Daniel L. RegerScott R. GoodeDavid W. Ball

www.cengage.com/chemistry/reger

Chapter 15Solutions of

Acids and Bases

• Acid: a substance that produces H3O+ when dissolved in water.

• Base: a substance that produces OH- when dissolved in water.

• Arrhenius acids and bases are limited to water solutions.

Arrhenius Acids and Bases

• Acid: proton donor.• Base: proton acceptor.• Bronsted-Lowry acid base

reaction: proton transfer from acid to base.

Bronsted-Lowry Acids and Bases style

• Conjugate Acid-Base Pairs: two species that differ by a proton.

• Acid : the species that contains the proton that is transferred.

• Conjugate base: the species formed by loss of the proton.HF(aq) + H2O(l) ⇌ F-(aq) + H3O

+(aq)

• HF and F- make up an acid-base conjugate pair. F- is the conjugate base of HF.

Conjugate Acid-Base Pairs

• HF(aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)

HF is the acid, F- is its conjugate base(or F- is a base and HF is its conjugate acid).

• The reverse reaction is F-(aq) + H3O

+(aq) ⇌ HF(aq) + H2O(l) H3O+ is the acid, H2O is its conjugate base.

Acid-Base Conjugate Pairs

Acid Base

HCl Cl -

H2SO4 HSO4-

HSO4- SO4

2-

NH4+ NH3

H3O+ H2O

H2O OH-

H2O is amphoteric - it acts as both an acid and a base.

Acid-Base Conjugate Pairs

• An Arrhenius systemacid + base water + salt

is limited to reactions in water.

• A Bronsted-Lowry system

HA + B ⇌ A- + BH+

is more general - a proton is transferred from the acid (HA) to the base (B).

Acid-Base Reactions

acid + base ⇌ conj. base + conj. acid

HF + NH3 ⇌ F- + NH4+

HCl + H2O ⇌ Cl- + H3O+

H2O + NH2- ⇌ OH- + NH3

Bronsted-Lowry Acid-Base Reactions

HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq) HCN is the acid, CN- is its conjugate base.

H2O is the base, H3O+ is its conjugate acid.

CH3COO-(aq)+H2O(l) ⇌ CH3COOH(aq)+OH-(aq) CH3COO- is the base, CH3COOH is its conjugate acid. H2O is the acid, OH- is its conjugate base.

Identifying Conjugate Pairs

Identify the acid-base conjugate pairs

HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2-(aq)

CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)

Test Your Skill

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)

• H3O+ and H2O are a conjugate acid-base pair.

• H2O and OH - are a conjugate acid-base pair.

Autoionization of Water

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)

• The equilibrium constant for this reaction is called Kw.

• Kw = [H3O+][OH -]

• Kw changes with temperature and is equal to 1.0 × 10-14 at 25o C.

Autoionization of Water

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)

• In pure water, [H3O+] = [OH -].

Kw = [H3O+][OH -] = 1.0 × 10-14

Kw = [H3O+]2 = 1.0 × 10-14

• [H3O+] = [OH -] = 1.0 × 10-7 M

Calculating Hydrogen and Hydroxide Ion Concentrations

• Adding an acid or a base to water causes the H3O+ and OH- concentrations to change.

• Calculate the hydroxide ion concentration of a solution in which the hydrogen ion concentration is 3.6 × 10-3 M.

Acidic or Basic Solutions

• Calculate the hydrogen ion concentration in a solution in which the hydroxide ion concentration is 0.025 M.

Test Your Skill

• pH = -log10[H3O+]Calculate the pH of a 0.0034 M solution of [H3O+].

• pH = -log(0.0034) = 2.47• pH values are generally given to two

decimal places. A pH of 2.47 has two significant figures; the 2 serves to locate the decimal point of the original number.

pH Scale

• Calculate the hydrogen ion concentration in a solution that has a pH of 3.52.

Calculating Hydrogen Ion Concentration from pH

• The p-notation can be used for other quantities.

• pOH = -log[OH-]

• pKw = -log Kw = -log(1.0 × 10-14) = 14.00

p-notation

[H3O+][OH-] = Kw = 1.0 × 10-

14

log([H3O+][OH-]) = log Kw

log[H3O+] + log[OH-] = log Kw

-log[H3O+] - log[OH-] = -log Kw

pH + pOH = pKw

pH + pOH = 14.00

Relating pH and pOH

pH + pOH = 14.00

Acid Neutral Base

pH <7 =7 >7

pOH >7 =7 <7

Relating pH and pOH

• HA(aq) + H2O(l) H3O+(aq) +A-(aq)

• Strong acids ionize completely in solution.

