jeffrey mack california state university, sacramento chapter 17 principles of chemical reactivity:...

Jeffrey MackCalifornia State University,

Sacramento

Chapter 17

Principles of Chemical Reactivity:

The Chemistry of Acids and Bases

• In Chapter 3, you were introduced to two definitions of acids and bases: the Arrhenius and the Brønsted–Lowry definition.

• Arrhenius acid: Any substance that when dissolved in water increases the concentration of hydrogen ions, H+.

• Arrhenius base: Any substance that increases the concentration of hydroxide ions, OH, when dissolved in water.

• A Brønsted–Lowry acid is a proton (H+) donor.• A Brønsted–Lowry base is a proton acceptor.

Acids & Bases: A Review

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

STRONG ACID:

HNO3(aq) + H2O(liq) H3O+(aq) + NO3-(aq)

HNO3 is about 100% dissociated in water.

Strong & Weak Acids/Bases

HNO3, HCl, H2SO4 and HClO4 are classified as strong acids.


• Strong Base: 100% dissociated in water.

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

Other common strong bases include KOH and Ca(OH)2.

CaO (lime) + H2O

Ca(OH)2 (slaked lime)CaO


• Weak base: less than 100% ionized in waterAn example of a weak base is ammonia

NH3(aq) + H2O(liq) NH4+(aq) + OH-(aq)


Weak acids are much less than 100% ionized in water.Example: acetic acid = CH3CO2H


• Proton donors may be molecular compounds, cations or anions.

3

3 2 3 3

4 2 3 3

22 3 3

HNO (aq) H O(l) NO (aq) H O (aq)

NH (aq) H O(l) NH (aq) H O (aq)

HCO (aq) H O(l) CO (aq) H O (aq)

The Brønsted–Lowry Concept of Acids & Bases Extended

• Proton acceptors may be molecular compounds, cations or anions.

3

3

3 2

2

2 25

3

2 6

23 2

NH (aq) H O(l) NH (aq) OH (aq)

Al H O OH (aq) H O(l)

Al H O (aq) OH (aq)

CO (aq) H O(l) HCO (aq) OH (aq)


Using the Brønsted definition, NH3 is a BASE in water and water is itself an ACID

Proton acceptor

Proton donor


• Acids such as HF, HCl, HNO3, and CH3CO2H (acetic acid) are all capable of donating one proton and so are called monoprotic acids.

• Other acids, called polyprotic acids are capable of donating two or more protons.


• A conjugate acid–base pair consists of two species that differ from each other by the presence of one hydrogen ion.

• Every reaction between a Brønsted acid and a Brønsted base involves two conjugate acid–base pairs

Conjugate Acid–Base Pairs

Conjugate Acid–Base Pairs

Water Autoionization and the Water Ionization Constant, Kw:

The water autoionization equilibrium lies far to the left side. In fact, in pure water at 25 °C, only about two out of a billion (109) water molecules are ionized at any instant.

Even in pure water, there is a small concentration of ions present at all times. [H3O+] = [OH] = 1.00 107

2 2 3H O(l) H O(l) H O (aq) OH (aq)

Water & the pH Scale

H2O can function as both an ACID and a BASE.

In pure water there can be AUTOIONIZATION.

Equilibrium constant for autoionization = Kw

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 °C


• In a neutral solution, [H3O+] = [OH]• Both are equal to 1.00 10 7 M• In an acidic solution, [H3O+] > [OH]

• [H3O+] > 1.00 10 7 M and [OH] < 1.00 10 7 M

• In a basic solution, [H3O+] < [OH]

• [H3O+] < 1.00 10 7 M and [OH] > 1.00 10 7 M


The pH Scale

• The pH of a solution is defined as the negative of the base (10) logarithm (log) of the hydronium ion concentration.

pH = log[H3O+]

• In a similar way, we can define the pOH of a solution as the negative of the base - 10 logarithm of the hydroxide ion concentration.

pOH = log[OH]pH + pOH = pKw = 14

The pH Scale

• The concentration of acid, [H3O+] is found by taking the antilog of the solutions pH.

