EQ acid-base Notes 2009

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

_{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 thefastest as it is the

reaction

forward slow down reverse

Rate

Time 0

forward reaction

reverse reaction Dynamic

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 rates

equal

observable (macroscopic) properties

constant

closed

_{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%

3. to the right (percent rxn )

quantitative >99%

A + B ⇌ C + D

>99%

or

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 H_2(g) and I_2(g) (simple molecules and takes place in gas phase no solvent

necessary!)

when different combinations of H_2(g), I_2(g), and HI_(g) were mixed and the concentrations

measured, it was discovered that all

cases

_{even though the equilibrium [ ] are} , the

different end quotient

equilibriumwas reached in

_{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

_{this law can be expressed mathematically:} For the reactionaA + bB ⇌ cC + dD

The law

is: K_c = [C]c [D]d [A]a [B]b

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

a, b, c, d = coefficients

equilibrium constant reactants

 _{is constant for a reaction at a given} …if you change the

temperature, K_c also changes

it is common to ignore the units for K_c and list it only as a numerical value (since depends on the powers of the various [ ] terms)

K_c

when determining K_c use only the species that are in or

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

temperature

the the value of K_c, the greater the tendency for the reaction to favor the

K_c 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 higher

if K_c is then the reaction

is favoured

if K_c is then the reaction is favoured

greater than 1, products

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) K_c = [NO₂(g)]2

[NO(g)]2[O

Example 2

Write the equilibrium law for the following reaction:

CaCO₃(s) ⇌ CaO(s) + CO₂(g) K_c = [CO₂(g)]

*** do not include solids in

K_c

Example 3

Write the equilibrium law for the following reaction:

2 H₂O(l) ⇌ 2 H₂(g) + O₂(g) K_c =

[H₂(g)]2[O

2(g)]

[H2O(l)] 2

*** do not include liquids in K_c if there is anything that is

aqueated(aq) as this also contains water by definition! You cannot tell which water is which---look at

Example 4

Phosphorus pentachloride gas can be

decomposed into phosphorus trichloride gas and chlorine gas.

a) Write the equilibrium law for this reaction.

K_c = [PCl₃(g) ][Cl₂(g)] [PCl₅(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 K_c.

K_c = [PCl₃(g) ][Cl₂(g)] [PCl₅(g)]

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

Example 5

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

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

K_c = [SO₃(g) ]2____

[SO₂(g)]2[O

2(g)] 85.0 = [SO₃(g) ]2

(0.500)2(0.500)

[SO_3(g) ]2 = 10.625

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

Example 1

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

Concentratio n

(mol/L)

Time (s) 0

O₂(g) SO₂(g )

SO₃(g )

10 20 30 75

50 25

K_c = [SO₃]2 [SO₂]2[O

2] = (75)2

(50)2(25) = 0.090

E. Le Châtelier’s Principle

 _{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 principle

equilibrium

change in property of the system,

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

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

reactants to the

products

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)

_{a change in temperature is the only stress that} the value of K_c!!!!!!! will change

if the shift is towards the side, K_c will

product increase

if the shift is towards the side, K_c will

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

_{if the number of moles are the on both} sides of the reaction, a change in volume

(pressure) has no effect

_{changes in volume and pressure have} on the value ofK_c no effect

_{in many equilibrium systems, the reactants will} have a different colour than the products

4. Colour Changes

Example

Use the following reaction to predict the

equilibrium shift and resulting colour change when the stresses are applied _{2 CrO}

42(aq) + 2 H3O+(aq) ⇌ 2 Cr2O72(aq) + 3 H₂O(_)

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

_{all of the changes that can happen to systems} in equilibrium can be shown graphically:

Example

State what change to the equilibrium takes place at each of the labelled parts of the

Concentration (mol/L)

Time (min) NH_3(g)

N_2(g)

H_2(g)

A B C D

Manipulations of An Equilibrium System

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) + energy

Equilibrium Time Stress A

B C D

addition of H₂(g)

addition of inert gas, addition of catalyst decrease in volume

F. ICE Tables

_{we can use a table set-up to calculate the}

equilibrium concentrations and/or K_c for any system

_{you must be able to calculate all} before you can use the equilibrium law

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 35C, the final concentration of H₂(g) is 0.214 mol/L. Find the value for K_c.

