ACIDS AND BASESACIDS AND BASES

Questions may involve any of the following:

description of acids and bases in terms of proton transfer

calculations involving Kw and pH

(Kw = [H3O+][OH] = 10-14 given)

comparison of properties of aqueous solutions of strong and weak acids and bases (properties restricted to conductivity, rate of reaction and pH)

recognition of acid-base properties of ionic compounds that contain only one ion that changes the pH of water.

Acids were originally classified in terms of the following behaviour in aqueous solution:

acids turn blue litmus red

acids have a pH less than 7

acids neutralise bases

acids react with metal carbonates which fizzes and produces CO2 gas.

acids react with active metals to produce hydrogen gas (but not with Cu, Ag or Au).

acids taste sour.

Bases were originally classified in terms of the following behaviour in aqueous solution:

turn red litmus blue,

have a pH greater than 7,

neutralise acids,

taste bitter,

feel soapy

Bronsted-Lowry acids and bases.

Bronsted acids are classified as those substances that donate a proton (H+) to a base, a proton acceptor. Proton donation to a water molecule forms H3O+ (hydronium) ions.

HA(aq) + H2O(l) H3O+(aq) + A (aq)

acid base proton

donor

proton

acceptor

Proton (H+) donation

Bronsted-Lowry acids and bases.

Similarly proton donation from water to a base produces OH- ions.

H2O(l) + B(aq) BH+(aq) + OH(aq)

acid base

proton

donor

proton

acceptor

Proton (H+) donation

Solution is now basic ie pH > 7 because of the formation of OH- ions

Conjugate acids and bases.

Conjugate acid-base pairs.

If 2 species differ by just 1 proton they are classed as a conjugate acid-base pair.

H2O(l) + B(aq) BH+(aq) + OH(aq)

acid 1 base 2 acid 2 base 1

The conjugate acid/base pairs in the above reaction are:

Acid Base

H2O / OH

BH+ / B

Remember water can act as an acid by donating an H+ or a base by accepting an H+

Conjugate acid-base pairs

If two species differ by just one proton they are classed as a conjugate acid-base pair. Examples of acid-base pairs are H2SO4/HSO4

, and NH4+/NH3.

Write the reaction of H2SO4 with water

H2SO4 + H2O HSO4- + H3O+

Acid Base

H2SO4 / HSO4-

H3O+ / H2O

The acid is always the species with the additional proton. It can also be said that HSO4

- is the conjugate base of H2SO4.

The conjugate acid/base pairs in the above reaction are:

Some substances can behave as both acids and bases. They are classed as amphiprotic

eg. H2O or HCO3.

HCO3- + OH H2O + CO3

2

acid1 base2 acid2 base1

HCO3 + H2O OH + H2CO3

base1 acid2 base2 acid1

Is HCO3– acting as an acid or a base in the above

reaction ?

Here HCO3– is acting as base because it is accepting a

proton in the above reaction

Exercises:

1. Complete the acid/base pairs for each of the following:

(a) acids - H2CO3 HSO4 H2O HCl

bases - HCO3 - SO4

2- OH - Cl -

(b) bases - HSO4 SO4

2 H2O CH3COO

acids - H2SO4 HSO4

- H3O+ CH3COOH

Which of the species listed above can be classified as amphiprotic?

Turn to page154 in your lab book – Acids and bases complete the questions

1. Water is very poor conductor of electricity. Explain why.

Pure water ionises to a very slight extent to produce the following concentrations

[H3O+] = 1 x 10 -7 molL-1

and

[OH-] = 1 x 10 -7 molL-1

such concentrations have a very small charge carrying capacity

2. Write a chemical equation to show what happens in a neutralisation reaction.

H3O+ (aq) + OH- (aq) 2H2O

3. Compound Species present

Is solution acidic, basic or neutral ?

Formula of contributing ion

Ethanoic acid

Sodium carbonate

Ammonium chloride

Sodium ethanoate

Sodium nitrate

Potassium chloride

Sodium hydrogen sulfate

H3O+, CH3COO -

acidic

acidic

basic

basic

neutral

neutral

acidic

H3O+

CO32-

NH4+

CH3COO -

_

_

HSO4-

Na+, CO32-,

(OH-) *

NH4+, Cl -

Na+, CH3COO -

Na+, NO3 -

K+, Cl -

Na+, HSO4-

4. Identify by symbol the conjugate base for the following

(a) HF

(b) H2O

(c) HSO3 –

(d) H2CO3

F -

OH –

SO32-

HCO3 -

5. Identify by symbol the conjugate acid for the following

(a) I -

(b) NO3-

(c) HSO4 –

(d) H2O

HI

HNO3

H2SO4

H3O +

6. Hydrogen Chloride and ammonia .

(a) HCl + NH3 NH4+ + Cl

acid1 base2 acid2 base1

(b) Water and the hydrogen sulfate ion

H2O + HSO4 H3O+ + SO4

2-

base1 acid2 acid1 base2

Discussion Questions for Strong and weak acid Expt

1.The balanced equations for the reactions involved are:

2CH3COOH + CaCO3 Ca(CH3COO)2 + H2O + CO2

2HCl + CaCO3 CaCl2 + H2O + CO2

2. How do the rates of reaction compare for acids of the same concentration?

The rates involving the HCl solutions were much greater than those involving CH3COOH solutions

Dissociation constant of water, Kw

Pure water is a very weak conductor of electricity showing it contains a very low concentration of ions. These ions are produced by the dissociation of water molecules as follows:

H2O + H2O H3O+ + OH

Dissociation constant of water, Kw

H2O + H2O H3O+ + OH

This equilibrium lies well to the left so that most of the water is in the molecular form.

Since each dissociation produces both a hydronium ([H3O+]) and a hydroxide (OH-) ion it follows that [H3O+] = [OH].

At 25 oC, the concentration of both species

(ie H3O+ and OH-) = 1 x 10-7 mol L-1.

Kw is the dissociation constant for pure water and it is given by the equation

Kw = [H3O+] x [OH]

= (1 x 10-7 ) x (1 x 10-7 )

= 1 x 10-14

In this course we will not assign Kw units

If an acid is added to pure water then the [H3O+]

increases and at the same time [OH-] decreases so that the product of the two concentrations remains unchanged.

In other words given an H3O+ or OH- concentration we can use the

Kw = [H3O+] [OH-] expression to find an unknown concentration

Remember Kw = 1x 10-14

This concept is very important – and useful

For example if [H3O+] = 2.5 x 10-3 mol L-1 find the [OH-]

O][H1.0x10

3

14

3

14

2.5x101.0x10

[OH] = = 4.0 x 10-12 mol L-1

Rearrange Kw = [H3O+] x [OH] to find the OH- concentration

Complete the following table using the previousKw formula.

[H3O+] [OH] acid or

base?

2.4 x 10-3    

0.0175    

  0.0036  

 5.4 x 10-10

 

4.17 x10-12

5.71 x 10 -13

2.78 x 10 -12

1.85 x 10 -5

acid

acid

base

acid

pH Scale

Because the H+ (or H3O+) concentration is a logrithmic scale and not a linear scale we use the pH scale to express the acidity of a solution

So the pH of an aqueous solution is given by the expression pH = -log10 [H3O+]

pOH Scale

This is the same for an OH- ion concentration too.

Likewise the pOH of an aqueous solution is given by the expression pOH = -log10 [OH-]

pH Scale

The pH of an aqueous solution is given by the expression pH = -log10 [H3O+]

In pure water, [H3O+] = 1 x 10-7 mol L-1,

so pH = -log10 (1 x 10-7) = 7.0

Any solution with a pH = 7.0 is a neutral solution

Since [H3O+] = [OH], ie. neither [H3O+] or [OH] is in excess.

The pH of an acidic solution is < 7, while the pH of a basic solution is >7.

Mathematically pH and pOH are related to each other

Remember in the dissociation of pure H2O

[H3O+] x [OH] = Kw

(1 x 10-7 ) x (1 x 10-7 ) = 1 x 10-14

-log10(1 x 10-7 ) +

-log10 [H3O+] + -log10 [OH-] = -log10[1 x 10-14]

-log10(1 x 10-7 ) = 14

7 + 7 = 14

pH + pOH = 14

This is handy to know!

pH Calculations (remember these!)

pH = -log [H3O+]

[H3O+] = 10-pH

pOH = -log [OH-]

[OH-] = 10-pOH

AND

Example calculation Finding pH

Find the pH of a O.1molL-1 HCl solution

For any strong acid in water the H3O+ concentration = the concentration of the acid

So pH = -log [H3O +]

pH = -log [0.1]

pH = 1

We can also find the pOH using the formula

pH + pOH = 14

ie 1 + 13 = 14 therefore the pOH of the solution is 13

Example Finding H3O+ concentration from pH

Calculate the H3O+ concentration from a pH of 3.2

pH = -log [H3O +]

rearranged:

[H3O+] = 10-pH

[H3O+] = 10-3.2

[H3O+] = 6.31 x 10-4 molL-1

Example Finding OH- concentration from pOH

Calculate the OH- concentration from a pOH of 8.4

pOH = -log [OH-]

rearranged:

[OH-] = 10-pOH

[OH-] = 10-8.4

[OH-] = 3.98 x 10-9 molL-1

[H[H33OO++]] [OH[OH]] pOHpOH pHpH

2.4 x 10-3

0.0175

0.0036

9.3

•Exercise: Complete the following table.