• Memorize the six common strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4.

100%

Strong Acids and Bases

• Calculate the pH of a 0.050 M HCl solution.

pH of a Strong Acid Solution

• Strong bases quantitatively produce hydroxide ions in water.

• The most common strong bases are the group IA and soluble IIA oxides and hydroxides.

Strong Bases

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

Li2O(s) + H2O(l) 2Li+(aq) + 2OH-(aq)

Strong Bases

• Calculate the pH of a 0.035 M Ba(OH)2 solution.

pH of a Strong Base Solution

What is the concentration of a solution of HCl if the pH is 3.75?

What is the concentration of a solution of KOH if the pH is 11.60?

Test Your Skill

• Weak acids and weak bases are those that do not ionize completely in water.

• A weak acid exists in equilibrium with its conjugate base.HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq)

• A weak base exists in equilibrium with its conjugate acid.

CH3NH2 (aq) + H2O(l) ⇌ CH3NH3

+(aq) + OH-(aq)

Weak Acids and Bases

• HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

• B(aq) + H2O ⇌ BH+(aq) + OH-(aq)

[HA]

]][AO[H3

aK

[B]

]][OH[BH bK

Weak Acids and Bases

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

• Both H2O and A- are bases and compete for the proton.

• If HA is a strong acid, A- is an extremely poor proton acceptor.

HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)

Proton bonded to Cl-

Proton bonded to H3O+

Competition for Protons

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

• If HA is a weak acid, A- is a weak base.• The stronger the weak acid, the weaker

the conjugate base.

HF(aq) + H2O ⇌ H3O+(aq) + F-(aq)

Proton bonded to F-

Proton bonded to H3O+

Competition for Protons

• Leveling Effect: In water, all acids stronger than H3O+ appear to be equally strong. By using a weaker base than water as the solvent, we can differentiate between the strong acids.

weakest strongest | |HNO3 < H2SO4 < HCl < HBr < HI < HClO4

The Influence of the Solvent

• Analytical Concentration: the total concentration of all forms of an acid; both the protonated form (the acid) and the unprotonated form (the conjugate base).

The Concentration of an Acid

• You will likely see two kinds of equilibrium calculations:

1. You measure equilibrium concentrations and calculate K or

2. you are given starting concentrations and K and calculate equilibrium concentrations.

• The approach is the same.

1. Write the balanced chemical equation.

2. Calculate equilibrium concentrations/iCe table.

3. Write the algebraic expression for K.

4. Substitute concentrations into expression.

5. Solve.

Equilibrium Calculations

• A 0.250 M solution of HF is determined to be 3.7% ionized. Determine Ka for HF.

Calculating Ka for a Weak Acid

• The pH of a 0.100 M solution of HOCl is 4.26. Calculate Ka for the acid.

HOCl + H2O ⇌ H3O+ + OCl-

Determining Ka from pH

• Calculate the pH of a 0.50 M

CH3COOH solution. Ka = 1.8 × 10-5

CH3COOH + H2O ⇌ H3O+ + CH3COO-

Determining Concentrations of Species in Weak Acid Solution

• Determine the pH of a 0.025 M solution of HCN, Ka = 7.2 × 10-10.

Test Your Skill

• Calculate the equilibrium concentrations and pH of a 0.100 M HF solution, Ka = 6.3 × 10-4. Use approximation.

Calculating pH of a Weak Acid Solution

• Is 7.9 10-3 << 0.050?• No, it is not. 5% of 0.050 is 2.5 10-3

and the calculated value is larger. The approximation fails.

• We need another method to compute the root.

Check Approximation

3-2

5-22

3-

224

1

224

2.

-31

106.7

1080.5

109.7100.0

)(103.6

100.0

))((103.6

. value,new a

calculate to ionapproximat an as it use and

107.9 value first the Call

y

y

y

y

yy

y

y

Successive Approximation

• Is y2 sufficiently close to y1?

• Restating the question, is 7.6 10-3 within 5% of 7.9 10-3?

• By within 5%, we mean is 7.6 10-3 within a range varying from 95% to 105% of 7.9 10-3?

• The answer is yes. Since the second approximation was quite close to the first, we can accept it.

Check Second Approximation

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

HA of ionconcentrat nalytical

]O[H

][A [HA]

][A ionized fraction

3

-

-

a

Fraction Ionized in Solution

• Calculate fraction ionized for 0.500 M HOCl, Ka = 4.0 × 10-8.

• HOCl + H2O ⇌ H3O+ + OCl-

Fraction Ionized in Solution

• A weak base reacts with water to form hydroxide ions.

