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Acid-Base Equilibria I - pH, Kw and Ka
January 2002 Number 25
1
ChemFactsheet
To succeed in this topic you need to be able to:
understand the writing of Kc expressions and their units (covered
in Factsheet No.21);
use your calculator to convert numbers into ‘logarithms to the
base 10 (log10) using the lg key;
turn log10 values back into numbers using the inverse of log10 i.e.
the 10x key (or second function above the lg key).
After working through this Factsheet you will be able to:
use the A2 level definitions of acids and bases (Br
nsted-Lowry
theory) and identify conjugate acid/base pairs;
define pH and perform calculations using its mathematical
expression;
understand how to use the ionic product of water, Kw, to calculate
the pH of a base;
understand the difference between the terms weak, strong,
concentrated and dilute as applied to acids and bases;
write Ka expressions for weak acids and perform calculations
using these expressions.
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 NH4++ Cl
acid 1 base 2 acid 2 base 1
e.g. H2SO4+H
2OH3O++ HSO4
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:
learning the definitions for acids/bases, pH, pOH, K
w
and K
a
performing calculations on them
The calculations all involve using the lg and 10
x
keys on the calculator
and being able to input ‘powers to the 10’ e.g. 1.2 x 10
-3
, 5 x 10
-6
- this
usually involves using the EXP key - eg 1.3 EXP -2 = 1.3 x 10
-2
(be
careful where you put the minus!)
* Many marks are lost in examination questions on this topic by
candidates who cannot use their calculators for these purposes. you
must practice this until you are competent!*
Br
nsted-Lowry Theory
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, HNO3) donate one H+, dibasic acids (e.g. H2SO4)
donate two H+ and tribasic acids (e.g. H3PO4) 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.
Using acid and base terms
Strong - acid or base which undergoes 100% dissociation as shown by
e.g. HNO3 H+ + NO3
NaOH Na+ + OH
N.B. At A2 level there are only
four strong acids: HCl, HNO3, H2SO4and H3PO4
two strong bases: NaOH and KOH
Weak - acid or base that only partially dissociates so it has the equilibrium
sign ž
e.g. CH3COOH ž CH3COO + H+
NH4OH ž NH4+ + OH
N.B. At A2 level organic acids e.g. methonoic, ethonoic, proponoic, etc
are weak, but NH3(aq) / NH4OH is the only weak base.
Concentrated/dilute
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.
pf3

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Acid-Base Equilibria I - pH, K

w

and K

a

January 2002 Number 25

C

hem

F

actsheet

To succeed in this topic you need to be able to:

  • •••• understand the writing of K c expressions and their units (covered in Factsheet No.21);
  • ••••^ use your calculator to convert numbers into ‘logarithms to the base 10 (log 10 ) using the lg key;
  • •••• turn log 10 values back into numbers using the inverse of log 10 i.e. the 10 x^ key (or second function above the lg key).

After working through this Factsheet you will be able to:

  • ••••^ use the A2 level definitions of acids and bases (Br ∅∅∅∅∅ nsted-Lowry theory) and identify conjugate acid/base pairs;
  • •••• define pH and perform calculations using its mathematical expression;
  • •••• understand how to use the ionic product of water, K w , to calculate the pH of a base;
  • •••• understand the difference between the terms weak, strong, concentrated and dilute as applied to acids and bases;
  • •••• write K a expressions for weak acids and perform calculations using these expressions.

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:

  • learning the definitions for acids/bases, pH, pOH, Kwand Ka
  • performing calculations on them The calculations all involve using the lg and 10x^ keys on the calculator and being able to input ‘powers to the 10’ e.g. 1.2 x 10 -3, 5 x 10-6^ - this usually involves using the EXP key - eg 1.3 EXP -2 = 1.3 x 10-2^ (be careful where you put the minus!)
  • Many marks are lost in examination questions on this topic by candidates who cannot use their calculators for these purposes. you must practice this until you are competent!*

Br ∅∅∅∅∅ nsted-Lowry Theory

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.

Using acid and base terms

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.

Concentrated/dilute

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

pH

You will know the term pH from the ‘pH scale’ and its link to Universal Indicator colours (Fig 1).

Acid Base Equilibria I

Fig 1. pH scale

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.

The pH of weak acids and K a

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^

+] × [A−]

[HA]

For acids, we use K a (where a = acid) instead of K c

K a = [H^

+] × [A−]

[HA]

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:

[HA] – [H +] ≈ [HA]

i.e. we can assume that the concentration of [HA] remains at its original value.

The equation now becomes,

K a = [H

+ ] 2

[HA]

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"

The pH of strong acids

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

The ionic product of water, K w

Water partially ionises, H 2 O û H+^ + OH -

Writing the K c expression K c = [H

+] × [OH−]

[H 2 O]

We can rearrange this to give:

K c × [H 2 O] = [H+^ ] × [OH−]

Since [H 2 O] is very large it can be considered a constant, we define: K w = [H+^ ] × [OH−] (the w stands for water)

This gives us K c × [H 2 O] = K w

At 25 o^ C, K w = 10-14^ mol dm -

Notice that if [H +^ ] = [OH−] the solution is neutral, then:

[H]^2 = K w = 10-14^ , so [H] = √ 10 -14^ = 10-

∴∴∴∴∴ 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.)