pH calculations
A guide for A level students
KNOCKHARDY PUBLISHING
2008
SPECIFICATIONS
pH calculations
INTRODUCTION
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may be used for classroom teaching if an interactive white board is available.
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pH calculations
CONTENTS
• What is pH? - a reminder
• Calculating the pH of strong acids and bases
• Calculating the pH of weak acids
• Calculating the pH of mixtures - strong acid and strong alkali
pH calculations
Before you start it would be helpful to…
• know the differences between strong and weak acid and bases
• be able to calculate pH from hydrogen ion concentration
• be able to calculate hydrogen ion concentration from pH
• know the formula for the ionic product of water and its value at 25°C
What is pH?
pH = - log10 [H+(aq)]
where [H+] is the concentration of hydrogen ions in mol dm-3
to convert pH into
hydrogen ion concentration
IONIC PRODUCT OF WATER
[H+(aq)] = antilog (-pH)
Kw =
[H+(aq)] [OH¯(aq)] mol2 dm-6
=
1 x 10-14 mol2 dm-6 (at 25°C)
Calculating pH - strong acids and alkalis
WORKED
EXAMPLE
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentration.
Calculate the pH of 0.02M HCl
H+ + Cl¯
HCl completely dissociates in aqueous solution
HCl
One H+ is produced for each HCl dissociating so
[H+]
= 0.02M
= 2 x 10-2 mol dm-3
pH
= - log [H+]
= 1.7
Calculating pH - strong acids and alkalis
WORKED
EXAMPLE
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentration.
Calculate the pH of 0.02M HCl
H+ + Cl¯
HCl completely dissociates in aqueous solution
HCl
One H+ is produced for each HCl dissociating so
[H+]
= 0.02M
= 2 x 10-2 mol dm-3
pH
= - log [H+]
= 1.7
Calculate the pH of 0.1M NaOH
Na+ + OH¯
NaOH completely dissociates in aqueous solution
NaOH
One OH¯ is produced for each NaOH dissociating
[OH¯] = 0.1M
The ionic product of water (at 25°C)
therefore
Kw
= [H+][OH¯]
= 1 x 10-1 mol dm-3
= 1 x 10-14 mol2 dm-6
[H+] = Kw / [OH¯] = 1 x 10-13 mol dm-3
pH
= - log [H+] = 13
Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
H+(aq) + A¯(aq)
(1)
Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
Applying the Equilibrium Law
Ka
=
H+(aq) + A¯(aq)
[H+(aq)] [A¯(aq)]
[HA(aq)]
mol dm-3
(1)
(2)
Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
Applying the Equilibrium Law
Ka
H+(aq) + A¯(aq)
=
[H+(aq)] [A¯(aq)]
mol dm-3
(1)
(2)
[HA(aq)]
The ions are formed in equal amounts, so
therefore
[H+(aq)]
=
Ka =
[A¯(aq)]
[H+(aq)]2
[HA(aq)]
(3)
Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
Applying the Equilibrium Law
Ka
H+(aq) + A¯(aq)
=
[H+(aq)] [A¯(aq)]
mol dm-3
(1)
(2)
[HA(aq)]
The ions are formed in equal amounts, so
therefore
[H+(aq)]
=
Ka =
[A¯(aq)]
[H+(aq)]2
[HA(aq)]
Rearranging (3) gives
therefore
[H+(aq)]2
=
[H+(aq)]
=
[HA(aq)] Ka
[HA(aq)] Ka
(3)
Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
Applying the Equilibrium Law
Ka
H+(aq) + A¯(aq)
=
[H+(aq)] [A¯(aq)]
mol dm-3
(1)
(2)
[HA(aq)]
The ions are formed in equal amounts, so
therefore
[H+(aq)]
=
Ka =
[A¯(aq)]
[H+(aq)]2
[HA(aq)]
Rearranging (3) gives
therefore
[H+(aq)]2
=
[H+(aq)]
=
pH
[HA(aq)] Ka
[HA(aq)] Ka
= [H+(aq)]
(3)
Calculating pH - weak acids
A weak acid is one which only partially dissociates in aqueous solution
A weak acid, HA, dissociates as follows
HA(aq)
Applying the Equilibrium Law
Ka
H+(aq) + A¯(aq)
=
[H+(aq)] [A¯(aq)]
mol dm-3
(1)
(2)
[HA(aq)]
The ions are formed in equal amounts, so
therefore
[H+(aq)]
=
Ka =
[A¯(aq)]
[H+(aq)]2
(3)
[HA(aq)]
Rearranging (3) gives
therefore
[H+(aq)]2
=
[H+(aq)]
=
pH
ASSUMPTION
[HA(aq)] Ka
[HA(aq)] Ka
= [H+(aq)]
HA is a weak acid so it will not have dissociated very much. You can assume that
its equilibrium concentration is approximately that of the original concentration.
