 7.1 Various Ways to Describe Acid Strength

Property Strong Acid Weak Acid

Ka value Ka is large Ka is small

Position of the Far to the right Far to the left

Dissociation equilibrium

Equilibrium concentration [H+] ~ [HA]o [H+]


4/15

The equilibrium position in an acid-

base reaction

The stronger acid and the stronger base will always

react to form the weaker conjugate base and theweaker conjugate acid

EP120a: Predict whether the forward or reverse

reaction is favored for the equilibrium

Because CH3COO- and HSO4

2- are the stronger base

and acid, the reverse reaction will be favored

2- - -

3 4 4 3CH COOH(aq) + SO (aq) HSO (aq) CH COO (aq)+R

Aqueous Basic Solution: contains an excess of

OH- ions over H3O+ ions.

Strong Base: Ionizes completely in aqueous solution. Example:

H2O(l) + NaOH (aq) -> Na+(aq) + OH-(aq)

H2O(l) + NH2-(aq) -> NH3(aq) + OH

-(aq)

The single arrow indicates that the reaction goes to completion.

EP 122: What is the concentration of H3O+ ions in a 0.10 M solution

of NaOH?

[H3O+] = Kw/[OH

-] = 1.0 x 10-14/0.10 = 1.0 x 10-13 M

The Definition of pH

pH = -log10[H3O+]

Significant Figures for Logarithms

The number of decimal places in the log is

equal to the number of significant figures in

the original number.

EP 123: What is the pH of (a) pure water, (b)

0.10M HCl, and (c) 0.1M NaOH?

(a)For pure water, [H+] = 1.0 x 10-7 M, so

pH = -log10[1.0 x10-7] = 7.00

(b) The pH of 0.10 M HCl is

pH = -log10[1.0 x 10-1] = 1.00

(c)The pH of 0.1 M NaOH is

pH = -log10(1.0 x 10-14/0.1) = -log10[1 x 10

-13] = 13.0

Classification of pH Values

pH < 7 Acidic Solution

pH = 7 Neutral Solution

pH > 7 Basic Solution

The pH Value of

Some Familiar

Aqueous

Solutions

The pH scale is a compact

way to represent solution

acidity. It involves base 10

logs (log), not natural logs(ln)


5/15

Methods for Measuring the pH of an

Aqueous Solution

(a) pH paper (b) Electrodes of a pH meter

EP 124:Calculating pH

Calculate the pH of an aqueous HNO3 solution that

has a volume of 250.mL and contains 0.465g HNO3.

[ ]

333 3

3

3+3

3

+

10

1 mol HNO0.465g HNO 7.38 10 mol HNO

63.01g HNO

7.38 10 mol HNOHNO 0.0295 H

0.250L

pH = -log H 1.530

M

=

= = =

=

The Stepwise Dissociation of Phosphoric Acid

Phosphoric acid is a weak acid, and normally only loses one proton

in solution, but it will lose all three when reacted with a strong base.

The ionization constants are given for comparison.

H3PO4 (aq) + H2O(l) H2PO4-(aq) + H3O

+(aq)

H2PO4-(aq) + H2O(l) HPO4

2-(aq) + H3O

+(aq)

HPO42-

(aq) + H2O(l) PO43-

(aq) + H3O+

(aq)

H3PO4 (aq) + 3 H2O(l) PO43-

(aq) + 3 H3O+

(aq)

Ka,1

Ka,2

Ka,3

3

1 7.2 10aK=

8

26.3 10aK

=

13

3 4.2 10aK=

221.9 10aK

=

Ka always decreases for successive H+

Two new definitions

1) pOH = -log10[OH-]

= - log10(Kw/[H+]) = - log10Kw - log10[H

+])

