a. dynamic equilibrium reactions are often we use the double arrow to show this relationship...

A. Dynamic Equilibrium

reactions are often

we use the double arrow to show this relationship

Equilibrium, Acids and Bases

eg)

reversible… which means that not only are the products formed but the reactants can be reformed

A + B C + D C + D A + B and

eg) A + B ⇌ C + D

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 forward reaction will initially be the fastest as it is the only reaction possible

as the products C and D are formed, the forward reaction will slow down and the reverse reaction will speed up

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

Rate

Time0

forward reaction

reverse reaction

Dynamic Equilibrium

equilibrium

a system is said to be in a state of when the rates of the forward and reverse reactions are and we observe no macroscopic ( ) changes (system is a system)

dynamic equilibrium

closedequal visible

B. Classes of Reaction Equilibria

the rate of a reaction depends on:

1. the EK (temperature) of the particles

2. orientation of collisions (shapes of particles)

3. # of collisions/second (concentration)

there are 4 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%

3. to the right (percent rxn ) quantitative >99%

A + B ⇌ C + D>99%

or

A + B C + D

4. to the left (percent rxn ) quantitative <1%

A + B ⇌ C + D<1%

or

A + B C + D

C. The Equilibrium Constant

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

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!)

specific quantitative relationship reactants

products

when different combinations of H2(g), I2(g), and HI(g) were

mixed and the concentrations measured, it was discovered that

H2(g) + I2(g) ⇌ 2 HI(g)

an equilibrium was reached in all cases:

even though the equilibrium [ ] are , the differentend quotient was the same each time (within experimental error)

this led to the empirical generalization known as the

this law can be expressed mathematically:

Law of Equilibrium

For the reaction aA + bB ⇌ cC + dD

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

[A]a [B]b

is the …it is constant for the reaction at a given temperature

it is common to ignore the units for Keq and list it only as a

numerical value (since depends on the powers of the various [ ] terms)

Keq equilibrium constant

when determining Keq use only the species that are in

or gas aqueous phase

***unless all states are the same, then use them all

the higher the value of Keq, the greater the tendency for the

reaction to favor the

Keq indicates the and not the

catalysts will not affect the [ ] at equilibrium…

forward direction (the products)

percent reaction rate of the reaction

they only increase the rate of the rxn

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)

Keq = [NO2(g)]2

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

Example 2

Write the equilibrium law for the following reaction:

CaCO3(s) ⇌ CaO(s) + CO2(g)

Keq = [CO2(g)]

*** do not include solids in Keq

Example 3

Write the equilibrium law for the following reaction:

2 H2O(l) ⇌ 2 H2(g) + O2(g)

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

*** do not include liquids in Keq

Example 4

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

a) Write the equilibrium law for this reaction.

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

[PCl5(g)]

PCl5(g) ⇌ PCl3(g) + Cl2(g)

b) If the [PCl5(g)] = 4.3 x 10-4 mol/L, the [PCl3(g) ] = 0.014

mol/L and the [Cl2(g)] = 0.014 mol/L then calculate Keq.

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

[PCl5(g)]

= (0.014 mol/L)(0.014 mol/L)(4.3 x 10-4 mol/L)

= 0.46

Example 5

Find the [SO3(g)] for the following reaction if Keq = 85.0 at

25.0C.2 SO2(g) + O2(g) ⇌ 2 SO3(g)

0.500 mol/L 0.500 mol/L ???

Keq = [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

Your Assignment: 1. pgs 1-2 in workbook 2. pg 431 #7; pg 434 #10

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 Keq?

Keq = [SO3]2 .

[SO2]2[O2]

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

= 0.090

Equilibrium is reached at approximately 20 seconds.

Your Assignment: pg 3 in workbook

E. ICE Tables

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

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

equilibrium [ ]

Example 1

Find the value for Keq for the following data:

2 HI(g) ⇌ H2(g) + I2(g)

Initial 2.00 mol/L 0 0

Change

Equil. 0.214 mol/L

+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

***Now you can use these equilibrium [ ]’s to calculate Keq

Keq = [H2(g)][I2(g)]

[HI(g)]2

= (0.214)(0.214) (1.572)2

= 0.0185

Example 2Phosphorus pentachloride gas decomposes into phosphorus trichloride gas and chlorine gas. If the [PCl5(g)]i = 8.1 x 10-3 mol/L

and the [PCl3(g)]i = 0.298 mol/L, calculate the equilibrium [ ] of all

chemical species and the Keq. The [Cl2g]eq = 2.00 x 10-3 mol/L.