• In a similar way, [OH] can be found from:

The pH Scale

Once [H3O+] is known, [OH] can be found from:

And vice versa.

The pH Scale

• In Chapter 3, it was stated that acids and bases can be divided roughly into those that are strong electrolytes (such as HCl, HNO3, and NaOH) and those that are weak electrolytes (such as CH3CO2H and NH3)

• In this chapter we will discuss the quantitative aspects of dissociation of weak acids and bases.

• The relative strengths of weak acids and bases can be ranked based on the magnitude of individual equilibrium constants.

Equilibrium Constants for Acids & Bases

• Strong acids and bases almost completely ionize in water (~100%):

Kstrong >> 1

(product favored)• Weak acids and bases almost completely

ionize in water (<<100%):Kweak << 1

(Reactant favored)


• The relative strength of an acid or base can also be expressed quantitatively with an equilibrium constant, often called an ionization constant. For the general acid HA, we can write:

Conjugate acid

Conjugate base


• The relative strength of an acid or base can also be expressed quantitatively with an equilibrium constant, often called an ionization constant. For the general base B, we can write:

Conjugate base

Conjugate Acid


Acids ConjugateBases

Increase strength

Increase strength

Ionization Constants for Acids/Bases

• The strongest acids are at the upper left. They have the largest Ka values.

• Ka values become smaller on descending the chart as the acid strength declines.

• The strongest bases are at the lower right. They have the largest Kb values.

• Kb values become larger on descending the chart as base strength increases.


• The weaker the acid, the stronger its conjugate base: The smaller the value of Ka, the larger the value of Kb.

• Aqueous acids that are stronger than H3O+ are completely ionized.

• Their conjugate bases (such as NO3) do not

produce meaningful concentrations of OH ions, their Kb values are “very small.”

• Similar arguments follow for strong bases and their conjugate acids.


Acid HCO3 HClO HF

Ka 4.8 1011 3.5 108 7.2 104

Base CO32 ClO F

Kb 2.1 104 2.9 107 1.4 1011


Ka Values for Polyprotic Acids

In general, each successive dissociation produces a weaker acid.

2 2 3

7a(1)

22 3

19a(2)

H S(aq) H O(l) H O (aq) HS (aq)

K 1 10

HS (aq) H O(l) H O (aq) S (aq)

K 1 10


Logarithmic Scale of Relative Acid Strength, pKa

• Many chemists use a logarithmic scale to report and compare relative acid strengths.

pKa = log(Ka)

The lower the pKa, the stronger the acid.

Acid HCO3 HClO HF

pKa 10.32 7.46 3.14


Relating the Ionization Constants for an Acid and Its Conjugate Base



3 2

a b 3 w2

H O HS H S OHK K H O OH K

H S HS



When adding equilibria, multiply the K values.


Acid–Base Properties of Salts

Anions that are conjugate bases of strong acids (for examples, Cl or NO3

.

These species are such weak bases that they have no effect on solution pH.

3 2NO (aq) H O(l) No Reaction


Anions such as CO3 that are the conjugate

bases of weak acids will raise the pH of a solution.

Hydroxide ions are produced via “Hydrolysis”.

23 2 3CO (aq) H O(l) HCO (aq) OH (aq)


Anions such as CO3 that are the conjugate

bases of weak acids will raise the pH of a solution.

Hydroxide ions are produced via “Hydrolysis”.A partially deprotonated anion (such as HCO3

) is amphiprotic. Its behavior will depend on the other species in the reaction.

23 2 3CO (aq) H O(l) HCO (aq) OH (aq)


Alkali metal and alkaline earth cations have no measurable effect on solution pH.

Since these cations are conjugate acids of strong bases, hydrolysis does not occur.

2Na (aq) H O(l) No Reaction


Basic cations are conjugate bases of acidic cations such as [Al(H2O)6]3+.

