A. Dynamic Equilibrium

reactions are often which means that not only are the

we use the double arrow to show this relationship

Equilibrium, Acids and Bases

reversible…

reactants can be reformed

eg) A + B ⇌ C + D

products formed but the

equilibrium theories and principles apply to a variety of phenomena in our world

eg) blood gases in scuba divingCO2 in carbonated beveragesbuffers in our blood

Video: equilibrium

the forward and reverse reaction will proceed at different rates…it depends on the concentration of the reactants and products

if we start with only the reactants A and B, the

only reaction possible

as the products C and D are formed, the reaction will and the reaction will

at some point, the rates of forward and reverse reactions become equal

forward will initially be thefastestas it is the

reaction

forwardslow down reversespeed up

EQUILIBRIUM ANALOGY – SOCCER PITCH

Equilibrium is like a soccer game.

- Lets say that one team starts with 20 players and the other team 0. - (kinda like reactants and products)

- When a team scores, one player switches teams. They should score quickly so its now 19 vs 1.

- Eventually it might be 11 vs 9. Either team might score and scoring will take more time.

- Sometimes a team of 1 might be better than the team of 19 if the 1 player is great and the 19 are little 2 year olds. This will affect who is favored as all reactions (situations) are different.

- -When scoring happens in either direction at the same rate (1:1) then we can say equilibrium has been reached. Or in other words, the teams should be tied or equal.

Rate

Time0

forward reaction

reverse reaction

Dynamic Equilibrium

equilibrium

a system is said to be in a state of when: dynamic equilibrium

1. the of the forward and reverse reactions are

2. the of the system, such as temperature, pressure, concentration, pH are

3. the system is a system at

ratesequal

observable (macroscopic) properties

constant

closedconstant temperature

there are three classes of chemical equilibria:

1. favoured (percent rxn )

reactants <50%

A + B ⇌ C + D

<50%

2. favoured (percent rxn )

products >50%

A + B ⇌ C + D

>50%

B. Classes of Reaction Equilibria

Trick: point to the favored side

3. to the right (percent rxn )

quantitative >99%

A + B ⇌ C + D

>99%

or

A + B C + D

C. The Equilibrium Constant

experiments have shown that under a given set of conditions (P and T) a specific quantitative relationship exists between the equilibrium concentrations of the reactants and products

one reaction that has been studied intensively is that between H2(g) and I2(g) (simple molecules and takes place in gas phase no solvent necessary!)

H2(g) + I2(g) ⇌ 2 HI(g)

when different combinations of H2(g), I2(g), and HI(g) were mixed and the concentrations measured, it was discovered that all cases

even though the equilibrium [ ] are , the

differentend quotient

equilibriumwas reached in

was the same each time (within experimental error)

this led to the empirical generalization known as the which says that there is a between the concentrations of the products and the concentrations of the reactants at equilibrium

Law of Equilibrium constant ratio

this law can be expressed mathematically:

For the reactionaA + bB ⇌ cC + dD

The law is: Kc = [C]c [D]d

[A]a [B]b

where: Kc =A, B =C, D =

a, b, c, d = coefficients

equilibrium constantreactantsproducts

balancing

is affected by temperature:

1. temperature change = change in Kc

Kc has no units:

2. It’s a numerical value. Ex: 2, or 0.0043

Kc

3. Use only gas or aqueous specie ***unless all states are the same, then use them all

Follow these rules…

Kc = 1+ Products are favored

catalysts will not affect the [ ] at equilibrium…

they only increase the rate of the rxn

Kc = 0 There are no products

Kc = 0 - 1 Reactants are favored

What does Kc tell us?

Example 1

Write the equilibrium law for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas.

2 NO(g) + O2(g) ⇌ 2 NO2(g) Kc = [NO2(g)]2

[NO(g)]2[O2(g)]

Example 2

Write the equilibrium law for the following reaction:

CaCO3(s) ⇌ CaO(s) + CO2(g)

Kc = [CO2(g)]

*** do not include solids in Kc

Example 3

Write the equilibrium law for the following reaction:

2 H2O(l) ⇌ 2 H2(g) + O2(g)

Kc = [H2(g)]2[O2(g)]

*** do not include liquids in Kc

Example 4

Phosphorus pentachloride gas can be decomposed into phosphorus trichloride gas and chlorine gas.

a) Write the equilibrium law for this reaction.