= - log10 (1.0 x 10-14) pH = 14.00 - pH

pOH = 14.00 pH

2) It is convenient to express equilibrium constants

for acid dissociation in the form

pKa = - log10Ka


6/15

The Relationship Between Ka and pKa

Acid Name (Formula) Ka at 25oC pKa

Hydrogen sulfate ion (HSO4-) 1.02 x 10-2 1.991

Nitrous acid (HNO2) 7.1 x 10-4 3.15

Acetic acid (CH3COOH) 1.8 x 10-5 4.74

Hypobromous acid (HBrO) 2.3 x 10-9 8.64

Phenol (C6H5OH) 1.0 x 10-10 10.00

Calculation of the pH of Weak Acid

Solutions - A Systematic Approach

EP 125: What is the pH of a solution of 1.00 M nitrous acid, Ka =4.0 x 10-4? What % of acid is ionized?

(1) Identify major species in solution:

HNO2 and H2O

(2) Which species can generate H+ ions?

HNO2(aq) = H+(aq) + NO2

-(aq) Ka = 4.0 x 10-4

H2O(aq) = H+(aq) + OH-(aq) Kw = 1.0 x 10

-14

(3) Is there a dominant species for H+(aq) production?

Because Ka >> Kw , can ignore contribution from water.

The equilibrium expression is

[ ]

+ -

24

2

H NO4.0 10

HNOaK

= =

The initial concentration are

[HNO2]0 = 1.00 M, [NO2-]0 = 0, [H

+]0 = 10-7 M ~ 0


-(aq)

Construct the reaction table:

001.00Initial

NO2-(aq)H+(aq)HNO2(aq)Conc.

(M)


-(aq)


xx-xChange

001.00Initial


(M)


-(aq)


xx1.00-xAt equlib

xx-xChange

001.00Initial


(M)


7/15

We rearrange this equation to yield a second

order polynomial:

a

acbbx

cbxax

xx

2

4

solutionsthehaswhich

0

formtheofispolynomialThe

0100.4100.4

2

2

442

=

=++

=+

[ ]2

+ - 24

2

H N O4.0 10

H N O 1.00

xK

x

= = =

For this equation, the solutions are:

x = 0.0198, -0.0202

Only the first solution is valid because it leads to

all positive concentrations.

[H+] = [NO2-] = 0.020 M

[HNO2] = 1.00 - 0.020 = 0.98 M

Question (a): What is the pH of this solution?

pH =-log([H+]) = -log(0.020) = 1.70

Question (b): Were we correct to neglect H+ from

the water

Yes because [H+]HNO2 >>[H+]w (0.020>> 10

-7)

Question (c): What % of the acid is ionized?

% ionized = ([H+]/[HNO2]0 ) x 100% =

(0.020/1.00) x 100% = 2.0 %

Although Ka is independent of molarity, the

percent dissociation depends on molarity

fraction ionized =

([H+]/[HNO2]0 ) Ka/[NO2-]

The pH of a Mixture of Weak Acids

EP 126: Calculate the pH of a mixture that is 1.00 M in

phenol (Ka = 1.6 x 10-10 and 5.00 M in acetic acid (Ka =

1.8 x 10-5).

Major Species in Solution: phenol (HPhe), acetic acid

(HAc) and H2OWhich Species Can Generate H+ ions?

HAc(aq) = H+(aq) + Ac-(aq) KHAc = 1.8 x 10-5

HPhe(aq) = H+(aq) + Phe-(aq) KHPhc = 1.8 x 10-10

H2O(aq) = H+(aq) + OH-(aq) Kw = 1.0 x 10

-14

Because KAc >> KHPhe >> Kw

Can ignore contribution from water and phenol.

HAc(aq) = H+(aq) + Ac-(aq)


At equlib

Change

005.00Initial

Ac -(aq)H+(aq)HAc(aq)Conc.

(M)


8/15



At equlib

xx-xChange

005.00Initial


(M)



xx5.00-xAt equlib

xx-xChange

005.00Initial


(M)

Focusing on the Acetic Acid equilibrium:

[ ]

+ - 25

H Ac1.8 10

H Ac 5.00

xK

x

= = =

9.00 x 10-5 1.8 x 10-5 x = x2

x2 + 1.8 x 10-5 x 9.00 x 10-5 = 0

x = .00948, -.00950 . The positive root is the correct

choice.

pH = -log(.0095) = 2.02

Position of phenol equilibrium

EP 127: What is the molarity of Phe- in the

equilibrium discussed in EP 126?