B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)

[B]

]][OH[BH bK

Solutions of Weak Bases

CH3NH2 + H2O ⇌ CH3NH3+ + OH-

Calculate the pH of a solution of 0.150 M methylamine, CH3NH2, Kb = 4.4 × 10-4.

pH of a Solution of a Weak Base

• Calculate the pH of a 0.35 M solution of hydroxylamine, NH2OH, Kb = 1.1 × 10-8.

Test Your Skill

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)

A-(aq) + H2O(l) ⇌ HA(aq) + OH-(aq)

[HA]

]][AO[H3a

K

][A

][HA][OH-b

K

Relating Ka and Kb

wKKK

KK

KK

ba

3ba

-3

ba

]][OHO[H

][A

][HA][OH

[HA]

]][AO[H

Relating Ka and Kb

• The conjugate base of a weak acid is a weak base:

CH3COOH + H2O ⇌ H3O+ + CH3COO- Ka = 1.8 × 10-5

CH3COO- + H2O ⇌ CH3COOH + OH- Kb = 5.6 × 10-10

HCN + H2O ⇌ H3O+ + CN- Ka = 6.2 × 10-10

CN- + H2O ⇌ HCN + OH- Kb = 1.6 × 10-5

Strengths of Weak Acid-Base Conjugate Pairs

• Ka for HNO2 is 4.6 x 10-4. Calculate Kb for the NO2

- ion. NO2

-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)

Calculating Ka and Kb for Acid-Base Conjugate Pairs

• Kb for pyridine is 1.8 × 10-9. Calculate Ka for the pyridinium ion.

Test Your Skill

Kb = 2.8 x 10-11 Kb = 1.4 x 10-5

Ka = 7.2 x 10-10 Ka = 3.5 x 10-4

acidStronger

acid

HCN HF

Weaker

baseStrongerWeaker

CN-F-

base

Ranking Acid-Base Strength

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

• strong acid very weak conjugate base

• strong base very weak conjugate acid

• weak acid weak conjugate base

Conjugate Partners of Strong Acids and Bases

• The pH of a salt solution is =7 (neutral) if it contains the cation of a strong

base (Na+) and the anion of a strong acid (Cl-) >7 (basic) if it contains the cation of a strong

base (K+) and the anion of a weak acid (F-) <7 (acidic) if it contains the cation of a weak

base (NH4+) and the anion of a strong acid

(NO3-).

pH of Salt Solutions

• Determine the pH of a 0.59 M solution of CH3OONa. Ka of CH3OOH = 1.8 × 10-5.

CH3OO-(aq) + H2O(l) ⇌ CH3OOH(aq) + OH-(aq)

• The acetate ion is a base. Calculate Kb for the acetate ion:

The pH of a Salt Solution

• When two or more acids are present in solution, the effects of the weaker acids may be ignored if Ka is smaller by a factor of 100 or more.

Mixtures of Acids

• To calculate the pH of a mixture of HCl (strong acid) and HOCl (Ka = 3.0 × 10-8), ignore the HOCl; treat the solution as a solution of HCl.

• To calculate the pH of a mixture of HCOOH (Ka = 1.8 × 10-4) and HCN (Ka = 7.2 × 10-10), ignore the HCN; treat the solution as a solution of HCOOH.

Mixtures of Acids

• Binary hydrides contain hydrogen and one other element.

• Acidity depends on the HA bond strength and the stability of A-.

Binary Hydrides

• Acidity increases with decreasing H-A bond dissociation energy

HF << HCl < HBr < HI

568.2 431.9 366.1 298.3 kJ/mol

Binary Hydrides

• Acidity increases with increasing electronegativity of A.

CH4 < NH3 < H2O < HF

2.5 3.0 3.5 4.0

Binary Hydrides

• Oxyacids contain H, O, and a third element X.

X–O–H• X is usually a non-metal or a

transition metal in a high oxidation state.

Oxyacids

• As the electronegativity of X increases, the X-O bond becomes more covalent and the O-H bond becomes more polar.

X–O–H X–O- + H+

Oxyacids

• Two factors increase acidity of the oxyacid: Greater electronegativity of central

atom.

HClO4 > HBrO4

Greater oxidation number of central atom (more oxygen atoms).

HClO4 > HClO3 > HClO2 > HClO

Oxyacids

• Lewis acid: electron pair acceptor.• Lewis base: electron pair donor.• Lewis acid-base reaction: formation

of a coordinate-covalent bond.

AlCl3 + Cl- AlCl4-

• Coordinate-covalent bond: covalent bond in which both electrons come from one atom.

Lewis Acids and Bases