THE CHEMISTRY OF ACIDS AND BASES

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CHAPTER 14

• THE CHEMISTRY OF ACIDS AND BASES

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"ACID"

• Latin word acidus, meaning sour.
• (lemon)

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"ALKALI"

• Water solutions feel slippery and
• taste bitter (soap).

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ACID-BASE THEORIES
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Arrhenius Definition

• acid--donates a hydrogen ion (H) in water
• base--donates a hydroxide ion in water (OH-)
• This theory was limited to
• substances with those "parts"
• ammonia is a MAJOR exception!

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Bronsted-Lowry Definition

• acid--donates a proton in water
• base--accepts a proton in water
• This theory is better it explains
• ammonia as a base! This is the main
• theory that we will use for our
• acid/base discussion.

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Lewis Definition

• acid--accepts an electron pair
• base--donates an electron pair
• This theory explains all traditional
• acids and bases a host of
• coordination compounds and is used
• widely in organic chemistry. Uses
• coordinate covalent bonds

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The Bronsted-Lowry Concept of Acids and Bases

• Using this theory, you should be
• able to write weak acid/base
• dissociation equations and identify
• acid, base, conjugate acid and
• conjugate base.

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Conjugate Acid-Base Pair

• A pair of compounds that differ
• by the presence of one H unit.
• This idea is critical when it comes
• to understanding buffer systems.

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Acids

• donate a proton (H)

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Neutral Compound

• HNO3 H2O ?? H3O NO3-
• acid base CA CB

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Cation

• NH4 H2O ?? H3O NH3
• acid base CA CB

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Anion

• H2PO4- H20 ?? H3O HPO42-
• acid base CA CB

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• In each of the acid examples---notice
• the formation of H3O
• This species is named the hydronium
• ion.
• It lets you know that the solution is
• acidic!

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Hydronium, H3O

• --H riding piggy-back on a water
• molecule.
• Water is polar and the charge of the
• naked proton is greatly attracted to
• Mickey's chin!)

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Bases

• accept a proton (H)

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Neutral Compound

• NH3 H2O ?? NH4 OH-
• base acid CA CB

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Anion

• CO32- H2O ?? HCO3- OH-
• base acid CA CB

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Anion

• PO43- H2O ? HPO42- OH-
• base acid CA CB

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• In each of the basic examples--
• notice the formation of OH- -- this
• species is named the hydroxide
• ion. It lets you know that the
• solution is basic!

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Exercise 1

• In the following reaction, identify
• the acid on the left and its CB on
• the right. Similarly identify the base
• on the left and its CA on the right.
• HBr NH3 ? NH4 Br-

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• What is the conjugate base of H2S?
• What is the conjugate acid of NO3-?

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• ACIDS ONLY DONATE
• ONE PROTON AT A
• TIME!!!

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• monoprotic--acids donating one H (ex. HC2H3O2)
• diprotic--acids donating two H's (ex. H2C2O4)
• polyprotic--acids donating many H's (ex. H3PO4)

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Polyprotic Bases

• accept more than one H
• anions with -2 and -3 charges
• (example PO43- HPO42-)

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Amphiprotic or Amphoteric

• molecules or ions that can behave as
• EITHER acids or bases
• water, anions of weak acids
• (look at the examples abovesometimes
• water was an acid, sometimes it acted as
• a base)

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Exercise 2 Acid Dissociation (Ionization)
Reactions

• Write the simple dissociation
• (ionization) reaction (omitting
• water) for each of the following
• acids.

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• a. Hydrochloric acid (HCl)
• b. Acetic acid (HC2H3O2)
• c. The ammonium ion (NH4)
• d. The anilinium ion (C6H5NH3)
• e. The hydrated aluminum(III) ion
• Al(H2O)63

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Solution

• A HCl(aq) -gt H(aq) Cl-(aq)
• B HC2H3O2(aq) ?
• H(aq) C2H3O2-(aq)
• C NH4(aq) ? H(aq) NH3(aq)

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Solution, cont.

• D C6H5NH3(aq) ? H(aq) C6H5NH2(aq)
• E Al(H2O)63(aq) ?
• H(aq) Al(H2O)5OH2(aq)

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Relative Strengths of Acids and Bases

• Strength is determined by the
• position of the "dissociation"
• equilibrium.