[ ]( ) ( )3+ -

10

3 3

108

3

9.48 10H Phe1.6 10

HPhe 1.00

1.00 1.00 9.48 10 9.48 10

1.6 101.6 10 M

9.48 10

y yK

y

y y

y

+ = = =

+

= =

Very little dissociation of HPhe.Very little H+ from HPhe.

EP 128: If the degree of dissociation of 0.10 M

propanoic acid is 1.1%, what is Ka?

We know [H+] = [Pr-], [HPr] = [HPr]0 [H+]

( )( )

+ - +

0

+

2

5

[H ][Pr ] [H ]; from dissociation: 0.011

[HPr] [HPr]

[H ] 0.011 0.10 .0011 [Pr ]

[HPr] 0.10 0.0011 0.099

0.00111.2 10

0.099

a

a

K

K

= =

= = =

= =

= =

Finding the Ka of a weak acid

from % dissociation. EP 129: Calculate the pH of a solution of 3.0 x10-3 M Ca(OH)2(aq)

( )

3 3

1412

3

12

[OH ] 2 3.0 10 6.0 10

1.0 10[H ] 1.7 10[OH ] 6.0 10

log 11.7 10 1.78

w

M M

K M

pH

+

= =

= = =

= =


9/15

Note: A base doesnt have to contain OH-, it just

needs to be able to accept H+, e.g. aqueous

ammonia. NH3(aq) + H2O(l) = NH4+(aq) + OH-(aq)

Kb refers to an equilibrium in which a base forms its

conjugate acid by removing H+ from water.

B(aq) + H2O(l) = BH+

(aq) + OH-

(aq)base acid conj. Acid conj. base

=

= + - +

- +

b +

a

a

b w

wK[BH ][OH ] [BH ]

K = [OH ][H ] =[B] [B][H ] K

K K K

[H2O] is essentially constant at 55.5M and is contained in Kb.

EP 130: What is the pH of 1.5 M

Dimethylamine (CH3)2NH (Kb = 5.9 x 10-4).

Equil.

Change

001.5Initial

OH-DMAH+DMAConc.(M)

(CH3)2NH(aq) +H2O(l) (CH3)2NH2+(aq) + OH-(aq)


Let xbe the amount of dimethylamine that has dissociated.

Equil.

+x+x-xChange

001.5Initial

OH-DMAH+DMAConc.

(M)


At equilibrium

xx1.5 - xEquil.

+x+x-xChange

001.5Initial

OH-DMAH+DMAConc.

(M)

24

4 4 2

2 4 4

-

1413

5.9 101.5

8.85 10 5.9 10

5.9 10 8.85 10 0;

OH 0.030 , 0.029

only positive root is meaningful

1.0 10[H ] 3.3 10 ; 12

0. 3.4

0 08

b

xK

x

x x

x x

x M M

pH

+

= =

=

+ =

= =

=

= =

Polyprotic Acids

Can give more than one proton per molecule of acid.

They do this in a step-wise manner.

Example: oxalic acid:

H2C2O4(aq) = H+(aq) + HC2O4

-(aq) Ka1 = 5.6 x 10-2

HC2O4-(aq) = H+(aq) + C2O4

2-(aq) Ka2 = 5.4 x 10-5


10/15

Polyprotic AcidsEP 131: Calculate the pH of 0.050 M Ascorbic Acid,

H2Asc (Ka1 = 1.0 x 10-5; Ka2 = 5.0 x 10

-12). Also

calculate the concentration of Asc2-.

H2Asc H+ + HAsc-

HAsc- H+ + Asc2-

Simultaneous equilibria. [H+] has same value in

both equilibria!

Can simplifying assumptions be made?