Kc = [PCl3(g) ][Cl2(g)] [PCl5(g)]

PCl5(g) ⇌ PCl3(g) + Cl2(g)

b) If the [PCl5(g)]eq = 4.3 x 10-4 mol/L, the [PCl3(g) ]eq = 0.014 mol/L and the [Cl2(g)]eq = 0.014 mol/L then calculate Kc.

Kc = [PCl3(g) ][Cl2(g)] [PCl5(g)]

= (0.014)(0.014) (4.3 x 10-4)

= 0.46

Example 5

Find the [SO3(g)] for the following reaction if Kc = 85.0 at 25.0C.

2 SO2(g) + O2(g) ⇌ 2 SO3(g)0.500 mol/L 0.500 mol/L ???

Kc = [SO3(g) ]2____[SO2(g)]2[O2(g)]

85.0 = [SO3(g) ]2

(0.500)2(0.500)[SO3(g) ]

2 = 10.625[SO3(g) ] = 3.26 mol/L

PRACTICE QUESTION

PRACTICE QUESTION Answer: C

PRACTICE QUESTIONS

Workbook time!!! - page 1

Answer: A

D. Graphical Analysis

a graph of vs. can be used to see when equilibrium has been reached…as soon as the concentrations , you can read this time off the graph

concentration time

don’t change any more

Example 1

Consider this rxn: 2 SO2(g) + O2(g) ⇌ 2 SO3(g)

Concentration (mol/L)

Time (s)0

O2(g)SO2(g)

SO3(g)

10 20 30

755025

At what time does equilibrium get reached and what is the value for Kc?

Kc = [SO3]2

[SO2]2[O2]

= (75)2 (50)2(25)

= 0.090

Equilibrium is reached at approximately 20 seconds.

PRACTICE

Workbook – p.4

E. Le Châtelier’s Principle – Video

states that when a chemical system at is disturbed by a the system adjusts in a way that

this takes place in a three-stage process

1. initial equilibrium state

2. shifting non-equilibrium

3.

state

new equilibrium state

Le Châtelier’s principleequilibrium

change in property of the system,opposes the change

Video: Le Chateliers Principle

PARTICLE MODEL OF MATTER

Review: add heat = movement increases in particles = more collisions = more chemical reactions

Use this to help explain the

following instead of memorizing it!!!

1. Concentration Changes

a system can be affected by a change in concentration, temperature and or volume (pressure)

an in the [ ] of the products or reactants favours

a in the [ ] of the products or reactants favours

the other side of the equation

the same side of the equation

increase

decrease

eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)

***Haber-Bosch process

↑ [N2(g)] will shift the equilibrium

↑ [NH3(g)] will shift the equilibrium

[NH3(g)] will shift the equilibrium

to the products

to the reactantsto the products

changes in concentration have on the value of

no effectKc

2. Temperature Changes energy is treated like a or

eg)

if cooled, the equilibrium shifts so

reactant product

reactants + energy ⇌ products

reactants ⇌ products + energy

if heated, the equilibrium shifts

more heat is produced (same side)

away from the heat so it cools down (opposite side)

a change in temperature is the only stress that the value of Kc!!!!!!! will change

if the shift is towards the side, Kc will

product

increase

if the shift is towards the side, Kc will

reactant

decrease

PRACTICE QUESTION

3. Volume and Pressure Changes with gases, volume and pressure are related

(volume , pressure )

the concentration of a gas is related to volume (pressure)…volume , concentration

an caused by a in volume causes a shift towards the side of the equation with

http://michele.usc.edu/java/gas/gassim.html

eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)

4 moles 2 moles

↑

↓ ↑

will shift to NH3(g)

increase in [ ] drop

fewer moles of gas

Answer: D

if the number of moles are the on both sides of the reaction, a change in volume (pressure) hasno effect

changes in volume and pressure have on the value of

no effectKc

same

PRACTICE QUESTION

in many equilibrium systems, the reactants will have a different colour than the products

4. Colour Changes

predictions can be made about the equilibrium shift and the resulting change in colour

Answer: D

Example

Use the following reaction to predict the equilibrium shift and resulting colour change when the stresses are applied 2 CrO4

2(aq) + 2 H3O+(aq) ⇌ 2 Cr2O7

2(aq) + 3 H2O(l)

yellow orange a) a crystal of Na2CrO4(s) is added

b) a crystal of K2Cr2O7(s) is added

c) a few drops of concentrated acid is added

d) water is removed

e) a few crystals of NaOH(s) are added

products, orange

products, orange products,

orange

reactants, yellow

reactants, yellow

Video: Equilibrium shift demo

all of the changes that can happen to systems in equilibrium can be shown graphically:

ExampleState what change to the equilibrium takes place at each of the labelled parts of the graph:

Concentration(mol/L)

Time (min)

NH3(g)

N2(g)

H2(g)

A B C D

Manipulations of An Equilibrium SystemN2(g) + 3 H2(g) ⇌ 2 NH3(g) + energy

Equilibrium Time StressABCD

addition of H2(g) addition of inert gas, addition of catalyst

decrease in volume increase in energy

PRACTICE

Practice – p.2

PRACTICE QUESTION

F. ICE Tables

we can use a table set-up to calculate the equilibrium concentrations and/or Kc for any system

you must be able to calculate all before you can use the equilibrium law

equilibrium [ ]

Answer: B

Example 1

Initial

+0.214 mol/L

+0.214 mol/L x 1/1

+0.214 mol/L

–0.214 mol/L x 2/1= –0.428

1.572 mol/L

Hydrogen iodide gas decomposes into hydrogen gas and iodine gas. If 2.00 mol of HI(g) is place in a 1.00 L container and allowed to come to equilibrium at 35C, the final concentration of H2(g) is 0.214 mol/L. Find the value for Kc.

2.00 mol/L 00

0.214 mol/L

2 HI(g) ⇌ H2(g) + I2(g)

Change

Equil.

Kc = [H2(g)][I2(g)] [HI(g)]2

= (0.214)(0.214) (1.572)2

= 0.0185

Example 2In a 500 mL stainless steel reaction vessel at 900C, carbon monoxide and water vapour react to produce carbon dioxide and hydrogen. Evidence indicates that this reaction establishes an equilibrium with only partial conversion of reactants to products. Initially, 2.00 mol of each reactant is placed in the vessel. Kc for this reaction is 4.20 at 900C. Calculate the concentration of each substance at equilibrium. CO(g) + H2O(g) ⇌ CO2(g) +

H2(g) 2.00/0.5L = 4.00 mol/L

0 0I

C

E

+x mol/L x 1/1

x mol/L

–x mol/L

4.00 x mol/L

2.00/0.5L = 4.00 mol/L–x mol/L +x mol/L x

1/1

4.00 x mol/L x mol/L

Kc = [CO2(g)][H2(g)][CO(g)][H2O(g)]

4.20 = (x)(x) (4.00 x)( 4.00 x)

4.20 = x2 (4.00 x)2

***Note, this is a perfect square so to solve for x simply square root both sides of the equation, then solve

2.05 = x (4.00 x)

2.05(4.00 x) = x 8.20 2.05x = x

8.20 = 3.05x x = 2.69 mol/L

[CO(g)] = 4.00 – x =

[H2O (g)] = 4.00 – x =

[CO2(g)] = x =

[H2(g)] = x =

4.00 – 2.69 =1.31 mol/L

1.31 mol/L4.00 – 2.69 =

2.69 mol/L

2.69 mol/L

Example 3Gaseous phosphorus pentachloride decomposes into gaseous phosphorus trichloride and chlorine gas at a temperature where Keq = 1.00 103. Suppose 2.00 mol of PCl5(g) in a 2.00 L vessel is allowed to come to equilibrium. Calculate the equilibrium [ ] of each species. PCl5(g) ⇌ PCl3(g) + Cl2(g)

2.00mol/2.00L = 1.00 mol/L

0 mol/L 0 mol/L

I

C

E

+x mol/L x 1/1

+x mol/L–x mol/L x 1/1

1.00 – x mol/L x mol/L x mol/L

Kc = [PCl3(g)][Cl2(g)]

[PCl5(g)]1.00 10-3 = (x)(x)

(1.00 - x)

***at this point, you would have to use the quadratic formula to solve for x

when the concentrations are greater than the equilibrium constant, we can make an that greatly simplifies our calculations

1000 X

approximation

if Kc is very small, the equilibrium doesn’t lie very far to the right and x is a very small

number

1.00 10-3 = (x)(x) (1.00 ) x2 = 1.00 10-3 x 1.00x = 0.0316

***in this example 1.00 – x can be assumed to be 1.00 since x is really small, so…

***now you can calculate the [ ]eq for each species …substitute x into the equilibrium values in the ICE table

[PCl5(g)]eq = 1.00 mol/L – 0.0316 mol/L = 0.967 mol/L [PCl3(g)]eq = 0 + 0.0316 mol/L = 0.0316 mol/L

[Cl2(g)]eq = 0 + 0.0316 mol/L = 0.0316 mol/L

Example 4Gaseous NOCl decomposes to form gaseous NO and Cl2. At 35C the equilibrium constant is 1.6 10-5. Calculate the equilibrium [ ] of each species when 1.0 mol of NOCl is placed in a 2.0 L covered flask.