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Strong acids/Strong bases

• dissociate completely in water
• have very large K values

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Weak acids/Weak bases

• dissociate only to a slight extent in
• water
• dissociation constant is very small

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Strong
Weak
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• Do Not
• confuse concentration
• with strength!

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Strong Acids

• Hydrohalic acids
• HCl, HBr, HI
• Nitric HNO3
• Sulfuric H2SO4
• Perchloric HClO4

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The more oxygen present in the polyatomic ion,
the stronger its acid WITHIN that group.
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Strong Bases

• Hydroxides OR oxides of IA and
• IIA metals
• Solubility plays a role (those that
• are very soluble are strong!)

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The stronger the acid, the weaker its CB. The
converse is also true.
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Weak Acids and Bases - Equilibrium
expressions

• The vast majority of acid/bases are
• weak.
• Remember, this means they do not
• ionize much.

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• The equilibrium expression for acids
• is known as the Ka (the acid
• dissociation constant).
• It is set up the same way as in
• general equilibrium.

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• Many common weak acids are
• oxyacids,
• like phosphoric acid and
• nitrous acid.

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Other common weak acids are organic acids,those
that contain a carboxyl group COOH group
like acetic acid and benzoic acid.
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For Weak Acid Reactions

• HA H2O ? H3O A-
• Ka H3OA- lt 1
• HA

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• Write the Ka expression for acetic
• acid using Bronsted-Lowry.
• (Note Water is a pure liquid and
• is thus, left out of the equilibrium
• expression.)

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• Weak bases (bases without OH-)
• react with water to produce a
• hydroxide ion.

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• Common examples of weak bases are
• ammonia (NH3), methylamine
• (CH3NH2), and ethylamine (C2H5NH2).
• The lone pair on N forms a bond with
• an H. Most weak bases involve N.

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• The equilibrium expression for
• bases is known as the Kb.

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For Weak Base Reactions

• B H2O ? HB OH-
• Kb H3OOH- lt1
• B

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• Set up the Kb expression for
• ammonia using Bronsted-Lowry.

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• Notice that Ka and Kb expressions
• look very similar.
• The difference is that a base
• produces the hydroxide ion in
• solution, while the acid produces the
• hydronium ion in solution.

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Another note on this point

• H and H3O are both equivalent
• terms here. Often water is left
• completely out of the equation since
• it does not appear in the equilibrium.
• This has become an accepted
• practice.
• (However, water is very important
• in causing the acid to dissociate.)

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Exercise 3 Relative Base Strength

• Using table 14.2, arrange the
• following species according to their
• strength as bases
• H2O, F-, Cl-, NO2-, and CN-

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Solution

• Cl- lt H2O lt F- lt NO2- lt CN-

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• WATER
• THE HYDRONIUM ION
• AUTO-IONIZATION
• THE pH SCALE

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• Fredrich Kohlrausch, around
• 1900, found that no matter how
• pure water is, it still conducts a
• minute amount of electric
• current. This proves that water
• self-ionizes.

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• Since the water molecule is
• amphoteric, it may dissociate
• with itself to a slight extent.
• Only about 2 out of a billion
• water molecules are ionized at
• any instant!

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H2O(l) H2O(l) ltgt H3O(aq) OH-(aq)

• The equilibrium expression used
• here is referred to as the Kw
• (ionization constant for water).

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• In pure water or dilute aqueous
• solutions, the concentration of water
• can be considered to be a constant
• (55.4 M), so we include that with the
• equilibrium constant and write the
• expression as
• KeqH2O2 Kw H3OOH-

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• Kw 1.0 x 10-14
• (Kw 1.008 x 10-14 _at_ 25 Celsius)
• Knowing this value allows us to
• calculate the OH- and H
• concentration for various situations.

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• OH- H solution is neutral (in
• pure water, each of these is 1.0 x 10-7)
• OH- gt H solution is basic
• OH- lt H solution is acidic

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Kw Ka x Kb

• another very beneficial equation

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Exercise 5 Autoionization of Water

• At 60C, the value of Kw is 1 X 10-13.
• a. Using Le Chateliers principle,
• predict whether the reaction
• 2H2O(l) ?? H3O(aq) OH-(aq)
• is exothermic or endothermic.

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Exercise 5, cont.

• b. Calculate H and OH- in a
• neutral solution at 60C.