2.00 mol/L 0 0

0.214 mol/L

2 HI(g) ⇌ H₂(g) + I₂(g)

Change

K_c = [H₂(g)][I₂(g)] [HI(g)]2

Example 2

In a 500 mL stainless steel reaction vessel at 900C, 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. K_c for this reaction is 4.20 at 900C. Calculate the concentration of each

substance at equilibrium. _{CO(g) + H}

2O(g) ⇌ CO2(g) + H₂(g)

2.00/0.5L = 4.00 mol/ L

0 0

I

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

K_c = [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

8.05  2.05x = x

8.05 = 3.05x

[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/L 4.00 – 2.69 =

Example 3

Gaseous phosphorus pentachloride decomposes into gaseous phosphorus trichloride and chlorine gas 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. _PCl

5(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

K_c = [PCl_3(g)] [Cl_2(g)]

[PCl_5(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 K_c is very small, the equilibrium doesn’t lie very far to the right and x is a very small 

1.00  10-3 = (x)(x)

(1.00 ) x2 = 1.00  10-3 x

1.00

x = 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

[PCl_5(g)]_eq = 1.00 mol/L – 0.0316 mol/L = 0.967 mol/L

Example 4

Gaseous NOCl decomposes to form gaseous NO and Cl₂. At 35C 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) + Cl_2(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

K_c = [NO_(g)]2[Cl

2(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

G. Ionization of Water

_{the equilibrium of water can be written as} follows:

_{the equilibrium law is:}

_{the equilibrium constant for water is} designated as

K_c = [H₃O+(aq) ][ OH

-(aq)]

K_w

 H₃O+(aq) OH-(aq)

+ H₂O(l) +

H₂O(l)

_{at 25}_{C, neutral water has} [H+

(aq)] = [OH-(aq)] = 1.0  10-7 mol/L

 K_w =

= (on pg 3 of Data Booklet) (1.0  10-7) (1.0  10-7)

K_w is constant and therefore can be used to determine the or the [ OH-(aq)]

[H₃O+(aq) ]

eg) if [H₃O+(aq)] = 1.0  104 mol/L then [OH_{(aq)] = 1.0} ₁₀14 =

1.0  104

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

Try These:

1. [H₃O+(aq)] = 1.0  109 mol/L [OH_{(aq)] =}

2. [H₃O+(aq)] = 1.0  101 mol/L [OH_{(aq)] =}

3. [H₃O+(aq)] =

[OH_{(aq)] = 1.0}  ₁₀2 mol/L

4. [H₃O+(aq)] = 6.3  109 mol/L [OH_{(aq)] =}

5. [H₃O+(aq)] = 8.1  103 mol/L [OH_{(aq)] =}

6. [H₃O+(aq)] =

[OH_{(aq)] = 2.8}  ₁₀7 mol/L

1.0  105 mol/L basic

1.0  1013 mol/L acidic

1.0  1012 mol/L

basic

1.6  106 mol/L basic

acidic

basic

1.2  1012 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 place

pH value sig digs

[H₃O+(aq)]

[H₃O+(aq)] = 10-pH

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

Example 1

Find the pH of a solution where the [H₃O+(aq)] = 4.7  10-11 _mol/L.

Example 2

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

pOH =  log[OH(aq)] =  log(2.4  103 ) = 2.619…

Example 3

Calculate 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 4

Calculate 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 g

M = 74.10 g/mol n = m  M

= 10.3 g  74.10 g/mol

= 0.139…mol

v = 0.500 L

n = 0.139…mol  2/1 = 0.278…mol

C = n  V

= 0.278…mol  0.500L

Example 4 (continued)

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

_{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

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!}

_{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+) _acid

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

eg) H₂O, HCO₃_{, HSO}

_{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 acid base acid conjugate base

base conjugate acid

_{the an acid, the its}

conjugate base

stronger weaker

_{the an acid, the its}

conjugate base

weaker stronge

_{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

more

_{acids that ionize in water}

to form H₃O+ (aq)

_{percent rxn =} 1. Strong Acids

100 %

quantitatively

the bigger the K_a (K_c for acids) the more the

product are favoured

s

_{top 6 acids on the table (pg 8-9 in Data Book)}

have a very large K_a

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

_{when calculating pH, the} so use

[SA] = [H₃O+

(aq)] pH = -log[H₃O+

(aq)]

Example

What is the pH of a 0.500 mol/L solution of HNO_3(aq)[H?

3O+(aq)] = [HNO3(aq)] = 0.500 mol/L

pH = -log[H₃O+

(aq)]

_{a weak acid is one that}

_{most ionize}

2. Weak Acids

K_a 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%

K_a value [WA]

100%

the K_a law is an and is devised the same way we did

eg) HA_(aq) + H₂O_(l) ⇌ H₃O+

(aq) + A-(aq)

K_a = [H₃O+

(aq)][A-(aq)] [HA_(aq)]

equilibrium law K_c

_{you will be required to figure out the}

before you can calculate the pH[H3O

+

_{you have the and the value but}

you don’t have the

K_a = [H₃O+

(aq)]2 [HA_(aq)]

_{since the mole ratio for}

is , they have the same [ ] (this is a !)

[WA] K_a

[A

-(aq)]

[H₃O+

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

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

(aq)]

[H₃O+

Example 1

What 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)

Example 2

What is the pH of a 1.0 mol/L acetic acid solution?