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

1.0mol/2.0L = 0.50 mol/L

0 mol/L 0 mol/L

I

C

E

+x mol/L x 2/1

+x mol/L–x mol/L x 2/1

0.50 – 2x mol/L 2x mol/L x mol/L

= +2x mol/L = –2x mol/L

Kc = [NO(g)]

2[Cl2(g)] [NOCl(g)]

2

1.6 10-5 = (2x)2(x) (0.50 -

2x)2

***using approximation,

0.50 – 2x = 0.50 1.6 10-5 = (4x 2)(x)

(0.50 )2 4x3 = 1.6 10-5 x 0.502

x3 = 4.0 10-6 / 4 x = 0.010 mol/L

[NOCl(g)]eq = 0.50 mol/L – (2)(0.010) mol/L = 0.48 mol/L [NO(g)]eq = 0 + (2)(0.010 mol/L) = 0.020 mol/L [Cl2(g)]eq = 0 + 0.010 mol/L = 0.010 mol/L

G. Ionization of Waterthe equilibrium of water can be written as

follows:

the equilibrium law is:

the equilibrium constant for water is designated as

Kc = [H3O+(aq) ][ OH-

(aq)]

Kw

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

at 25C, neutral water has [H+

(aq)] = [OH-(aq)] = 1.0 10-7 mol/L

Kw = = (on pg 3 of Data Booklet)

(1.0 10-7) (1.0 10-7) 1.0 10-14

Kw is constant and therefore can be used to determine the or the [ OH-

(aq)]

[H3O+(aq) ]

eg) if [H3O+(aq)] = 1.0 104 mol/L

then [OH(aq)] = 1.0 1014 = 1.0 104

if [H3O+(aq)] = [OH(aq)], then solution is

if [H3O+(aq)] > [OH(aq)], then solution is

if [H3O+(aq)] < [OH(aq)], then solution is

neutral

acidic

basic

1.0 1010 mol/L

Try These:1. [H3O+(aq)] = 1.0 109 mol/L [OH(aq)] =

2. [H3O+(aq)] = 1.0 101 mol/L [OH(aq)] =

3. [H3O+(aq)] = [OH(aq)] = 1.0 102 mol/L

4. [H3O+(aq)] = 6.3 109 mol/L [OH(aq)] =

5. [H3O+(aq)] = 8.1 103 mol/L [OH(aq)] =

6. [H3O+(aq)] = [OH(aq)] = 2.8 107 mol/L

1.0 105 mol/Lbasic

1.0 1013 mol/Lacidic

1.0 1012 mol/L basic

1.6 106 mol/Lbasic

acidic

basic

1.2 1012 mol/L

3.6 108 mol/L

H. Review of pH and pOH

the number of digits following the in the is equal to the number of in the

pH = log[H3O+(aq)]

decimal placepH value sig digs[H3O

+(aq)]

[H3O+(aq)] = 10-pH

pOH = log[OH-(aq)] [OH-(aq)] = 10-

pOH

pH + pOH = 14

Example 1Find the pH of a solution where the [H3O

+(aq)] = 4.7 10-11 mol/L.

pH = log[H3O+(aq)]

= log(4.7 10-11) = 10.33

Example 2Find the pH of a solution where the [OH-(aq)] = 2.4 10-3 mol/L.

pOH = log[OH(aq)] = log(2.4 103 ) = 2.619…

pH = 14 – pOH = 14 – 2.619… = 11.38

Example 3Calculate the [H3O

+(aq)] if the pH of the solution is 5.25.

[H3O+(aq)] = 10-pH

= 10-5.25

= 5.6 10-6 mol/L 

Example 4Calculate the pH of a solution where 10.3 g of Ca(OH)2(s) is dissolved in 500 mL of water.

Ca(OH)2(s) Ca2+(aq) + 2 OH-(aq) m = 10.3 gM = 74.10 g/moln = m M = 10.3 g 74.10 g/mol

= 0.139…mol

v = 0.500 L n = 0.139…mol 2/1 = 0.278…molC = n V = 0.278…mol 0.500L = 0.556…mol/L

Example 4 (continued)

pOH = log[OH(aq)] = log(0.556… ) = 0.254…

pH = 14 – pOH = 14 – 0.254… = 13.745

this theory looks at the role of the acid or base

an acid is a

like in electrochemistry where e are transferred…now we transfer H+

I. Brønsted-Lowry Definition of Acids & Bases (1923)

chemical species (anion, cation or molecule) that loses a proton

a base is a chemical species that gains a proton

Video: definition of A + B

HCl(aq) + H2O(l) ⇌

NH3(aq) + H2O(l) ⇌

H+

H3O+

(aq) + Cl-(aq)

NH4+

(aq)+ OH-

(aq)

H+

water does not have to be involved!

