Unit 11: Acid-Base EquilibriumChapter 16 and 17

Problem Set• Chapter 16: 17, 21, 37,

43, 45, 61, 65, 69, 77, 79, 101, 107• Chapter 17: 19, 23, 27,

31, 41,

WarmUp

1) nitric acid 2) hydrofluoric acid 3) hydrobromic

acid 4) chloric acid 5) acetic acid 6) carbonic acid 7) tetraboric acid(tetraborate =

B4O72-)

1) HCl 2) H2S 3) H3PO4

4) HI 5) HNO2

6) H3P 7) H2SO4

Write the Formula • Name the Acid

WarmUp

1) nitric acid 2) hydrofluoric acid 3) hydrobromic

acid 4) chloric acid 5) acetic acid 6) carbonic acid 7) tetraboric acid(tetraborate =

B4O72-)

1) HCl 2) H2S 3) H3PO4

4) HI 5) HNO2

6) H3P 7) H2SO4

Write the Formula • Name the Acid

Arrhenius Definition• Acids produce hydrogen ions in

aqueous solution.• Bases produce hydroxide ions

when dissolved in water.• Limits to aqueous solutions.• Only one kind of base.• NH3 ammonia could not be an

Arrhenius base.

Bronsted-Lowry Definitions

• And acid is an proton (H+) donor and a base is a proton acceptor.

• Acids and bases always come in pairs.• HCl is an acid..• When it dissolves in water it gives its

proton to water.• HCl(g) + H2O(l) H3O+ + Cl-

• Water is a base makes hydronium ion

Neutralization Reactions• Remember: Acid + Base → Water +

Salt• The salt’s name and formula will be

based off of the cation of the base and the anion of the acid

• Example: HCl + NaOH ↔ H2O + NaCl• Neutralization reactions are one way

that is used to produce pure salts.• Amphoteric Substances: can act as

either an acid or a base

Titration• Method used to perform

neutralization and to determine concentrations of an unknown acid or base.

• Titrant – added standard solution

• End Point – indicator color change

• Equivalence Point – [H+] = [OH-]

Acid Base Calculations• In neutral solutions, [H+] and [OH-]

are equal• [H+] = [OH-] = 1 x 10-7 M• For other solutions, when [H+]

increases, then [OH-] will decrease and vise versa

• [H+] x [OH-] = 1.0 x 10-14 • Kw = [H+] x [OH-] = 1.0 x 10-14 • This is the Ion-Product Constant for

Water• Acidic Solutions: [H+] > [OH-] • Basic Solutions: [H+] < [OH-]

Using pH Scale• pH: Power of Hydrogen• pH = 7 Neutral [H+] = 1 x 10-7 • pH < 7 Acidic [H] > 1 x 10-7

• pH > 7 Basic [H+] = 1 x 10-7 • pH = -log [H+]• Example Calculate pH if [H+] = 1 x 10-7 M

pH = -log (1 x 10-7 M)= -(log 1 + log 10-7)= -(0 +(-7))pH = 7 * The exponent shows you the pH!*

The pOH Scale• pOH : Power of Hydroxide• pOH = 7 Neutral[OH-] = 1 x 10-7 • pOH < 7 Basic [OH-] > 1 x 10-7

• pOH > 7 Acidic [OH-] < 1 x 10-7

• pOH = -log [OH-]*Same process as before, only now

you are dealing with [OH-]Other Key Equations

• pH + pOH = 14• pH = 14 – pOH• pOH = 14 – pH

Calculations: Complete the following chart

pH pOH [H] [OH] Acid-Base?

2.34.5

3.4x10-

5

4.6x10-9

7

How do we know if something is a strong or weak acid/base?

We must calculate a value for the ionic dissociation for that substance.

ACIDKa = acid dissociation constant: the ratio

of the concentration of the dissociated form of the acid to the concentration of the undissociated form.

BASEKb = base dissociation constant: the ratio

of the product of the conjugate acid and OH concentrations to the concentration of the conjugate base

Strong and Weak Acids and Bases • Dependent on how they dissolve in

water• Strong Acids completely ionize in

aqueous solutions• Weak acids only partially ionize in

aqueous solutions• Strong bases dissociate completely

into metal ions and hydroxide ions in aqueous solutions

• Weak bases react with water to form OH and the conjugate acid of the base

**Memorize Strong Acids and Bases**

Memorize the 8 strong acids… all others are weak

• HCl hydrochloric acid

• HNO3 nitric acid• HBr hydrobromic

acid • HIO4 periodic acid

• HI hydroiodic acid • H2SO4 sulfuric acid• HClO4 perchloric

acid • HClO3 chloric acid

Writing the Ka Expression

• Example HCl + H2O ↔ H3O+ + Cl-

• First you need to write a Keq expression

• Second, water can be eliminated because its concentration is a constant.

