5.3 Acids, Bases and Buffers

Question 1

Brønsted–Lowry acid

Answer 1

a species that can donate a proton
HCl(aq) → H+(aq) + Cl-(aq)
When an acid is added to water, it releases H+ ions (protons) into solution

Question 2

Brønsted–Lowry base

Answer 2

a species that can accept a proton

NH3 (aq) + H+(aq) → NH4+(aq)

Question 3

common base

Answer 3

Common bases are metal oxides and metal hydroxides
e.g. MgO NaOH
Ammonia is also a base

Question 4

monoprotic acids

Answer 4

Many acids are called monoprotic acids. This means that they only donate one mole of protons per mole of acid;
e.g. HCl, HNO3, CH3COOH

Question 5

diprotic acids

Answer 5

Some acids are diprotic acids. This means that they can donate two moles of protons per mole of acid; e.g. H2SO4, HOOCCOOH.
H2SO4(aq) + 2H2O(l) -> 2H3O+(aq) + SO42-(aq

Question 6

triprotic acids

Answer 6

Triprotic acids donate three moles of protons per mole of acid; e.g. H3PO4.

Question 7

ammonia

Answer 7

Ammonia is a gas that dissolves in water to form a weak alkaline solution.
Dissolved NH3 reacts with water
NH3(aq) + H2O ⇌ NH4+(aq) + OH-(aq)
acts as a base because N in NH3 is accepting a proton

Question 8

alkali

Answer 8

An alkali is any chemical compound that gives a solution of pH greater than 7.0 when dissolved in water.
Sodium hydroxide NaOH
Potassium hydroxide KOH
Sodium hydroxide Mg(OH)2
is a soluble base that releases OH- ions in aqueous solutions

Question 9

conjugate acid-base pairs

Answer 9

Two species differing by H

Question 10

conjugate acid

Answer 10

is a species formed from a Brønsted-Lowry base by the addition of a proton.

Question 11

conjugate base

Answer 11

is a species formed from a Brønsted-Lowry acid by the loss of a proton

Question 12

amphoteric compounds

Answer 12

is a molecule or ion that can react both as an acid as well as a base.
e.g. amino acids, which have amine and carboxylic acid groups, and self-ionisable compounds such as water
HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)
base 1 acid 1

Question 13

Kw

Answer 13

constant called the ionic product for water, Kw, is defined as:
Kw = [H+][OH–]
units= mol2 dm–6

Question 14

what is meant by the strength of an acid?

Answer 14

the extent of dissociation

Question 15

strong acids

Answer 15

A strong acid has a low pH (usually 0 or 1). This means that the concentration of H+ is high. This is because the acid is fully dissociated into its ions.

Question 16

weak acid

Answer 16

A weak acid has a higher pH (but still less than 7). This means that the concentration of H+ is lower than for a strong acid. This is because the acid is not fully dissociated into its ions.

Question 17

find pH of 250 cm3 of an aqueous solution containing 3.65 g hydrochloric acid

Answer 17

n(HCl)= 3.65g / (1 + 35.5) = 0.1 mols
conc(HCl)= 0.1 / (250/1000) = 0.4 mols dm-3
pH = –log10 [0.4]
pH=0.398

Question 18

weak acid and Ka

Answer 18

HA(aq) + H2O(l) H3O+(aq) + A–(aq)
HA(aq) ⇌ H+(aq) + A–(aq)

Ka= [H+][A-]
[HA]
unit= mol dm–3

Question 19

approximation 1

Answer 19

[H+(aq)] = [A−(aq)]
[H+] = square root of Ka X [HA]

Question 20

approximation 2

Answer 20

The equilibrium [HA] is smaller than the undissociated [HA]
The dissociation of weak acids is small you can assume [HA]start &raquo_space; [H+] and you can neglect decrease in concentration of the HA dissociation
assume [HA]start&raquo_space; [H+] and equilibrium [HA] = undissociated [HA]