HCl(g) + NH3(g) ⇌ NH4Cl(s)

H+

a Brønsted-Lowry acid doesn’t necessarily have to produce an acidic solution…it depends on what accepts the proton

an acid/base reaction is a chemical reaction in which a is transferred from an to a forming a and a

this theory explains how some chemical species can be used to neutralize both acids and bases

eg) HCO3-(aq) + H3O

+(aq) ⇌ H2O(l) + H2CO3(aq)

HCO3-(aq) + OH-

(aq) ⇌ H2O(l) + CO32-

(aq)

proton (H+) acidbase new acid new base

a substance that appears to act as a Brønsted-Lowry acid in some rxns and a Brønsted-Lowry base in other rxns is said to be amphiprotic or amphoteric

eg) H2O, HCO3, HSO4

, HOOCCOO etc

a pair of substances that differ only by a proton is called a …the is on one side of the reaction and the is on the other in general, the reaction can be shown as follows:

J. Conjugate Acids and Bases

HA(aq) + H2O(l) ⇌ H3O+

(aq) + A-

(aq)

conjugate acid-base pair

acidbase

acid conjugate base

base conjugate acid

the an acid, the its conjugate base

stronger weaker

the an acid, the its conjugate base

weaker stronger

Do questions workbook p.8

PRACTICE QUESTION

two different acids (or bases) can have the same [ ] but have different strengths

the stronger the acid, the electricity it conducts, the the pH and the it reacts with other substances

K. Strengths of Acids and Bases

eg) 1 M CH3COOH(aq) and 1 M HCl(aq) will react in the same way but not to the same degree

morelower faster

acids that ionize in water to form H3O

+(aq)

percent rxn =

1. Strong Acids

100%

quantitatively

the bigger the Ka (Kc for acids) the more the are favoured product

stop 6 acids on the table (pg 8-9 in Data Book) have a very large Ka …note the H3O

+ is the strongest acid on the chart (leveling effect)…all strong acids react to form H3O

+(aq) so it is the strongest

when calculating pH, the so use

[SA] = [H3O+

(aq)] pH = -log[H3O+

(aq)]

ExampleWhat is the pH of a 0.500 mol/L solution of HNO3(aq)? [H3O

+(aq)] = [HNO3(aq)] = 0.500

mol/L

pH = -log[H3O+

(aq)] = -log(0.500 mol/L) = 0.301

a weak acid is one that

most ionize

2. Weak Acids

Ka value is

to calculate pH, you need to use the …you cannot use just the because it is not

only partially ionizes in water to form H+

(aq) ions

small (<1)

<50%

Ka value [WA]

100% dissociated

the Ka law is an and is devised the same way we did

eg) HA(aq) + H2O(l) ⇌ H3O+

(aq) + A-(aq)

Ka = [H3O+

(aq)][A-(aq)]

[HA(aq)]

equilibrium lawKc

you will be required to figure out the before you can calculate the pH

[H3O+

(aq)]

you have the and the value but you don’t have the

Ka = [H3O+

(aq)]2

[HA(aq)]

since the mole ratio for is , they have the same [ ] (this is a !)

[WA] Ka [A-

(aq)]

[H3O+

(aq)]:[A-(aq)] 1:1

dissociation

now you can solve for x to get the [H3O+

(aq)]

[H3O+

(aq)] = (Ka)([WA])

Example 1What is the pH of a 0.10 mol/L acetic acid solution?

CH3COOH(aq) + H2O(l) ⇌ H3O+

(aq) + CH3COO-

(aq)

***check in DB…weak acid!!!!!

[H3O+

(aq)] = (Ka)([WA])

= (1.8 x 10-5 mol/L )(0.10 mol/L) = 1.34… 10-3 mol/L

pH = -log[H3O+

(aq)] = -log(1.34… 10-3 mol/L) = 2.87

Example 2What is the pH of a 1.0 mol/L acetic acid solution? CH3COOH(aq) + H2O(l) ⇌ H3O

+(aq) + CH3COO-

(aq) [H3O+

(aq)] = (Ka)([WA])

= (1.8 x 10-5 mol/L )(1.0 mol/L) = 4.24… 10-3 mol/L

pH = -log[H3O+

(aq)] = -log(4.24… 10-3 mol/L) = 2.37

Example 3A 0.25 mol/L solution of carbonic acid has a pH of 3.48. Calculate Ka.