+ - 25

1

2

5 5 2

2 5 7

4

[H ][HAsc ]1.0 10

[H Asc] 0.050

0.05 10 1.0 10

1.0 10 5.0 10 0;[H ] 7 3.15.0 10 ;

a

xK

x

x x

Hx

pxx

+

= = =

=

+ =

== =

( )( )( )

4+ 2-12

2 - 4

2 12

7.0 10[H ][Asc ]5.0 10

[HAsc ] 7.0 10

[Asc 5.0 10]

a

yK

y M

= = =

= =

Calculate [H+] from Ka1,

substitute this value in seconddissociation equilibrium.

This OK only because K

values very different

In some cases, further dissociation of a

polyprotic acid can not be neglected

EP 132: Calculate the pH of a 0.10 M sulfuric acid

solution. H2SO4 (first dissociation complete; Ka2 =

1.2 x 10-2).

Major species in solution are H+(aq), HSO4-(aq)

and H2O. HSO4-(aq) can dissociate further.

HSO4-(aq) = H+(aq) + SO4

2-(aq)

Examine this equilibrium. Initial concentrationsare [H+(aq)]=0.10M, [HSO4

-(aq)]=0.10M.

At equilibrium

x0.10 + x0.10 - xEquil.

+x+x-xChange

00.100.10Initial

SO42-(aq)H+(aq)HSO4

-

(aq)Conc.

(M)

( )+ 2- 242 -

4

3

3 2

2

0.10[H ][SO ]1.2 10

[HSO ] 0.10

9.8 10

[H ] 0.010 9.8 10 2.0 10 ;

If had neglected 2nd d

1.70

2.00

issociation,

[H ] 1.0 10 ;

a

x

pH

pH

xK

x

x

x

+

+

+= = =

=

= + =

== =

=

Zumdahl shows in Example 7.9 that can neglect dissociation

of HSO4- for a 1.0M sulfuric acid solution.

What species are present in a solution of a

polyprotic acid depends on whether the pH is

determined only by the acid, or by the

presence of another source of H+.

+ 7

2 3 3 1

+ 2 11

3 3 2

2 +

2 3 3 3

H CO (aq) H (aq) + HCO (aq) 4.3 10

HCO (aq) H (aq) + CO (aq) 4.8 10

can have H CO (aq), HCO (aq), CO (aq), and H (aq)

in solution

a

a

K

K

=

=

R

R


11/15

[ ]

[ ]

[ ]

3

3 2

2 3 3 3

2

32 3

3 3

+ 2

32 3 2

+13 3

3 +

2

+1

HCOfraction HCO

H CO + HCO CO

1

COH CO

+ 1HCO HCO

H COH CO

HCO HCO H

1fraction HCO

H+ 1

H

a

a

a

a

K

K

K

K

= +

= +

= =

= +

Fraction of HCO3- (and other species) determined by pH

which can be changed by adding a strong acid such as

HCl

[ ]

+

2 3 3

+

3

1

2 3

2 +3 3

2 +

3

2

3

H CO HCO H

HCO H

H CO

HCO CO H

CO H

HCO

a

a

K

K

+

=

+

=

R

R

Acid-Base Properties of SaltsSalt an ionic compound that dissolves in H2O to give

ions. Sometimes the ions can behave as acids or bases.

(a) Strong base + weak acid yields a basic salt (pH> 7).

NaOac

(b)Strong acid + weak base yields an acidic salt (pH< 7).

NH4Cl

(c) Strong acid + strong base yields a neutral salt (pH= 7).

NaCl

(d)a-c are general rules that have some exceptions.

In each case, do ions react with water to produce

H+ or OH-?

EP 133: What is the pH of a 0.25 M Sodium

Acetate (NaAc) solution?

Let xbe the amount of acetic acid that is formed by thefollowing reaction: Ac-(aq) + H2O(l) = HAc(aq) + OH

-(aq).

Equil.

Change

000.25Initial

OH-HAc +Ac-Conc.