Acidic cations fall into two categories: (a) metal cations with 2+ and 3+ charges and (b) ammonium ions (and their organic derivatives).

All metal cations are hydrated in water, forming ions such as [M(H2O)6]n+.

3

2 26

2

2 35

4

Al H O (aq) H O(l)

Al H O (OH ) (aq) H O (aq)

Ka 7.9 10


Salt pH of (aq) solution

CaCl2

NH4Br

NH4F

KNO3

KHCO3

Acid–Base Properties of Salts: Practice

Acid–Base Properties of Salts: Practice

Salt pH of (aq) solution

CaCl2 Neutral

NH4Br Acidic

NH4F Basic

KNO3 Neutral

KHCO3 Basic

• According to the Brønsted–Lowry theory, all acid–base reactions can be written as equilibria involving the acid and base and their conjugates.

• All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base.

Acid Base Conjugate base of the acid + Conjugate acid of the base

Predicting the Direction of Acid–Base Reactions

• When a weak acid is in solution, the products are a stronger conjugate acid and base. Therefore equilibrium lies to the left.

• All proton transfer reactions proceed from the stronger acid and base to the weaker acid and base.


• Will the following acid/base reaction occur spontaneously?

3 4 3 2 2 3 3 2H PO (aq) CH CO (aq) H PO (aq) CH CO H(aq)



Ka = 7.5 105 Ka = 1.8 105

Kb = 5.6 1010 Kb = 1.3 1012




• Equilibrium lies to the right since all proton transfer reactions proceed from the stronger acid and base to the weaker acid and base.

Ka = 7.5 105 Ka = 1.8 105

Kb = 5.6 1010 Kb = 1.3 1012

Stronger Acid + Stronger Base Weaker Base + Weaker Acid



Strong acid (HCl) + Strong base (NaOH)

Net ionic equation

Mixing equal molar quantities of a strong acid and strong base produces a neutral solution.

3HCl (aq) NaOH (aq) H O (aq) NaCl (aq)

3 2H O (aq) OH (aq) 2H O(l)

Types Acids–Base Reactions

Weak acid (HCN) + Strong base (NaOH)

Mixing equal amounts (moles) of a strong base and a weak acid produces a salt whose anion is the conjugate base of the weak acid. The solution is basic, with the pH depending on Kb for the anion.

2HCN (aq) + OH (aq) CN (aq) + H O (l)

2CN (aq) + H O (l) HCN (aq) + OH (aq)


Strong acid (HCl) + Weak base (NH3)

Mixing equal amounts (moles) of a weak base and a strong acid produces a conjugate acid of the weak base. The solution is basic, with the pH depending on Ka for the acid.

3 3 2 4H O (aq) + NH (aq) H O (l) + NH (aq)

4 2 3 3NH (aq) + H O (l) H O (aq) + NH (aq)


Weak acid (CH3CO2H) + Weak base (NH3)

Mixing equal amounts (moles) of a weak acid and a weak base produces a salt whose cation is the conjugate acid of the weak base and whose anion is the conjugate base of the weak acid. The solution pH depends on the relative Ka and Kb values.

3 2 3 3 2 4CH CO H (aq) + NH (aq) CH CO + NH (aq)


Weak acid + Weak base

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

cation and anion.


Types Acids–Base Reactions Summary

Determining K from Initial Concentrations and pH

[H2S] [H3O+] [HS]

Initial 0.10

Change

Equilibrium

Calculations with Equilibrium Constants


[H2S] [H3O+] [HS]

Initial 0.10 0 0

Change 0.10 - x + x + x

Equilibrium 0.10 - x x x



[H2S] [H3O+] [HS]

Initial 0.10 0 0

Change 0.10 - x + x + x


2

3 2a

2

H O NO x x xK

HNO 0.10 x 0.10 x



[H3O+] = [NO2] = 6.76 103

234

a 3

6.76 10K 4.9 10

0.10 6.76 10


2

3 2a

2

H O NO x x xK

HNO 0.10 x 0.10 x


[H2S] [H3O+] [HS]