H2CO3(aq) + H2O(l) ⇌ H3O+

(aq) + HCO3

-(aq) [H3O

+(aq)] = 10-pH

= 10-3.48

= 3.31… 10-4 mol/L

Ka = [H3O +

(aq)]2

[H2CO3 (aq)] = (3.31…x10-4 mol/L)2

0.25 mol/L = 4.4 x 10-7 mol/L

the can be written as a above the in a chemical reaction:

eg) CH3COOH(aq) + H2O(l) ⇌ H3O+

(aq) + CH3COO-(aq) Ka =

1.8 x 10-5

% reaction (% ionization) % ⇌

1.3%

the % reaction be calculated using [H3O+] and

[WA]

% ionization = [H3O+

(aq)] 100

[WA(aq)]

Example 1Calculate the % ionization for a 0.500 mol/L solution of hydrosulphuric acid if the [H+

(aq)] is 5.0 10-4 mol/L.

% ionization = [H3O +

(aq)] 100 [WA(aq)]

= 5.0 10-4 mol/L 100

0.500 mol/L = 0.10 %

Example 2The pH of a 0.10 mol/L solution of methanoic acid is 2.38. Calculate the % ionization.

% ionization = [H3O +

(aq)] 100 [WA(aq)]

= (0.00416…mol/L) x (100) 0.10 mol/L

= 4.2 %

[H3O+] = 10-pH

= 10-2.38

= 0.00416… mol/L

according to Arrhenius, bases are substances that increase the of a solution

3. Strong Bases

all are strong bases

strength depends on … is a stronger base than at the same [ ] because it produces

hydroxide [ ]

ionic hydroxides

percent rxn = 100%

# of hydroxide ions Ba(OH)2(aq) NaOH (aq)

2 OH-(aq)

where x is the number of ions (think about the dissociation equation!)

ExampleCalculate the pH of a 0.0600 mol/L solution of Ca(OH)2(aq).

[OH-(aq)] = x[SB(aq)]

= x[Ca(OH)2(aq)] = 2(0.0600 mol/L) = 0.120 mol/L

[OH-(aq)] = x[SB(aq)]

hydroxide

pOH = log(0.120)

= 0.9208…pH = 14 – 0.9208… = 13.079

do not in water…just like weak acids

4. Weak Bases

is the dissociation constant or equilibrium constant for bases

dissociate completely

B + H2O(l) ⇌ BH+(aq) + OH-

(aq)

Kb

Kb = [BH+(aq)][OH-

(aq)] [WB(aq)]

you can calculate Kb using for the

Ka Kb = KW = 1.00 1014

eg) Calculate Kb for SO42

(aq).

Kb = KW

Ka

= 1.00 1014

1.0 102

= 1.0 1012 mol/L

Ka base

[OH-(aq)] = (Kb)([WB])

once you have Kb, you can then solve for [OH

(aq)] using the equilibrium law (just like with weak acids!) Kb = [BH+

(aq)][OH-(aq)]

[WB(aq)] Kb = [OH-

(aq)]2

[WB(aq)]

now you can solve for [OH(aq)]

Example Find the pH of a 15.0 mol/L NH3(aq) solution.

Ka = 5.6 10-10 mol/L

Kb = Kw

Ka

= 1.00 10-14

5.6 10-10

= 1.78… 10-5 mol/L

NH3(aq) + H2O(l) ⇌ NH4+

(aq) + OH-(aq)

[OH-(aq)] = (Kb)([NH3(aq)])

[OH-(aq)] = (1.78… 10-5)(15.0)

[OH-(aq)] = 1.63… x 10-2 mol/L

pOH = -log[OH-(aq)]

= -log(1.63…x 10-2

mol/L) = 1.78… pH = 14 – pOH = 14 – 1.78… = 12.214

SUMMARY OF ACIDS AND BASES

Species Strong Acids (SA)

Weak Acids (WA)

Strong bases (SB)

Weak Bases (WB)

Definition 100% ionized Less than 100% ionized

100% ionized Less than 100% ionized

How to identify it

Top 6 acids on left side of booklet

Below top 6 acids on left side of booklet

All bases that contain hydroxide (OH-)

All other bases found on the right side of the booklet

Equation to use

Other equations needed

PRACTICE QUESTION

acids are listed in order of strength on the left side and bases are listed in order of strength on the right side

L. Predicting Acid-Base Equilibria

when predicting reactions, the substance with the will react with the substance that

decreasing

increasing

greatest attraction for protons (the strongest base) gives up its proton most easily (strongest acid)

we will assume that only is transferred per reaction

one proton

to predict the acid-base reaction, follow the following steps:

Steps

1.