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Solution

• A endothermic
• B H OH- 3 X 10-7 M

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The pH Scale
Used to designate the H in most aqueous
solutions where H is small.
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• pH - log H
• pOH - log OH-
• pH pOH 14
• pH 6.9 and lower (acidic)
• 7.0 (neutral)
  
• 7.1 and greater (basic)

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• Use as many decimal places as
• there are sig.figs. in the problem!
• The negative base 10 logarithm of
• the hydronium ion concentration
• becomes the whole number
• therefore, only the decimals to the
• right are significant.

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Exercise 6 Calculating H and OH-

• Calculate H or OH- as required for
• each of the following solutions at
• 25C, and state whether the solution
• is neutral, acidic, or basic.
• a. 1.0 X 10-5 M OH-
• b. 1.0 X 10-7 M OH-
• c. 10.0 M H

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Solution

• A H 1.0 X 10-9 M, basic
• B H 1.0 X 10-7 M, neutral
• C OH- 1.0 X 10-15 M, acidic

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Exercise 7 Calculating pH and pOH

• Calculate pH and pOH for each of
• the following solutions at 25C.
• a. 1.0 X 10-3 M OH-
• b. 1.0 M H

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Solution

• A pH 11.00
• pOH 3.00
• B pH 0.00
• pOH 14.00

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Example

• Order the following from strongest base to
  weakest base. Use table 14.2.
• H2O NO3-1 OCl-1 NH3

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Exercise 8 Calculating pH

• The pH of a sample of human blood
• was measured to be 7.41 at 25C.
• Calculate pOH, H, and OH- for
• the sample.

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Solution

• pOH 6.59
• H 3.9 X 10-8
• OH- 2.6 X 10-7 M

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Exercise 9 pH of Strong Acids

• Calculate the pH of
• a. 0.10 M HNO3
• b. 1.0 X 10-10 M HCl

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Solution

• A pH 1.00
• B pH 7.00

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Exercise 10 The pH of Strong Bases

• Calculate the pH of a 5.0 X 10-2 M
• NaOH solution.

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Solution

• pH 12.70

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Calculating pH of Weak Acid Solutions

• Calculating pH of weak acids
• involves setting up an equilibrium.

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Always start by

• 1) writing the equation
• 2) setting up the acid equilibrium
• expression (Ka)
• 3) defining initial concentrations, changes, and
  final concentrations in terms of X

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• 4) substituting values and variables
• into the Ka expression
• 5) solving for X
• (use the RICE diagram learned in
• general equilibrium!)

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Example

• Calculate the pH of a 1.00 x 10-4 M
• solution of acetic acid.

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• The Ka of acetic acid is 1.8 x 10-5
• HC2H3O2 ? H C2H3O2-
• Ka HC2H3O2- 1.8 x 10-5
• HC2H3O2

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• Reaction HC2H3O2 ? H C2H3O2-
• Initial 1.00 x 10-4 0
  0
• Change -x x
  x
• Equilibrium 1.00 x 10-4 - x x x

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Often, the -x in a Ka expression can be treated
as negligible.

• 1.8 x 10-5 (x)(x) _
• 1.00x10-4 - x
• 
  
• 1.8 x 10-5 ? (x)(x) _
• 1.00 x 10-4
• x 4.2 x 10-5

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• When you assume that x is
• negligible, you must check the
• validity of this assumption.

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• To be valid, x must be less than
• 5 of the number that it was to be
• subtracted from.
• dissociation "x" x 100
• original

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• In this example, 4.2 x 10-5 is greater
• than 5 of 1.00 x 10-4.
• This means that the assumption that
• x was negligible was invalid and x
• must be solved for using the
• quadratic equation or the method of
• successive approximation.

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Use of the Quadratic Equation
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ax2 bx c 0
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Using the values a 1, b 1.8x10-5, c
-1.8x10-9
and
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Since a concentration can not be negative

• x 3.5 x 10-5 M
• x H 3.5 x 10-5
• pH -log 3.5 x 10-5 4.46

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• Another method which some people
• prefer is the method of successive
• approximations. In this method, you
• start out assuming that x is
• negligible, solve for x, and repeatedly
• plug your value of x into the
• equation again until you get the
• same value of x two successive times.