8.1 x 10-3 mol/L 0.298 mol/L 0 mol/L

2.00 x 10-3 mol/L

I

C

E

+2.00 x 10-3 mol/L+2.00 x 10-3 mol/L x 1/1

0.300 mol/L

–2.00 x 10-3 mol/L x 1/1

6.1 x 10-3 mol/L

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

[PCl5(g)]

= (0.300)(2.00 x 10-3) (6.1 x 10-3) = 9.8 x 10-2

you will get questions where only initial [ ] are given

use to represent the change in concentration

once you calculate x, you can +/- from the [ ]i to get the [ ]eq

“x”

Example 1PCl5(g) decomposes into PCl3(g) and Cl2(g) at a temperature where Keq

= 1.00 10-3. 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.


2.00mol/2.00L = 1.00 mol/L 0 mol/L 0 mol/LI

C

E

+x mol/L x 1/1+x mol/L

0 + x mol/L

–x mol/L x 1/1

1.00 – x mol/L 0 + x mol/L = x mol/L = x mol/L

Keq = [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 Xapproximation

if Keq 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 2Gaseous 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/LI

C

E

+x mol/L x 2/1 +x mol/L

0 + 2x mol/L

–x mol/L x 2/1

0.50 – 2x mol/L 0 + x mol/L = 2x mol/L = x mol/L

= +2x mol/L = –2x mol/L

Keq = [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

Your Assignment: pgs 4-5 in workbook

F. Le Châtelier’s Principle

Le Châtelier’s principle states that

this takes place in a three-stage process

1.

2.

3.

when a chemical system at equilibrium is disturbed by a change in property of the system, the system adjusts in a way that opposes the change

initial equilibrium state

shifting non-equilibrium state

new equilibrium state

1. Concentration Changes

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

an increase in the [ ] of the products or reactants favours

eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)

↑ [N2(g)] will shift the equilibrium

↑ [NH3(g)] will shift the equilibrium

[NH3(g)] will shift the equilibrium

a decrease in the [ ] of the products or reactants favours the opposite side

the same side

to the products

to the reactants

to the products

this process is used in industry to produce desired products

also happens in our bodies in our blood

eg) the for making ammonia from nitrogen and hydrogen

Haber Process

eg) Hb + O2(g) ⇌ HbO2

***as the [O2(g)] increases in our lungs there is a shift

towards the product (HbO2)

2. Temperature Changes energy is treated like a or

eg)

if cooled, the equilibrium shifts

reactant product

reactants + energy ⇌ products

reactants ⇌ products + energy

if heated, the equilibrium shifts

to the side with energy

to the side without energy

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

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

an in [ ] caused by a in volume (increase in pressure) causes a shift towards

gases

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

increase decrease the side of the equation with fewer moles

eg) N2(g) + 3 H2(g) ⇌ 2 NH3(g)

4 moles 2 moles

if the number of moles are the same on both sides of the reaction, then there is no shift in the equilibrium

Your Assignment: 1. pgs 6-7 in workbook 2. pg 440 #12-15

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

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

the equilibrium law is:

the equilibrium constant for water is designated as

H2O(l) ⇌ H+(aq) + OH-

(aq)

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

(aq)]

Kw Kw = [H+

(aq)][ OH-(aq)]

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 always constant and therefore can be used to

determine the or the [H+(aq)] [ OH-

(aq)]

it is unlikely that H+(aq) (which basically is a ) would

exist alone in proton

water

it is attracted to the negative end of a water molecule forming the (also called the “hydrated proton”)

therefore interchangeable with

hydronium ion, H3O+

(aq)

H3O+

(aq) is H+(aq)

in 1909, Soren Sorenson devised the

it is used because the [H3O+

(aq)] is very small and

cumbersome to write logarithmic scale based on whole numbers that are powers

of 10 ***this means that a change of 1 on the pH scale is a

pH scale

10-fold change in [H3O+

(aq)]

0 1 3 7 8 14

strong acid strong base

H. pH and pOH

HCl(aq) acid rainpure H2O

NaOH(aq)blood

the pH of a solution can be found by using the (which is the same thing as the )

[H3O+

(aq)] [H+

(aq)]

the number of digits following the decimal place in the pH value is equal to the number of sig digs in the [H3O

+

(aq)]

pH = log[H3O+

(aq)]

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.