Question 21

limitations to approximation 1

Answer 21

At 25 C [H+] from dissociation of water = 10-7. If the pH > 6 then [H+] from the dissociation of water will be significant compared with the dissociation of HA.
Approximation breaks down for very weak acids or diluted solutions

Question 22

limitations to approximation 2

Answer 22

This approximation holds for weak acids with small Ka values. It breaks down when [H+] becomes significant and there is a real difference between [HA]eqm and [HA]start – [H+]eqm
This approximation is not justified for stronger weak acids with Ka> 10-2 mol dm-3 and for very dilute solutions

Question 23

Ka and pKa

Answer 23

The extent to which an acid dissociates is shown by the value of Ka. The larger the value of Ka, the stronger the acid.
if Ka is less than 1, the acid is weak
if Ka is greater than 1, the acid is strong
The values of Ka span a wide range. To make them easier to interpret, a new term pKa is used
pKa = –log10 Ka

Question 24

calculating Ka from pKa

Answer 24

pKa = –log10 Ka
so
Ka = 10–pka

Question 25

buffer

Answer 25

A system that minimises pH changes on addition of small amounts of an acid or a base.
It will remove any H+ or OH- that you might add to it, otherwise the pH will change

Question 26

composition of buffers

Answer 26

They are composed of weak acids and their conjugate bases:
HA ⇌ H+ + A-

HA and A- act as two independent reservoirs that are capable of reacting with added alkalis or acids, respectively:
The buffer is most effective when [HA] = [A-].
The buffer is no longer effective when one of the components is completely used up.

The pH of a buffer does vary slightly when reacting with any added species, so shouldn’t be considered totally constant.

Question 27

how buffers work

Answer 27

As buffers have two independent reservoirs, a reaction with one will cause a shift in the equilibrium to negate the change
-> adding alkali
HA ⇌ A- + H+

Question 28

addition of an acid to a buffer

Answer 28

[H+] increases
Conjugate base (A-) reacts
Equilibrium shifts left
[H+] removed

Question 29

addition of an alkali to a buffer

Answer 29

[OH-] increases
H+ reacts (H+ + OH- → H2O)
Equilibrium shifts right
HA dissociates restoring [H+]

Question 30

forming buffers- addition

Answer 30

Adding solutions of a weak acid and its conjugate salt.
As the dissociation of weak acids is very low, the concentration of the acid can assume to remain the same.
As salts are ionic they completely dissociate in solution so provide a concentration for the conjugate base.

Question 31

forming buffers- neutralisation

Answer 31

Partial neutralisation of a weak acid.
Add an alkali, e.g. NaOH(aq) to excess weak acid.
This will neutralise some of the acid to form the conjugate base and leaving some acid left over.

Question 32

making a buffer from an acid and a base

Answer 32

When a small quantity of a strong base is added to a weak acid, a buffer solution is formed.
Some of the weak acid reacts with the strong base to form the salt of the weak acid. For example:
CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)
If the acid is in excess, the solution contains some of the salt of the weak acid, along with some of the weak acid, so the solution is a buffer.

Question 33

method 1 of calculating pH of buffers

Answer 33

weak acid and a salt 
-calculate the concentration of weak acid, HA 
-calculate the concentration of the salt, A- 
use: 
[H+] = Ka X [HA]
               [A-]
then use: 
pH=-log10[H+]

Question 34

method 2 of calculating pH of buffer

Answer 34

n weak acid 
n base 
n acid in excess 
n salt formed 
total volume 
conc of acid [HA]
conc of salt [A-] 
use rearrangement of formula for Ka
[H+] = ka X [HA] 
              [A-]
then use: 
pH=-log10[H+]

Question 35

how does a buffer act when added to acid

Answer 35

Equilibrium shifts to the left
CH3COOH ⇌ CH3COO– + H+
[CH3COOH] increases and [CH3COO¯] decreases

Question 36

how does a buffer act when added to an alkali

Answer 36

Equilibrium shifts to the right
CH3COOH ⇌ CH3COO– + H+
[CH3COOH] decreases and [CH3COO¯] increases