Note:

List all species (ions, atoms, molecules) initially present.

strong acids ionize into H3O+

and the anion weak acids are NOT dissociated

don’t forget to include water

2. Identify all possible

acids and bases.

dissociate ionic compounds

4. To write the reaction, transfer one proton from the acid to the base to predict the conjugate acid and conjugate base.

3. Identify the and …like redox rxns the and the

strongest acid (SA) strongest base (SB)

SA is top left SB is bottom right.

5. Predict the position of the equilibrium.

Note: if acid is above base, then >50% (favours products) ⇌

if base is above acid, then <50% (favours reactants) ⇌

Example 1Predict the acid-base reaction that occurs when sodium hydroxide is mixed with vinegar.

Na+(aq)

List: OH-

(aq) CH3COOH(aq) H2O(l)

AB A/BSB SA

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

(aq) ⇌

Example 2Predict the acid-base reaction when ammonia is mixed with HCl(aq).

NH3(aq

List: H3O

+

(aq) Cl-

(aq) H2O(l)

B B A/BSASB

NH3(aq + H3O+

(aq) H2O(l) + NH4+

(aq)

A

⇌

M. Monoprotic vs. Polyprotic Acids and Bases

if an acid can transfer more than one proton, it is called ( if 2 protons, if 3 protons)

an acid capable of donating only one proton is called monoprotic

polyprotic

eg) HCl(aq), HNO3(aq), HOCl(aq) etc.

diprotictriprotic

eg) Label each of the following acids as monoprotic or polyprotic:

1. H2SO4(aq)

2. HOOCCOOH(aq)

3. HCOOH(aq)

4. CH3COOH(aq)

5. H2PO4-(aq)

6. NH4+

(aq)

polyprotic

polyprotic

polyprotic

monoprotic

monoprotic

monoprotic

a base that can accept more than one proton is called a polyprotic base

eg) can accept up to 3 H+ to form and respectively

( or ) diprotic triprotic

PO43-

(aq) HPO4

2-(aq), H2PO4

-(aq), H3PO4(aq)

a base capable of accepting only one proton is called a monoprotic base

eg) Label each of the following as monoprotic or polyprotic acids, monoprotic or polyprotic bases: 1. HSO4

-(aq)

2. H2PO4-(aq)

3. HPO42-

(aq)

4. HCO3-(aq)

5. H2O(l)

monoprotic acid; monoprotic base polyprotic acid; monoprotic base

monoprotic acid; monoprotic base

monoprotic acid; polyprotic base monoprotic acid; monoprotic base

only is transferred at a time and always from strongest acid to strongest base

reactions involving polyprotic acids or polyprotic bases substances involve the same principles of reaction prediction

one proton

Example 1Potassium hydroxide is continuously added to oxalic acid until no more reaction occurs.

K+

(aq)

List: OH-(aq) HOOCCOOH(aq) H2O(l)

AB A/BSB SA

OH-(aq) + HOOCCOOH(a

q)

⇌ H2O(l) + HOOCCOO-(aq)

HOOCCOO-(aq)

A/B

SA

OH-(aq) + HOOCCOO-

(aq)⇌ H2O(l) + OOCCOO2-(aq)

OOCCOO2-

(aq) B

Net Reaction: Add all reactions together (only if all quantitative), cancelling out any species that occur in the same quantity on both the reactant and product sides and summing any species that occur more than once on the same side

OH-(aq) + HOOCCOOH(a

q)

H2O(l) + HOOCCOO-(aq)

OH-(aq) + HOOCCOO-

(aq) H2O(l) + OOCCOO2-(aq)

2 OH-(aq) + HOOCCOOH(a

q)

2 H2O(l) + OOCCOO2-

(aq)

H3O+

(aq) + HPO42-

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

H3O+

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

A/B ASB SA

A/B

Na+(aq)List: HPO4

2-(aq) H2O(l)H3O

+(aq) I-

(aq)

B

H2PO4-(aq)

A/BSB

H3PO4(aq)

A

2 H3O+

(aq) + HPO42-

(aq) 2 H2O(l) + H3PO4(aq)

Example 2Sodium hydrogen phosphate is titrated with hydroiodic acid. If we assume all steps are quantitative, give the net reaction.