Initial 1.00

Change

Equilibrium


[H2S] [H3O+] [HS]

Initial 1.00 0 0

Change - x + x + x


2

3 7a

2

H O HS x x xK 1.0 10

H S 1.00 x 1.00 x




27

Since x << 1.00, the equation reduces to:

x1.0 10

1.00

x = 3.2 104 pH = 3.50

2

3 7a

2

H O HS x x xK 1.0 10

H S 1.00 x 1.00 x



In general, the approximation that

[HA]equilibrium = [HA]initial x [HA]initial

is valid whenever [HA]initial is greater than or

equal to 100 Ka.

If this is not the case, the quadratic equation

must by used.


Determining pH after an acid/base reaction:

Calculate the hydronium ion concentration and pH of the solution that results when 22.0 mL of 0.15 M acetic acid, CH3CO2H, is mixed with 22.0 mL of 0.15 M NaOH.



Determining pH after an acid/base reaction:


Solution: From the volume and concentration of each solution, the moles of acid and base can be calculated. Knowing the moles after the reaction and the equilibrium constants, the concentration of H3O+ and pH can be calculated.


All of the acetic acid is converted to acetate ion.

3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)

3 23

1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH

10 mL 1 L


All of the acetic acid is converted to acetate ion.

3 2 3 2 2CH CO H (aq) OH (aq) CH CO (aq) H O (l)

3 23

1 L 0.15 mols22.0 mL 0.0033 mols of CH CO H & OH

10 mL 1 L

3

3 2

0.0033 mols 10 mLCH CO 0.075 M

44.0 mL 1 L


[CH3CO2-] [CH3CO2H] [OH-]

Initial 0.075 0 0

Change

Equilibrium

3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)


3 2 2 3 2CH CO (aq) H O (l) CH CO H (aq) OH (aq)


Initial 0.075 0 0

Change - x + x + x




Initial 0.075 0 0

Change - x + x + x



Since Kb100 > [CH3CO2-]initial, the quadratic equation

is not needed.


Initial 0.075 0 0

Change - x + x + x


• Because polyprotic acids are capable of donating more than one proton they present us with additional challenges when predicting the pH of their solutions.

• For many inorganic polyprotic acids, the ionization constant for each successive loss of a proton is about 104 to 106 smaller than the previous step.

• This implies that the pH of many inorganic polyprotic acids depends primarily on the hydronium ion generated in the first ionization step.

• The hydronium ion produced in the second step can be neglected.

Polyprotic Acuids & Bases

Sulfurous acid, H2SO3, is a weak acid capable of providing two H+ ions.

(a) What is the pH of a 0.45 M solution of H2SO3?

(b) What is the equilibrium concentration of the sulfite ion, SO3

2- in the 0.45 M solution of H2SO3?

Polyprotic Acids & Bases




2- in the 0.45 M solution of H2SO3?2

–2 3 3

2 3

[HSO ][H O ] x1.2 10

[H SO ] 0.45 – x






Since 100 Ka is not << 0.45M, the quadratic equation must be used

2–2 3 3

2 3

[HSO ][H O ] x1.2 10

[H SO ] 0.45 – x







(a)

x = [H3O+] = 0.0677 M

pH = 1.17





(a) in part a we found that x = [H3O+] = 0.0677 M

(b)


Halide Acid Strengths• Experiments show that the acid strength increases in the

order: HF << HCl < HBr < HI.• Stronger acids result when the HX bond is readily broken

(as signaled by a smaller, positive value of H for bond dissociation) and a more negative value for the electron attachment enthalpy of X.

Molecular Structure, Bonding, & Acid–Base Behavior

Comparing Oxoacids: HNO2 and HNO3

• In all the series of related oxoacid compounds, the acid strength increases as the number of oxygen atoms bonded to the central element increases.