CH₃COOH_(aq) + H₂O_(l) ⇌ H₃O+

(aq) + CH3COO -(aq) _[H

3O+(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)

Example 3

A 0.25 mol/L solution of carbonic acid has a pH of 3.48. Calculate K_a.

H₂CO_3(aq) + H₂O_(l) ⇌ H₃O+

(aq) + HCO₃

-(aq)

[H₃O+

(aq)] = 10-pH = 10-3.48

= 3.31…  10-4 mol/L

K_a = [H₃O +

(aq)]2 [H₂CO_{3 (aq)}]

= (3.31…x10-4 mol/L)2 0.25 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

Example 1

Calculate 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

Example 2

The 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

_{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

 _{where x is the} number of ions (think about the dissociation equation!)

Example

Calculate the pH of a 0.0600 mol/L solution of Ca(OH)_2(aq).

[OH

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

= x[Ca(OH)_2(aq)]

= 2(0.0600 mol/L) = 0.120 mol/L

[OH

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

pOH =

log(0.120)

= 0.9208… pH = 14 –

0.9208…

_{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)

K_b

K_b = [BH+

(aq)][OH-(aq)]

you can calculate K_b using for the K_a  K_b = K_W = 1.00  1014

eg) Calculate K_b for SO₄2

(aq). K_b = K_W

K_a

= 1.00  1014 1.0  102 = 1.0  1012 mol/L

[OH

-(aq)] = (Kb)([WB])

once you have K_b, you can then solve for

[OH

(aq)] using the equilibrium law (just like with weak acids!) _K

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

[WB_(aq)] K_b = [OH

-(aq)]2

[WB_(aq)]

Example

Find the pH of a 15.0 mol/L NH_3(aq) solution.

K_a = 5.6  10-10 mol/L K_b = K_w

K_a

= 1.00  10-14

5.6  10-10

= 1.78…  10-5 mol/L

NH_3(aq) + H₂O_(l) ⇌ NH₄+

[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}

_{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

_{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)

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 1

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

Na+ (aq)

List: OH

-(aq) CH3COOH(aq) H2O(l) A

B A/B

SB SA

OH

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

Example 2

Predict the acid-base reaction when ammonia is mixed with HCl_(aq).

NH_3(aq

List: H₃O+

(aq)

Cl

-(aq) H2O(l)

B _{B A/B}

SA SB

NH_3(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.}

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

1. H₂SO_4(aq)

2. HOOCCOOH_(aq) 3. HCOOH_(aq)

4. CH₃COOH_(aq) 5. H₂PO₄

-(aq) 6. NH₄+

(aq)

polyprotic polyprotic

polyprotic 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

PO₄

3-(aq) HPO₄

2-(aq), H2PO4-(aq), H3PO4(aq)

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

bases: 1. HSO₄

-(aq) 2. H₂PO₄

-(aq) 3. HPO₄

2-(aq) 4. HCO₃

-(aq) 5. H₂O_(l)

monoprotic acid; monoprotic base

polyprotic acid; monoprotic base

monoprotic acid; monoprotic base

monoprotic acid; polyprotic 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

Example 1

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

K+ (aq)

List: OH

-(aq) HOOCCOOH(aq) H2O(l) A

B A/B

SB SA

OH

-(aq) + HOOCCOOH(a q)

⇌ H2O(l) + HOOCCOO-(aq)

HOOCCOO -(aq) A/B

SA

OH

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

OOCCOO

 _{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)

H₃O+

(aq) + HPO42-(aq) H2O(l) + H2PO4-(aq) H₃O+

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

A/B _A

SB _SA

A/B Na+

(aq)

List: HPO₄

2-(aq) I-(aq) H3O+(aq) H2O(l) B

H₂PO₄ -(aq) A/B

SB

H₃PO_4(aq) A

2 H₃O+

(aq) + HPO42-(aq) 2 H2O(l)+ H3PO4(aq) Example 2

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

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

 _{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

 _{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)}

 _{you need to be able to interpret pH curves:} 1. Strong Monoprotic with Strong

Monoprotic _{ 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

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

pH 0 7 14 volume pH 0 7 14 volume

WA titrated with SB WB titrated with SA

EP

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

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 _blue

eg) litmus

when it reacts with anacid or baseand are usually themselves

 _{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

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 solutions containing}

relatively large amounts of such as a and the

buffers

conjugate pairs

weak acid _{salt of the conjugate 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

pK_a = -log(8.9 × 10-8) = 7.1

pK_a = -log(1.8 × 10-5) = 4.7

pK_a = -log(4.7 × 10-11) = 10.3

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

pK_a = –logK_a

H₂S – HS

-CH₃COOH – CH₃COO

2-_{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 base acid

Example

Using 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 – B A B A/B

S A S

B

CH₃COO

b) a small amount of NaOH_(aq) is added

CH₃COOH_(aq) Na+

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

A – B B A/B

S A

S B

CH₃COO -(aq) + OH

-(aq) ⇌