O. Titrations

the information from the titration can be plotted on a graph, buffer regions can be analyzed and stoichiometric calculations can be performed

titrations are used to determine the pH of the of acid-base reactions endpoint

1. pH Curves

graph of vs

a is a graph showing the

the is the point (usually shown by a change in indicator colour) when the reaction has gone to completion

the is the of titrant required for the reaction to go to completion

pH curve continuous change of pH during an acid-base reaction

pH (y-axis) volume of titrant added to sample (x-axis)

endpoint

equivalence point

volume

all pH curves have 4 major features:

they contain a relatively flat region called the buffer region

the initial pH of the curve must be the pH of the sample

the co-ordinate of the equivalence point must be correct in terms of pH and volume

the “over-titration” must be asymptotic with the pH of the titrant number of equivalence points must match the number of quantitative reactions occurring

titrant selection: if the sample is an acid, titrant should

be a such as

if the sample is a base, titrant should be

strong base NaOH(aq) or KOH(aq)

HCl(aq)

you need to be able to interpret pH curves:

1. Strong Monoprotic with Strong Monoprotic pH of at the equivalence point 7

0

7

14

volume0

7

14

volume

SA titrated with SB SB titrated with SA

EP

2. Weak Monoprotic with Strong Monoprotic if weak acid, then pH of at equivalence

point if weak base, then pH of at equivalence point bottom “flat” region is

>7

<7

not as flat as with strong acid/strong base

0

7

14

volume 0

7

14

volume

WA titrated with SBWB titrated with SA

3. Polyprotic with Strong Monoprotic more than 1 equivalence

point

pH

0

7

14

volume

pH

0

7

14

volume

WA(poly) titrated with SB

WB(poly) titrated with SA

EP2EP1

EP1

EP2

PRACTICE QUESTION

2. Indicators

an indicator is a substance that changes colour

they exist in one of two conjugate forms that are reversible and distinctly different in color

HIn(aq) + H2O(l)

⇌ In-(aq) + H3O

+(aq)

acid base conjugate base

conjugate acid

red blueeg) litmus

when it reacts with anacid or baseand are usuallythemselves weak acids

INDICATOR COLORS

recall that to show the equivalence point of an acid-base titration, choose an indicator:

1.

2.

whose includes the of the titration

that will react …this means the indicator is a weaker acid or base than the sample

colour change range

right after the sample reacts

equivalence point

3. Buffers : video

they are used to and

are chemicals that, when added to water, protect the solution from large pH changes when acids or bases are added to them

calibrate pH meters

control the rate of pH sensitive reactions (eg. in the blood)

typical buffers are solutions containing relatively large amounts of such as a and the

buffers

conjugate pairsweak acid salt of the conjugate base

eg)H2CO3 and NaHCO3

BUFFERS IN A GRAPH

eg) Choose a buffer that would be useful for each of the following solutions:

2. pH of 4.5

1. pH of 7.0

3. pH of 10.0

Ka = 10-7.0 = (1.0 x 10-7)

Ka = 10-4.5 = (3.16 x 10-5)

Ka = 10-10 =(1.0 x 10-10)

can be selected by using …this tells you the pH at which the buffer is most useful. To determine the buffer for a particular pH use

pH = –logKa

H2S – HS-

CH3COOH – CH3COO-

HCO3- – CO3

2-

Ka = 10-pH

the in the conjugate pair of the buffer protects against any added base

buffers can be by the addition of

acid

the in the conjugate pair of the buffer protects against any added acid

base

overwhelmedtoo much acid or base

In the equation below (an acid was added to a base) – the buffer solution includes the base and its conjugate pair

PRACTICE QUESTION

ExampleUsing an acetic acid – sodium acetate buffer system, show what happens when:

a) a small amount of HCl(aq) is added CH3COOH(aq) Na+

(aq) CH3COO-(aq) H3O

+(aq) Cl-

(aq) H2O(l)

A BAB A/B–SA

SB

CH3COO-(aq) + H3O

+(aq)⇌ CH3COOH(aq)+ H2O(l)

b) a small amount of NaOH(aq) is added CH3COOH(aq) Na+

(aq) CH3COO-(aq) OH-

(aq) H2O(l)

A BB A/B–SA

SB

CH3COO-(aq)+ OH-

(aq) ⇌CH3COOH(aq) + H2O(l)