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Exercise 11 The pH of Weak Acids

• The hypochlorite ion (OCl-) is a
• strong oxidizing agent often found
• in household bleaches and
• disinfectants. It is also the active
• ingredient that forms when
• swimming pool water is treated with
• chlorine.

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• In addition to its oxidizing abilities,
• the hypochlorite ion has a relatively
• high affinity for protons (it is a
• much stronger base than Cl-, for
• example) and forms the weakly
• acidic hypochlorous acid (HOCl,
• Ka 3.5 X 10-8).

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• Calculate the pH of a 0.100 M
• aqueous solution of hypochlorous
• acid.

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Solution

• pH 4.23

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Determination of the pH of a Mixture of Weak Acids

• Only the acid with the largest Ka
• value will contribute an appreciable
• H.
• Determine the pH based on
• this acid and ignore any others.

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Exercise 12 The pH of Weak Acid Mixtures

• Calculate the pH of a solution that
• contains
• 1.00 M HCN (Ka 6.2 X 10-10) and
• 5.00 M HNO2 (Ka 4.0 X 10-4).

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Exercise 12, cont.

• Also, calculate the concentration of
• cyanide ion (CN-) in this solution at
• equilibrium.

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Solution

• pH 1.35
• CN- 1.4 X 10-8 M

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Exercise 13 Calculating Percent Dissociation

• Calculate the percent dissociation of
• acetic acid (Ka 1.8 X 10-5) in
• each of the following solutions.
• a. 1.00 M HC2H3O2
• b. 0.100 M HC2H3O2

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Solution

• A 0.42
• B 1.3

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Exercise 14 Calculating Ka from Percent
Dissociation

• Lactic acid (HC3H5O3) is a waste
• product that accumulates in muscle
• tissue during exertion, leading to
• pain and a feeling of fatigue.

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Exercise 14, cont.

• In a 0.100 M aqueous solution,
• lactic acid is 3.7 dissociated.
• Calculate the value of Ka for this
• acid.

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Solution

• Ka 1.4 X 10-4

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• Determination of the pH of a weak
• base is very similar to the
• determination of the pH of a weak
• acid.
• Follow the same steps.

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• Remember, however, that x is the
• OH- and taking the negative log
• of x will give you the pOH and not
• the pH!

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Exercise 15 The pH of Weak Bases I

• Calculate the pH for a 15.0 M
• solution of NH3 (Kb 1.8 X 10-5).

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Solution

• pH 12.20

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Exercise 16 The pH of Weak Bases II

• Calculate the pH of a 1.0 M solution
• of methylamine (Kb 4.38 X 10-4).

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Solution

• pH 12.32

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Calculating pH of polyprotic acids

• Acids with more than one ionizable
• hydrogen will ionize in steps.
• Each dissociation has its own Ka
• value.

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• The first dissociation will be the
• greatest and subsequent
• dissociations will have much
• smaller equilibrium constants.

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• As each H is removed, the
• remaining acid gets weaker and
• therefore has a smaller Ka.

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• As the negative charge on the acid
• increases, it becomes more difficult
• to remove the positively charged
• proton.

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Example

• Consider the dissociation of
• phosphoric acid.
• H3PO4(aq) H2O(l) ltgt
• H3O(aq) H2PO4-
  (aq)
• Ka1 7.5 x 10-3

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• H2PO4-(aq) H2O(l) ltgt
• H3O(aq) HPO42-(aq)
• Ka2 6.2 x 10-8

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• HPO42-(aq) H2O(l) ltgt
• H3O(aq) PO43-(aq)
• Ka3 4.8 x 10-13

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• Looking at the Ka values, it is
• obvious that only the first
• dissociation will be important in
• determining the pH of the solution.

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• Except for H2SO4, polyprotic acids
• have Ka2 and Ka3 values so much
• weaker than their Ka1 value that
• the 2nd and 3rd (if applicable)
• dissociation can be ignored.

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• The H obtained from this 2nd
• and 3rd dissociation is negligible
• compared to the H from the 1st
• dissociation.

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• Because H2SO4 is a strong acid in its
• first dissociation and a weak acid in
• its second, we need to consider both
• if the concentration is more dilute
• than 1.0 M.
• The quadratic equation is needed to
• work this type of problem.