• Write the Ka expression for this acid: CH3COOH + H2O ↔ H3O+ + CH3COO-

Significance of Ka

• Ka indicated the fraction of acid in the ionized form

• Weak acids have small Ka values• Strong acids have large Ka

values because their ionization is more complete

Writing the Kb Expression

• Example NH3 + H2O ↔ NH4+ + OH-

• Remember:

• Kb tells us how weak bases compete for OH- from strong bases

• The smaller the value for Kb, the weaker the base

Relationships• KW = [H+][OH-]• -log KW = -log([H+][OH-])• -log KW = -log[H+]+ -log[OH-]• pKW = pH + pOH• KW = 1.0 x10-14

• 14.00 = pH + pOH• [H+],[OH-],pH and pOH

Given any one of these we can find the other three.

Example• Calculate the pH of 2.0 M acetic

acid HC2H3O2 with a Ka 1.8

x10-5

• Calculate pOH, [OH-], [H+]

A mixture of Weak Acids• The process is the same.• Determine the major species.• The stronger will predominate.• Bigger Ka if concentrations are

comparable• Calculate the pH of a mixture 1.20

M HF (Ka = 7.2 x 10-4) and 3.4 M HOC6H5 (Ka = 1.6 x 10-10)

Salt Hydrolysis• Most salt solutions are neutral, but

some can be acidic or basic• Salts like sodium chloride and

potassium sulfate are neutralWHY?

• Neutral salts are made from strong acids and strong bases

• Salt Hydrolysis – the cations or anions of the dissolved salt remove H+ from, or donate H+ to water

• A solution is either acidic or basic based on H+ transfer.

Salt Cations/Anions Effect on pH

In water:• If cation and anion do not react,

then pH = neutral• If cation does not react, but anion

does, then hydroxide is produced, pH = basic

• If cation reacts but anion does not, then H are produced, pH = acidic

If both react: OH- and H3O+ both produced

1. Anions from strong acids will not change pH (ex: Br-)

2. Anions from a weak acid will increase pH (ex: CN-)

3. Cations from weak bases will decrease pH (ex: CH3NH3

+)4. Group 1A ions will increase pH5. Group 2A ions will decrease pH

What is the pH of a 0.140M solution of sodium acetate?

C2H3O2- + HOH↔HC2H3O2 + OH-

But what is Kb? KbxKa = Kw Kb = Kw/Ka = 1x10-14/1.8x10-5

Solve for x x = 8.77x10-6 =[OH-]pOH=5.06, pH = 8.94 so the solution is basic.

I 0.140 M 0 0C -x +x +xE 0.140 – x x x

The Common Ion Effect• When the salt with the anion of a

weak acid is added to that acid,• It reverses the dissociation of the

acid.• Lowers the percent dissociation of the

acid.• The same principle applies to salts

with the cation of a weak base..• The calculations are the same as last

chapter.

Types of Acids• Monoprotic Acied – one acidic hydrogen• Polyprotic Acids- more than 1 acidic

hydrogen (diprotic, triprotic).• Oxyacids - Proton is attached to the oxygen

of an ion.• Organic acids contain the Carboxyl group -

COOH with the H attached to O• Generally very weak.• H2SO3 ↔ H+ + HSO3

- Ka1 = 1.7 x 10-2

• HSO3- ↔ H+ + SO3

2- Ka2 = 6.4 x 10-8

• Why is K smaller?• It is always easier to remove the first

proton

Structure and Acid-Base Properties

• Any molecule with an H in it is a potential acid.

• The stronger the X-H bond the less acidic (compare bond dissociation energies).

• The more polar the X-H bond the stronger the acid (use electronegativities).

• The more polar H-O-X bond -stronger acid.

Strength of oxyacids• The more oxygen hooked to the

central atom, the more acidic the hydrogen.

• HClO4 > HClO3 > HClO2 > HClO

• Remember that the H is attached to an oxygen atom.