(M)The Na+ is a

spectator ion

and can be

ignored in

terms of its

effect on pH

Ac-(aq) + H2O(l) = HAc(aq) + OH-(aq)

Equil.

+x+x-xChange

000.25Initial

OH-HAc +Ac- =Conc.

(M)

Ac-(aq) + H2O(l) = HAc(aq) + OH-(aq)

xx0.25 - xEquil.

+x+x-xChange

000.25Initial

OH-HAc +Ac- =Conc.

(M)


12/15

1410

5

1.0 105.6 10

1.8 10

wb

a

KK

K

= = =

- 2

10-

10 10 2

2 10 10

5 5

[HAc][OH ]5.6 10 [Ac ] 0.25

1.4 10 5.6 10

5.6 10 1.4 10 0;

1.2 10 , 1.2 10

x

x

x x

x x

x

= =

=

+ =

=

1

5

410

5

If assum

1

e

.0 10[H ] 8.3 10 ;

1.

0.25 0.25, 1.2 10 ; 9.08

2 109.08p

x

H

x pH

+

= =

= =

=

EP 134: What is the pH of a 0.15 M

ammonium chloride (NH4Cl) solution?

Let xbe the amount of H3O+ that is formed by the following

reaction: NH4+

(aq) + H2O(l) = NH3(aq) + H3O+

(aq)

Equil.

Change

000.15Initial

H+(aq)NH3(aq)NH4+(aq)Conc.

(M)

The Cl- is a

spectator ion

and can be

ignored in

terms of its

effect on pH

5

3given that 1.8 10 for NHbK

=

NH4+(aq) + H2O(l) = NH3(aq) + H

+(aq)

Equil.

xx-xChange

000.15Initial


(M)

NH4+(aq) + H2O(l) = NH3(aq) + H

+(aq)

xx0.15-xEquil.

xx-xChange

000.15Initial


(M)

+ 210 3

+

14 5

10

4

for reaction is 1.0 10 1.8 10

5

[NH ][H ]5.6 10

[NH ]

0

0

. 1

.15

6

a w bK K K

x

x

= =

=

=

b

2 210

6

Because K is so small, 1 and 0.15 0.15.

Therefore 5.6 100.15 0.15

[H ] 9.2 1 .00 4; 5pH

x x

x x

x

x

+

=

=

= =

Acid Solutions in Which Water

Contributes to [H+]

Previously, we have assumed that the acid is

the dominate source of H+ ions.

However, in some cases the contribution of

H+ ions from water must be taken into

account.

This situation occurs when pH between 6 and

7 i.e. near neutral.

Examples include low concentrations of weak

(or even strong) acids, and acids with pKas

near 7.


13/15

Posing the Weak Acid Problem We have the following four unknowns to solve

for: [H+], [OH-], [HA] and [A-]

We have the following four equations which

govern the concentrations of the above

species:

[ ]

[ ] [ ]0

H AK acid-base equilibrium

HA

K H OH water hydrolysis

H A OH charge balance

HA HA A

a

w

+

+

+

=

=

= +

= + material balance

The Solution to the Weak Acid ProblemWe now proceed to solve the four simultaneous equations for [H+]

by successive elimination of variables. The result is the following

cubic equation for [H+].

[H+]3 + Ka[H+]2 (Kw +Ka[HA]0)[H

+] KaKw = 0

Alternately

Finding [H+] using cubic equation can be done in several ways

Use an equation solver (This is the way to go.)

Use method of successive approximations

[ ]

2+

2+

+0

H

HHA

H

w

a

w

KK

K

=

pH of a Very Weak Acid

EP 135: Find [H+] and the pH of a 1.00 L solutioncontaining 3.00 x 10-5 g of vitamin B1, thiamine

hydrochloride (HC12H17ON4SCl2, mw = 337.27 g/mol ).

For this compound, Ka = 3.4 x 10-7. Use the method of

successive approximations.