Kw = [H3O+

(aq)] [OH-(aq)]

1.0 10-14 = [H3O+

(aq)] (2.4 10-3)

[H3O+

(aq)] = 4.16… 10-12 mol/L

pH = log[H3O+

(aq)]

= log(4.16… 10-12 mol/L) = 11.38

Example 3Calculate 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 3 (continued)

Kw = [H3O+

(aq)] [OH-(aq)]

1.0 10-14 = [H3O+

(aq)] (0.566…)

[H3O+

(aq)] = 1.79… 10-14 mol/L

pH = log[H3O+

(aq)]

= log(1.79… 10-14 mol/L) = 13.745

[H3O+

(aq)] = 10-pH

you could also be given the pH and asked to calculate the[H3O

+(aq)]

use the in the pH value to determine the in your answer

number of decimal places sig digs

Example 1Calculate 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

just as pH deals with [H+(aq)], deals with [OH-

(aq)]

***p just means -log

at SATP…

pOH = log[OH-(aq)]

pOH

pH + pOH = 14

to calculate the pOH or the [OH-(aq)], use the same formulas

as pH but substitute the OH-(aq) values:

[OH-(aq)] = 10-pOHand

if you dilute a solution, the pH will and the pOH will

to find how many times stronger one solution is over another,

increase decrease

subtract the pH’s then go 10x where x is the difference

eg) How many times more acidic is a solution with a pH of 3.45 over a solution with a pH of 6.20?

6.20 – 3.45 = 2.7510x = 102.75 = 562 X more acidic

Acids Bases electrolyte

electrolyte

taste sour taste bitter

litmus turns red (pink) litmus turns blue pH<7

pH>7

neutralize bases neutralize acids

reacts with metals (H2(g))

and carbonates (CO2(g))

vinegar, lemon juice tums, ammonia

acids dissociate into

bases dissociate into

limitation:

2. Arrhenius Definition(1887)

H+(aq) ions

OH-(aq) ions

some species predicted to be acids by Arrhenius are actually bases, some things are unpredictable

eg) HCO3-(aq)

it is unlikely that H+(aq) (which basically is a proton) would

exist alone in water…it would form the hydronium ion, H3O

+(aq)

the modified definition gives us the following:

limitation:

3. Modified Arrhenius Definition

when writing these acid and base formulas you can run into the situation where two equations can be written for the same substance and you must have water present

acids HA(aq) + H2O(l) ⇌ H3O+

(aq) + A-(aq)

bases (other than hydroxides)

B(aq) + H2O(l) ⇌ BH+(aq) + OH-

(aq)

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+

4. Brønsted-Lowry Definition (1923)

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

a base is a chemical species that gains a proton

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 proton (H+) is transferred from an acid to a base forming a new acid and a new base

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)

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) bicarbonate ions, hydrogen sulphate ions, water

Your Assignment: pg 11 #1-4 in workbook

a pair of substances that differ only by a proton is called a …the is on one

side of the rxn 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

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

CH3COOH(aq) + H2O(l) ⇌ H3O+

(aq) + CH3COO-(aq)

CH3COOH is the and CH3COO- is the

H2O is the and H3O+ is the

acid conjugate base

base conjugate acid

Your Assignment: pg 11 #1,3,4,5 in workbook

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 (Keq for acids) the more the

are favoured

products

top 6 acids on the table (pg 11 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

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

2. 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%

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

(aq)

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

(aq)

# 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 [OH-

(aq)] of a 0.600 mol/L solution of Ca(OH)2(aq).

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

= x[Ca(OH)2(aq)]

= 2(0.600 mol/L) = 1.20 mol/L

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

hydroxide

a weak acid is one that

most ionize

3. 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 is devised the same

way we did

eg) CH3COOH(aq) + H2O(l) ⇌ H3O+

(aq) + CH3COO-(aq)

Ka = [H3O+

(aq)][CH3COO-(aq)]

[CH3COOH(aq)]

equilibrium law andKeq

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


(aq) + CH3COO-(aq)

Ka = [H3O+

(aq)][CH3COO-(aq)]

[CH3COOH(aq)]

Ka = (x)(x)

[CH3COOH(aq)]

Ka = x2

[CH3COOH(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

x x

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?


(aq) + CH3COO-(aq)

***check in DB…weak acid!!!!!

[H3O+

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

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

= 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?