• The oxygens are electronegative• Pull electrons away from

hydrogen

Strength of oxyacids

Electron Density

Cl

O H

Strength of oxyacids

Electron Density

Cl

O HO

Strength of oxyacids

Cl

O H

O

O

Electron Density

Strength of oxyacids

Cl

O H

O

O

O

Electron Density

Calculate the Concentration

• Of all the ions in a solution of 1.00 M Arsenic acid H3AsO4

Ka1 = 5.0 x 10-3

Ka2 = 8.0 x 10-8

Ka3 = 6.0 x 10-10

Sulfuric acid is special

• In first step it is a strong acid.• Ka2 = 1.2 x 10-2

• Calculate the concentrations in a 2.0 M solution of H2SO4

• Calculate the concentrations in a 2.0 x 10-3 M solution of H2SO4

Hydrated metals

• Highly charged metal ions pull the electrons of surrounding water molecules toward them.

• Make it easier for H+ to come off.

Al+3 OH

H

Acid-Base Properties of Oxides

• Non-metal oxides dissolved in water can make acids.

• SO3 (g) + H2O(l) H2SO4(aq)• Ionic oxides dissolve in water to

produce bases.• CaO(s) + H2O(l)

Ca(OH)2(aq)

Lewis Acids and Bases

• Most general definition.• Acids are electron pair

acceptors.• Bases are electron pair donors.

B FF

F

:NH

HH

Lewis Acids and Bases

• Boron triflouride wants more electrons.

B FF

F

:NH

HH

Lewis Acids and Bases

• Boron triflouride wants more electrons.

• BF3 is Lewis base NH3 is a Lewis Acid.

BFF

FN

H

HH

Lewis Acids and Bases

Al+3 ( )H

HO

Al ( )6

H

HO

+ 6

+3

Buffers• Solutions in which the pH remains

relatively constant when small amounts of acid of base are added.

• Acidic Buffer – composed of a weak acid and one of its salts

• Basic Buffer – composed of a weak base and one of its salts

• Buffer Capacity – the amount of acid or base that can be added to a buffer solution before a significant change in pH occurs

Buffers have the ability to acquire or give a way a H+, which allows it

to maintain a standard pH

• Example: Ethanoic Acid/Ethanoate system

(Acetic Acid / Acetate)CH3COO- + H+ ↔ CHCOOHCH3COOH + OH- ↔ CHCOO- + H2OBoth reactions occur simultaneously

to maintain pH

pH of Buffers

• To determine pH of Buffers, we use the Henderson-Hasselbalch Equation

• Calculate the pH of a solution that is .50 M HAc and .25 M NaAc (Ka = 1.8 x 10-5)

Calculate the pH of the following mixtures

1. 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

2. 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)

Titration Curves

pH

mL of Base added

7

• Strong acid with strong Base• Equivalence at pH 7

pH

mL of Base added

>7

Weak acid with strong Base Equivalence at pH >7

pH

mL of Base added

7

Strong base with strong acid

Equivalence at pH 7

pH

mL of Base added

<7

Weak base with strong acid

Equivalence at pH <7

Summary• Strong acid and base just

stoichiometry.• Determine Ka, use for 0 mL base• Weak acid before equivalence point

–Stoichiometry first–Then Henderson-Hasselbach

• Weak acid at equivalence point Kb• Weak base after equivalence -

leftover strong base.

Summary• Determine Ka, use for 0 mL

acid.• Weak base before equivalence

point.–Stoichiometry first–Then Henderson-Hasselbach

• Weak base at equivalence point Ka.

• Weak base after equivalence - leftover strong acid.

Indicators• Weak acids that change color when

they become bases.• weak acid written HIn• Weak base• HIn H+ + In- clear red• Equilibrium is controlled by pH• End point - when the indicator

changes color.

Indicators• Since it is an equilibrium the color

change is gradual.• It is noticeable when the ratio of

[In-]/[HI] or [HI]/[In-] is 1/10• Since the Indicator is a weak acid, it

has a Ka.• pH the indicator changes at is.• pH=pKa +log([In-]/[HI]) = pKa

+log(1/10) • pH=pKa - 1 on the way up

Indicators• pH=pKa + log([HI]/[In-])

= pKa + log(10)• pH=pKa+1 on the way down• Choose the indicator with a pKa one

less than the pH at equivalence point if you are titrating with base.

• Choose the indicator with a pKa one greater than the pH at equivalence point if you are titrating with acid.