2.624.17x10-

12

1.765.75x10-

13

2.78x10-

12

11.6

5.01x10-10 2.00x10-5 4.7

12.24

2.4

11.38

Complete the pH problems on page 166 lab manual

Calculate the pH of the following [H3O+]:

a) 0.1 molL-1

b) 0.001molL-1

c) 1 x 10 -7 molL-1

d)3 x 10 -7 molL-1

e) 5 x 10 -4 molL-1

pH = -log [H3O+] = - log 0.1 = 1

pH = -log [H3O+] = - log 1x10 -3 = 3

pH = -log [H3O+] = - log 1x10 -7 = 7

pH = -log [H3O+] = - log 3x10-7 = 6.5

pH = -log[H3O+] = -log 5 x10-4 = 3.3

2. Using the relationship:

Kw=[H3O+][OH-] = 10-14,

calculate the pH of solutions with the following [OH-]

a) 0.1 molL-1

b) 0.001molL-1

c) 1 x 10 -7 molL-1

d)3 x 10 -7 molL-1

e) 5 x 10 -4 molL-1

Strong Acids

These are Acids that dissociate (ie react fully) in H2O to form a lot of H3O+ ions in solution to form solutions with very low pH ie 1- 2.

Examples HCl, HNO3 and H2SO4

HCl reacting with water is written as

HCl(aq) + H2O(l) H3O+(aq) + Cl (aq)

How could you tell by just looking at the equation that this is a strong acid?

Weak Acids

These are Acids that don’t dissociate (react) fully in H2O and produce a low number of H3O+ ions in solution to give higher pH values than strong acids eg pH 3 - 5.

Examples CH3COOH (ethanoic acid) HCOOH (methanoic acid) and NH4+ (ammonium ion from a NH4Cl salt)

CH3COOH reacting with water is written as

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq)

How could you tell by just looking at the equation that this is a weak acid?

Demo of Weak and strong acidsDemo of Weak and strong acids

Why the difference in conductivity of the Why the difference in conductivity of the two acids?two acids?

Turn to page 158 in Lab book: The Relative Strength of Acids

Complete the experiment recording your observations in the table on page 159 and then complete the discussion questions on page 159

Discussion Questions pg 159

3. Explain the different rates of reaction obtained when solutions of HCl and CH3COOH of the same concentration, reacted with CaCO3

Rates of these reactions depend on [[H3O+] ions in solution. The stronger acid HCl has completely dissociated and contains a higher initial [H3O+] than the weak acid CH3COOH

Discussion Questions pg 159

This next question involves a very important concept you need to understand

It may pay to remember it

4. 10 mL of 1 molL-1 HCl and 10 mL of 1 molL-1 CH3COOH are mixed separately with NaOH solution.

Which acid will require the greater amount of NaOH for complete reaction? Explain.

4. Both acids will need the amount of NaOH. In the case of CH3COOH adding OH-

ions will move the equilibrium to the RHS

CH3COOH + H2O H3O+ + CH3COO-

This is because H3O+ ions are removed by the addition of OH- ions and eventually, CH3COOH will give up all its potential H+ ions.

SAME

Removed by the addition of OH- ions

further explanation

10 mL of 1 molL-1 HCl and 10 mL of 1 molL-1 CH3COOH are mixed separately with NaOH solution.

Both acids will produce the same amount of H3O+ ions in solution with the NaOH

The (strong acid) HCl produces the H3O+ ions almost immediately and reacts quickly with the OH- ions.

The (weak acid) CH3COOH will produce the same number of H3O+ ions at a slower rate taking more time to react with the OH- ions.

5. How does a strong acid differ from a weak acid chemically?

Strong acids react faster than weak acids because initially there is a greater [H3O+] available for reaction.

Example calculation

Find the pOH of a 0.012molL-1 NaOH solution

For any strong base in water the OH- concentration = the concentration of the base

So pOH = -log [OH-]

pOH = -log [0.012]

pOH = 1.92

We can also find the pH using the formula

pH + pOH = 14

ie 12.08 + 1.92 = 14

therefore the pOH of the solution is 13