(aq) + CH3COO-(aq)

[H3O+

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

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

= 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 = 4.4 x 10-7 mol/L

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


(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

do not in water…just like weak acids

4. Weak Bases

is the dissociation constant or equilibrium constant for bases

you need to use two things to calculate [OH-(aq)]:

dissociate completely

B + HOH(l) ⇌ BH+(aq) + OH-

(aq)

Kb

1. use the Kb expression from the dissociation equation

2. use the fact that Ka Kb = 1.00 10-14

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

Example You have a 15.0 mol/L NH3(aq) solution.

a) Find Kb.

Ka = 5.6 10-10 mol/L

Kb = Kw

Ka

= 1.00 10-14

5.6 10-10

= 1.8 10-5 mol/L

b) Find the pH.

NH3(aq) + H2O(l) ⇌ NH4+

(aq) + OH-(aq)

Kb = [NH4+

( aq)][OH-(aq)]

[NH3(aq)]

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

[OH-(aq)] = (1.8 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

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 will

react with the substance that

decreasingincreasing

the 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.

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

strongest acid (SA) strongest base (SB) SA is top left SB is bottom

right.

dissociate ionic compounds

4. To write the reaction,

5. Predict the position of the equilibrium.

transfer one proton from the acid to the base to predict the conjugate acid and conjugate base.

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

if H3O+ and the base is stronger than (below or

including) F-, then stoichiometric

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

if OH- and the acid is stronger than (above or including) H2PO4

-, then stoichiometric

Example 1

Predict 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)

*** Reaction has OH-(aq), and CH3COOH(aq) is higher than

H2PO42-

(aq) rxn is stoichiometric

Example 2

Predict 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. Polyprotic and Polybasic Substances

if an acid can transfer more than one proton, it is called ( if 2 protons, if 3 protons)…the conjugate base also appears on the acid list on pg 11

an acid capable of donating only one proton is called monoprotic

polyprotic

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

diprotic triprotic

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 polybasic

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

( or ) dibasic tribasic

PO43-

(aq) HPO42-

(aq), H2PO4

-(aq), H3PO4(aq)

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

1. HSO4-(aq)

2. H2PO4-(aq)

3. HPO42-

(aq)

4. HCO3-(aq)

5. H2O(l)

monoprotic acid; monobasic

polyprotic acid; monobasic


monoprotic acid; polybasic


a base capable of accepting only one proton is called monobasic

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

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

one proton

Example 1

Potassium 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(aq) ⇌ 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(aq) ⇌ H2O(l) + HOOCCOO-

(aq)

OH-(aq) + HOOCCOO-

(aq) ⇌ H2O(l) + OOCCOO2-(aq)

2 OH-(aq) + HOOCCOOH(aq) ⇌ 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 2

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

use the same rules of stoich that we always have

O. Acid-Base Stoichiometry

write the first using the acid-base reaction prediction rules (List, identify A, B, SA, SB etc)

if the question says “at the second endpoint”, then you know the reaction is dealing with something that is

…determine all reaction steps then use the to do the stoich calculation

net reaction

polyprotic or polybasic net reaction

Example 1

A 25.0 mL NH3(aq) solution is titrated with 0.100 mol/L HCl(aq).

Calculate the concentration of the NH3(aq) using the following

data: final buret reading: 16.30 mLinitial buret reading: 0.60 mL

NH3(aq) H3O+

(aq) Cl-(aq) H2O(l)

B BA A/BS S

NH3(aq) + H3O+

(aq) ⇌ H2O(l) + NH4+

(aq) v = 16.30 mL – 0.60 mL = 15.70 mL = 0.01570 LC = 0.100 mol/Ln = CV = (0.100 mol/L)/(0.01570L) = 0.001570 mol

V = 0.0250 LC = ?n = 0.001570 mol × 1/1C = n/V = (0.001570 mol)

0.0250 L = 0.0628 mol/L

Example 2

What is the volume of 0.500 mol/L KOH(aq) needed to react

with 10.0 mL of 0.500 mol/L nitric acid?