Strategy

(1)Calculate molarity of vitamin B1

(2) [H+] must be between 1 x 10 -7 and molarity of vitamin B1

plus +1 x 10-7.

[HA]0 = (3.00 x 10-5 / 337.27) / 1.00 L = 8.89 x 10-8 M

[H+]3 + Ka[H+]2 (Kw +Ka[HA]0)[H

+] KaKw = 0

let y = [H+]

y3 + 3.4 x 10-7y2 (1.0 x 10-14 + (3.4 x 10-7)(8.89 x 10-8))y (3.4 x

10-7)(1.0 x 10-14) = 0

This expression may be simplified to:

y3 + (3.4 x 10-7)y2 (4.0 x 10-14)y 3.4 x 10-21 = 0

Using an equation solver [H+]= - 4.2 x 10-7, - 6.0 x 10-7, 1.4 x 10-7

+3.0 x 10-2116 x 10-8same

+1.6 x 10-2114 x 10-8same

+0.41 x 10-2114 x 10-8same

-0.67 x 10-2113 x 10-8same

-1.6 x 10-2112 x 10-8same

-2.4 x 10-2111 x 10-8same

-3.0 x 10-2110 x 10-8same

resultyy3 + (3.4 x 10-7)y2 (4.0 x 10-14)y 3.4x 10-21

For correct value of y, value of function = 0.

+0.41 x 10-2114 x 10-8same

-0.14 x 10-2113.5 x 10-8same

-0.67 x 10-2113 x 10-8y3 + (3.4 x 10-7)y2 (4.0 x 10-14)y 3.4 x 10-21

[H+] correct to 2 sig figs is 1.4 x 10-7

pH = 6.85


14/15

When can this complicated expression be

reduced to a simpler one?

[ ] [ ]

[ ] [ ] [ ]

2+ 6 +

2 2+ +

2 2+ +

+ +0 0

2+ + - + -

+ +

0 0

If H 10 , H 100

H H

H HHA HA

H H

H H A H A

HAHA H HA H

w

w

a

w

wK

KKK

K > >

=

= = =

>>

This is result you would obtain neglecting water dissociation

It is valid for[H+]>10-6

An approximate solution for [H+] in the range

10-7


15/15

We can simplify the four equations, because Ka infinity, so

[HA] = 0, and [A-] = [HA]0, thus we are left with

[ ]0

K H OH water hydrolysis

H HA OH charge balance

w+

+

=

= +

We now proceed to solve the two equations for [H+] by elimination

of [OH-]. The result is

[H+]2 - [HA]0[H+] Kw = 0

Finding [H+] requires that we find the two roots of the above

quadratic equation and then selecting the one root that is consistent

with physical reality, i.e. leads to all positive concentrations.

pH of a Low Concentration of a Strong Acid

EP 136: Find the pH of 2.0 x 10-8 M solution of HCl. This is asufficiently dilute solution that the autoionization of water cannot

be neglected. However, since the acid is strong, all of the acidmay be assumed dissociated.

[H+]2 - [HA]0[H+] Kw = 0

Where [HA]0 = 2.0 x 10-8 and Kw = 1.0 x 10

-14

Thus the equation to solve is y2 - (2 x 10-8)y 1.0 x 10-14 = 0

y = 1.1 x 10-7, -9.1 x 10-7. We keep the positive root.

pH = -log(1.1 x 10-7) = 6.96

Answers to Numerical Problems

EP121: 1.0 x 10-13M EP122: 1.0 x 10-13M

EP123 a) 7.00, b) 1.00, c) 13.0 EP124: 1.530

EP125: 1.70, 2.0% EP 126: 2.02

EP127: 1.6 x 10-8M EP 128: 1.2 x 10-5M

EP129: 11.78 EP 130: 12.48

EP 131: 3.15, 5.0 x 10-12 M EP 132: 1.70

EP 133: 9.08 EP 134: 5.04

EP 135: 1.4 x 10-7M, 6.85 EP 135a: 6.79

EP 136: 6.96

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Notes Solutions Chapter 07