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Atomic Structure Protons, Neutrons, Electrons: Mass, charge, and location. Isotopes: Same protons, different neutrons. Example: Carbon-12 vs Carbon-14.
Typology: Summaries
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After working through this Factsheet you will be able to:
Generally, the conjugate base of a weak acid will show weak basic character, whilst the conjugate base of a strong acid will not show basic character.
Usually there are two acid/base pairs:
e.g. HCl + NH 3 → NH 4 +^ + Cl− acid 1 base 2 acid 2 base 1
e.g. H 2 SO 4 + H 2 O → H 3 O +^ + HSO 4 − acid 1 base 2 acid 2 base 1
This is ‘chase the H +’ because it is not seen as a separate entity in the equation although it is being donated and accepted.
The 1 and 2 used have no significance except to identify which acid goes with which base.
For practice on conjugate acid-base pairs go to question 1. Exam Hint :-Questions tend to be based on:
At GCSE level you were taught that acids needed water to be present to show their acidic properties, and that they produced H +^ ions. The Br∅∅∅∅∅nsted- Lowry Theory takes this one step further:
Acids are species that donate protons/H+ Bases are species that accept protons/H+
Monobasic acids (e.g. HCl, HNO 3 ) donate one H+, dibasic acids (e.g. H 2 SO 4 ) donate two H+^ and tribasic acids (e.g. H 3 PO 4 ) donate three H+
Acids and bases are linked by the H +^ as this example shows:-
HCl → H+^ + Cl− acid conjugate base
The HCl and Cl-^ are a conjugate acid/base pair.
Note that this new definition of bases includes species such as Cl−^ that would not have been regarded as bases at GCSE, and do not show basic character.
Strong - acid or base which undergoes 100% dissociation as shown by →
e.g. HNO 3 → H+^ + NO 3 − NaOH → Na+^ + OH−
N.B. At A2 level there are only four strong acids: HCl, HNO 3 , H 2 SO 4 and H 3 PO 4 two strong bases: NaOH and KOH
Weak - acid or base that only partially dissociates so it has the equilibrium sign û
e.g. CH 3 COOH û CH 3 COO−^ + H+ NH 4 OH û NH 4 +^ + OH−
N.B. At A2 level organic acids e.g. methonoic, ethonoic, proponoic, etc are weak, but NH 3 (aq) / NH 4 OH is the only weak base.
These terms refer to the ratio of moles to volume i.e. mol dm-3^ of the solutions of acid and bases,
i.e. 2 mol dm-3^ (2M) would be considered ‘dilute’ and 6 mol dm-3^ (6M ) ‘concentrated’.
There is no ‘cut-off’ value between ‘dilute’ and ‘concentrated’ as descriptions of acids and alkalis – and it will not give you any problems.
So it is possible to have a concentrated weak acid and a dilute strong acid.
Chem Factsheet
You will know the term pH from the ‘pH scale’ and its link to Universal Indicator colours (Fig 1).
Acid Base Equilibria I
p H 1 2 3 4 5 6 7 8 9 10 11 12 13 14
strong acid
strong base
weak neutral acid
weak base
At A2 level the numbers 1 – 14 are a logarithmic scale linked to the mathematical expression below.
Definition of pH pH is the minus log to the base 10 of the hydrogen ion concentration.
pH = - log 10 [H+]
For practice on pH and [H+] calculations using the calculator (lg and 10x keys) go to question 2 and 3.
The reason K w is introduced to be able to calculate the pH of bases which lie in the range of pH 8 – 14 at 25o^ C
e.g. Q. What is the pH of 0.1 mol dm-3^ NaOH? A. NaOH → Na +^ + OH− 0.1 → 0.1 + 0. K w = [H +^ ] × [OH−] = 10 - [H +^ ] × 0.1 = 10 - [H +^ ] = 10
− 14 0.1 =^10 -
p H = -log 10 [H+^ ] = - log 10 [10-13^ ] ∴∴∴∴∴ pH = 13
For further practice on calculations involving pH of strong bases go to question 4.
A weak acid only partially dissociates, so the concentration of hydrogen ions, [H +^ ] cannot be calculated from the concentration, mol dm-3, as it can for strong acids.
Since a weak acid only partially dissociates there is an equilibrium set up:
HA û H+^ + A−
and you can write the K c expression,
K c = [H^
For acids, we use K a (where a = acid) instead of K c
K a = [H^
Note that K a is not defined for strong acids , because there will be no undissociated acid left.
Since 1 HA → 1 H+^ + 1 A−^ every time HA dissociates (according to the moles in the equation): [H+^ ] = [A−]
If the degree of dissociation is very small (which is the case for weak acids) then [H +^ ] is very small compared to [HA]. This means we can use:
i.e. we can assume that the concentration of [HA] remains at its original value.
The equation now becomes,
K a = [H
Units = mol dm -
What are logs (or logarithms)? Logs are related to powers: log 10 100 = log 10 102 = 2 log 10 0.1 = log 10 10 -1^ = - So if log 10 (0.63) = -0.201, then 0.63 = 10 -0.
On your calculator, the key for "log 10 " will be called "log" or "lg"
Strong acids undergo 100% dissociation so the concentration (mol dm -3^ or molarity, M ) is all that is needed to calculate its pH. e.g. Q. What is the pH of 0.1 mol dm -3^ HCl? A. HCl → H +^ + Cl− 0.1 → 0.1 + 0. p H = -log 10 [H +] = - log 10 [0.1] ∴∴ ∴∴∴ pH = 1
e.g. Q. What is the pH of 0.05 mol dm -3^ H 2 SO 4? A. H 2 SO 4 → 2H +^ + SO 42 − 0.05 → 0.1 + 0. p H = -log 10 [H +] = - log 10 [0.1] ∴∴ ∴∴∴ pH = 1
Water partially ionises, H 2 O û H+^ + OH -
Writing the K c expression K c = [H
We can rearrange this to give:
Since [H 2 O] is very large it can be considered a constant, we define: K w = [H+^ ] × [OH−] (the w stands for water)
At 25 o^ C, K w = 10-14^ mol dm -
Notice that if [H +^ ] = [OH−] the solution is neutral, then:
∴∴∴∴∴ pH = -log 10 (10-7^ ) = 7
You will see the link between ‘neutral’ and a pH of 7 (although this is only true at 25oC - at higher temperatures K w increases, and so does the pH of water.)