• Thus, nitric acid (HNO3) is a stronger acid than nitrous acid (HNO2).


Why Are Carboxylic Acids Brønsted Acids?• There is a large class of organic acids, all like acetic

acid (CH3CO2H) have the carboxylic acid group, CO2H

• They are collectively called carboxylic acids.


Why Are Carboxylic Acids Brønsted Acids?• The carboxylate anion is stabilized by

resonance.


Why Are Carboxylic Acids Brønsted Acids?• The acidity of carboxylic acids is enhanced if

electronegative substituents replace the hydrogen atoms in the alkyl (–CH3 or –C2H5) groups.

• Compare, for example, the pKa values of a series of acetic acids in which hydrogen is replaced sequentially by the more electronegative element chlorine.


• Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl.

• Cl withdraws electrons from the rest of the molecule.

• This makes the O—H bond highly polar. The H of O—H is very positive.

• Trichloroacetic acid is a much stronger acid owing to the high electronegativity of Cl.

• Cl withdraws electrons from the rest of the molecule.

• This makes the O—H bond highly polar. The H of O—H is very positive.

Acetic acid Trichloroacetic acid

Ka = 1.8 x 10-5 Ka = 0.3

• The concept of acid–base behavior advanced by Brønsted and Lowry in the 1920’s works well for reactions involving proton transfer.

• However, a more general acid– base concept, was developed by Gilbert N. Lewis in the 1930’s.

• A Lewis acid is a substance that can accept a pair of electrons from another atom to form a new bond.

• A Lewis base is a substance that can donate a pair of electrons to another atom to form a new bond.

The Lewis Concept of Acids & Bases

A + B: BAAcid Base Adduct

• The product is often called an acid–base adduct. In Section 8.3, this type of chemical bond was called a coordinate covalent bond.

• Lewis acid-base reactions are very common. In general, they involve Lewis acids that are cations or neutral molecules with an available, empty valence orbital and bases that are anions or neutral molecules with a lone electron pair.


Lewis acid a substance that accepts an electron pair

Lewis base a substance that donates an electron pair


• New bond formed using electron pair from the Lewis base.

• Coordinate covalent bond

• Notice geometry change on reaction.

Reaction of a Lewis Acid & Lewis Base

The formation of a hydronium ion is an example of a Lewis acid / base reaction

HH

H

BASE

••••••

O—HO—H

H+

ACID

The H+ is an electron pair acceptor.Water with it’s lone pairs is a Lewis acid donor.


Lewis AcidBase Reactions

Metal cations often act as Lewis acids because of open d-orbitals.

Lewis Acids & Bases

The combination of metal ions (Lewis acids) with Lewis bases such as H2O and NH3 leads to Coordinate Complex ions.

Lewis Acids & Bases

Aqueous solutions of Fe3+, Al3+, Cu2+, Pb2+, etc. are acidic through hydrolysis.

This interaction weakens this bond

Another H2O pulls this H away as H+

[Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5(OH)]2+(aq) + H3O+(aq)

Lewis Acids & Bases

• Because oxygen is more electronegative than C, the CO bonding electrons in CO2 are polarized away from carbon and toward oxygen.

• This causes the carbon atom to be slightly positive, and it is this atom that the negatively charged Lewis base OH can attack to give, ultimately, the bicarbonate ion.

Molecular Lewis Acids

• Ammonia is the parent compound of an enormous number of compounds that behave as Lewis and Brønsted bases. These molecules all have an electronegative N atom with a partial negative charge surrounded by three bonds and a lone pair of electrons.

• This partially negative N atom can extract a proton from water.

Molecular Lewis Acids

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

2+ ++

Lewis Acids & Bases

Reaction of NH3 with Cu2+(aq)

• The heme group in hemoglobin can interact with O2 and CO.