NO3-(aq) H3O

+(aq) K+

(aq) OH-(aq)

B –A BSS

OH-(aq)+H3O

+(aq) ⇌ 2 H2O(l)

V = 0.0100 LC = 0.500 mol/Ln = CV = (0.500 mol/L)(0.0100L) = 0.00500 mol

H2O(l) A/B

V = ?C = 0.500 mol/Ln = 0.00500 mol X 1/1V = n C = 0.00500 mol 0.500 mol/L = 0.0100 L

Example 3

What is the pH of a mixture of 50.0 mL of 0.30 mol/L HCl(aq)

with 30.0 mL of 0.40 mol/L NaOH(aq)? Cl-

(aq) H3O+

(aq) Na+(aq) OH-

(aq) B –A BSS

OH-(aq)+H3O

+(aq) ⇌ 2 H2O(l)

V = 0.0500 LC = 0.30 mol/Ln = CV = (0.30 mol/L)(0.0500 L) = 0.015 mol

H2O(l) A/B

V = 0.0300 LC = 0.40 mol/Ln = CV = (0.40 mol/L)(0.0300 L) = 0.012 mollimiting (used up) excess (some left over…

use to find pH)

nH3O+ left = 0.015 mol – 0.012 mol

= 0.0030 molVtotal = 0.0500 L + 0.0300 L = 0.0800L

C = n/V = 0.0030 mol/0.0800 L = 0.0375 mol/L

pH = -log(0.0375 mol/L) = 1.43

OH-(aq)+H3O

+(aq) ⇌ 2 H2O(l)

Your Assignment: pgs 18-19

P. 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

o

o

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

o

o

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: o if the sample is an acid, titrant should be a

such as

o if the sample is a base, titrant should be

strong base NaOH(aq) or KOH(aq)

HCl(aq)

(strong, monoprotic, minimizes secondary rxns)

you need to be able to interpret pH curves:

1. Strong Monoprotic with Strong Monoprotic

o pH of at the equivalence point 7

pH

0

7

14

volume

pH

0

7

14

volume

SA titrated with SB SB titrated with SA

EP EP

2. Weak Monoprotic with Strong Monoprotic o if weak acid, then pH of at equivalence point o if weak base, then pH of at equivalence point o bottom “flat” region is

>7

<7

not as flat as with strong acid/strong base

pH

0

7

14

volume

pH

0

7

14

volume

WA titrated with SB WB titrated with SA

EP

EP

3. Polyprotic with Strong Monoprotic o o middle buffer region at pH

more than 1 equivalence point

7

pH

0

7

14

volume

pH

0

7

14

volume

WA(poly) titrated with SB WB(poly) titrated with SA

EP2

EP1

EP1

EP2

2. Indicators for Titrations

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

1.

2.

whose colour change range includes the equivalence point of the titration

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

L. Acid/Base Indicators

an indicator is a substance that changes colour when it reacts with an acid or base and are usually weak acids themselves

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

several indicators can be used to determine the approximate pH of a solution

eg) Solution 1: indigo carmine is blue pH is

thymol blue is blue pH is

thymolphthalein is blue pH is

Solution 1 has a pH between and

11.4

9.6 10.6 10.6 11.4

eg) Solution 2: methyl violet is blue pH is

orange IV is yellow pH is

methyl orange is red pH is

Solution 2 has a pH between and

1.6

2.8 3.2

2.8 3.2

Your Assignment: pg 11 #7,8 in workbook

3. Buffers

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 conjugate pairs such as a weak acid and the salt of the conjugate base

buffers

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

buffers can be by the addition of

can be selected by using …this tells you the pH at which the buffer is most useful

pKa = –logKa

acid

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

base

overwhelmedtoo much acid or base

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

H2S – HS- pKa = -log(8.9 × 10-8) = 7.1

CH3COOH – CH3COO- pKa = -log(1.8 × 10-5) = 4.7

HCO3- – CO3

2- pKa = -log(4.7 × 10-11) = 10.3

Example 1Using 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–SASB

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)

Example 2One important buffer is described by the following equilibrium:

H2PO4-(aq) + H2O(l) ⇌ HPO4

2-(aq) + H3O

+(aq)

A sample of this buffer contains 3.2 10-3 mol/L H2PO4-(aq) and

9.8 10-4 mol/L HPO42-

(aq) at equilibrium. Calculate the pH of

the sample. Ka = [HPO42-

(aq)][H3O+

(aq)]

[H2PO4-(aq)]

6.2 10-8 = (9.8 10-4) [H3O+

(aq)]

3.2 10-3 [H3O

+(aq)] = 2.02… 10-7 mol/L

pH = -log[H3O+

(aq)]

= -log(2.02…10-7 mol/L) = 6.69

Your Assignment: pg 21

Q. Society and Technological Connections

using pages 488-491 in your text and your data booklet, describe the following:

1. definition and formation of acid deposition

2. environmental impact and measures being taken by industries to reduce emissions

3. technology of acid deposition

4. social aspects of acid deposition

acid deposition has been an environmental concern for many years

a. dynamic equilibrium reactions are often we use the double arrow to show this relationship...

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