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Exercise 17 The pH of a Polyprotic Acid

• Calculate the pH of a 5.0 M H3PO4
• solution and the equilibrium
• concentrations of the species
• H3PO4, H2PO4-, HPO42-, and PO43-

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Solution

• pH 0.72
• H3PO4 4.8 M
• H2PO4- 0.19 M
• HPO42- 6.2 X 10-8 M
• PO43- 1.6 X 10-19 M

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Exercise 18 The pH of a Sulfuric Acid

• Calculate the pH of a 1.0 M H2SO4
• solution.

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Solution

• pH 0.00

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Exercise 19 The pH of a Sulfuric Acid

• Calculate the pH of a 1.0 X 10-2 M
• H2SO4 solution.

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Solution

• pH 1.84

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ACID-BASE PROPERTIES OF SALTS

• HYDROLYSIS

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• Salts are produced from the reaction
• of an acid and a base. (neutralization)
• Salts are not always neutral. Some
• hydrolyze with water to produce
• acidic and basic solutions.

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Neutral Salts

• Salts that are formed from the
• cation of a strong base and the
• anion of a strong acid form
• neutral solutions when dissolved
• in water.
• A salt such as NaNO3 gives a
• neutral solution.

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Basic Salts

• Salts that are formed from the
• cation of a strong base and the
• anion of a weak acid form basic
• solutions when dissolved in
• water.

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• The anion hydrolyzes the water
• molecule to produce hydroxide
• ions and thus a basic solution.

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• K2S should be basic since S-2 is
• the CB of the very weak acid HS-,
• while K does not hydrolyze
• appreciably.

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• S2- H2O ? OH- HS-
• strong base weak acid

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Acid Salts

• Salts that are formed from the
• cation of a weak base and the
• anion of a strong acid form
• acidic solutions when dissolved in
• water.

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• The cation hydrolyzes the water
• molecule to produce hydronium
• ions and thus an acidic solution.

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• NH4Cl should be weakly acidic,
• since NH4 hydrolyzes to give an
• acidic solution, while Cl- does not
• hydrolyze.
• NH4 H2O ? H3O NH3
• strong acid weak base

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• If both the cation
• and the anion
• contribute to the
• pH situation,
• compare Ka to Kb.
• If Kb is larger, basic!
• The converse is also true.

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• The following will help predict
• acidic, basic, or neutral.
• However, you must explain using
• appropriate equations as proof!!!

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• 1. Strong acid Strong base
• Neutral salt

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• 2. Strong acid Weak base
• Acidic salt

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• 3. Weak acid Strong base
• Basic salt

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• 4. Weak acid Weak base
• ???
• (must look at K values to decide)

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Exercise 20 The Acid-Base Properties of Salts

• Predict whether an aqueous solution
• of each of the following salts will be
• acidic, basic, or neutral. Prove with
• appropriate equations.
• a. NH4C2H3O2
• b. NH4CN
• c. Al2(SO4)3

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Solution

• A neutral
• B basic
• C acidic

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Exercise 21 Salts as Weak Bases

• Calculate the pH of a 0.30 M NaF
• solution.
• The Ka value for HF is 7.2 X 10-4.

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Solution

• pH 8.31

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Exercise 22 Salts as Weak Acids I

• Calculate the pH of a 0.10 M NH4Cl
• solution.
• The Kb value for NH3 is 1.8 X 10-5.

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Solution

• pH 5.13

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Exercise 23 Salts as Weak Acids II

• Calculate the pH of a 0.010 M AlCl3
• solution.
• The Ka value for Al(H2O)63 is
• 1.4 X 10-5.

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Solution

• pH 3.43

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The Lewis Concept of Acids and Bases

• acid--can accept a pair of
• electrons to form a coordinate
• covalent bond
• base--can donate a pair of
• electrons to form a coordinate
• covalent bond

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• Yes, this is the dot guy and the
• structures guy.

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BF3 the most famous of all!!
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Exercise 24

• Tell whether each of the following is
• a Lewis acid or base.
• Draw structures as proof.
• a) PH3 c) H2S
• b) BCl3 d) SF4

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Exercise 25 Lewis Acids and Basis

• For each reaction, identify the Lewis
• acid and base.
• a. Ni2(aq) 6NH3(aq)? ?
• Ni(NH3)62(aq)
• b. H(aq) H2O(aq) ??H3O(aq)

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Solution

• A Lewis acid nickel(II) ion
• Lewis base ammonia
• B Lewis acid proton
• Lewis base water molecule

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