• The Fe ion in hemoglobin is a Lewis acid

• O2 and CO can act as Lewis bases

Heme group

Lewis Acid–Base Interactions in Biology

Slides from Chang Book

99

pH – A Measure of Acidity

pH = -log [H+]

[H+] = [OH-][H+] > [OH-][H+] < [OH-]

Solution Isneutralacidicbasic

[H+] = 1 x 10-7

[H+] > 1 x 10-7

[H+] < 1 x 10-7

pH = 7pH < 7pH > 7

At 250C

pH [H+]

100

percent ionization =

Ionized acid concentration at equilibrium

Initial concentration of acidx 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0

x 100% [HA]0 = initial concentration

% Ionization =

101

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6

102

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF] = 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka x2

0.50 = 7.1 x 10-4

0.50 – x 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72[HF] = 0.50 – x = 0.48 M

103

When can I use the approximation?

0.50 – x 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.019 0.019 M0.50 Mx 100% = 3.8%

Less than 5%

Approximation ok.

What is the pH of a 0.05 M HF solution (at 250C)?

Ka x2

0.05 = 7.1 x 10-4 x = 0.006 M

0.006 M0.05 Mx 100% = 12%

More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation or method of successive approximations.

Ka *100 << [HA]0

104

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka x2

0.122= 5.7 x 10-4

0.122 – x 0.122Ka << 1

x2 = 6.95 x 10-5 x = 0.0083 M

0.0083 M0.122 Mx 100% = 6.8%

More than 5%

Approximation not ok.

105

Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0 -b ± b2 – 4ac 2ax =

x = 0.0081 x = - 0.0081

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

106

Acid-Base Properties of Salts

Neutral Solutions:

Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid (e.g. Cl-, Br-, and NO3

-).

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Basic Solutions:

Salts derived from a strong base and a weak acid.

NaCH3COOH (s) Na+ (aq) + CH3COO- (aq)H2O

CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)

107


Acid Solutions:

Salts derived from a strong acid and a weak base.

NH4Cl (s) NH4+ (aq) + Cl- (aq)

H2O

NH4+ (aq) NH3 (aq) + H+ (aq)

Salts with small, highly charged metal cations (e.g. Al3+, Cr3+, and Be2+) and the conjugate base of a strong acid.

Al(H2O)6 (aq) Al(OH)(H2O)5 (aq) + H+ (aq)3+ 2+

108


Solutions in which both the cation and the anion hydrolyze:

• Kb for the anion > Ka for the cation, solution will be basic

• Kb for the anion < Ka for the cation, solution will be acidic

• Kb for the anion Ka for the cation, solution will be neutral

109

What is the molarity of an NH4NO3(aq) solution that has a pH = 4.80?

StrategyAmmonium nitrate is the salt of a strong acid (HNO3) and a weak base (NH3). In NH4NO3(aq), NH4

+ hydrolyzes and NO3– does not. The ICE format

must be based on the hydrolysis equilibrium for NH4+(aq). In that format

[H3O+], derived from the pH, will be a known quantity, and the initial concentration of NH4

+ will be the unknown.Solution

We begin by writing the equation for the hydrolysis equilibrium and theequation for Ka in terms of Kw and Kb.

As usual, we can calculate [H3O+] from the pH of the solution.

log[H3O+] = –pH = –4.80

[H3O+] = 10–4.80 = 1.6 x 10–5 M

110

Example 15.15 continued

Solution continuedIf we assume that all the hydronium ion comes from the hydrolysis reaction, we can set up an ICE format in which x represents the unknown initialconcentration of NH4

+.

We now substitute equilibrium concentrations into the ionization constantexpression for the hydrolysis reaction.

We can assume that the ammonium ion is mostly nonhydrolyzed and that the change in [NH4

+] is much smaller than the initial [NH4+], so that 1.6 x 10–

5 << x and we can replace (x – 1.6 x 10–5) by x. Then we can solve for x.

The solution is 0.46 M NH4NO3.

jeffrey mack california state university, sacramento chapter 17 principles of chemical reactivity:...

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