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
HX(aq)
H+(aq)
+
X¯(aq)
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
Dissociation constant for a weak acid
H+(aq)
HX(aq)
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
Dissociation constant for a weak acid
H+(aq)
HX(aq)
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in
equal amounts and then rearrange equation
[H+(aq)]
=
[HX(aq)] Ka
mol dm-3
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
Dissociation constant for a weak acid
H+(aq)
HX(aq)
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in
equal amounts and the rearrange equation
[H+(aq)]
=
[HX(aq)] Ka
mol dm-3
ASSUMPTION
HA is a weak acid so it will not have dissociated very much. You can assume that its
equilibrium concentration is approximately that of the original concentration
WORKED
EXAMPLE
Calculating pH - weak acids
Calculate the pH of a weak acid HX of concentration 0.1M ( Ka = 4x10-5 mol dm-3 )
HX dissociates as follows
H+(aq)
HX(aq)
Dissociation constant for a weak acid
Ka
=
+
X¯(aq)
[H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
[H+(aq)]
Substitute for X¯ as ions are formed in
equal amounts and the rearrange equation
=
[HX(aq)] Ka
mol dm-3
ASSUMPTION
HA is a weak acid so it will not have dissociated very much. You can assume that its
equilibrium concentration is approximately that of the original concentration
[H+(aq)]
ANSWER
=
0.1 x 4 x 10-5 mol dm-3
=
4.00 x 10-6
mol dm-3
=
2.00 x 10-3
mol dm-3
pH = - log [H+(aq)]
=
2.699
CALCULATING THE pH OF MIXTURES
The method used to calculate the pH of a mixture of an acid and an alkali depends on...
• whether the acids and alkalis are STRONG or WEAK
• which substance is present in excess
STRONG ACID and STRONG BASE - EITHER IN EXCESS
pH of mixtures
Strong acids and strong alkalis (either in excess)
1. Calculate the initial number of moles of H+ and OH¯ ions in the solutions
2. As H+ and OH¯ ions react in a 1:1 ratio; calculate unreacted moles species in excess
3. Calculate the volume of solution by adding the two original volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess the ion in mol dm-3
6. Convert concentration to pH
If the excess is
H+
pH = - log[H+]
If the excess is
OH¯
pOH = - log[OH¯]
then
pH + pOH = 14
or use Kw = [H+] [OH¯] = 1 x 10-14 at 25°C
[H+] = Kw / [OH¯]
pH = - log[H+]
then
therefore
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
25cm3 of
20cm3 of
0.1M NaOH
0.1M HCl
2.5 x 10-3
moles
2.0 x 10-3
moles
moles of OH ¯
= 0.1 x 25/1000
= 2.5 x 10-3
moles of H+
= 20 x 20/1000
= 2.0 x 10-3
WORKED
EXAMPLE
pH of mixtures
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
25cm3 of
20cm3 of
0.1M NaOH
0.1M HCl
2.5 x 10-3
moles
2.0 x 10-3
moles
The reaction taking place is…
or in its ionic form
HCl + NaOH
H+ +
OH¯
NaCl
H2O
+
H2O
(1:1 molar ratio)
WORKED
EXAMPLE
pH of mixtures
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
25cm3 of
20cm3 of
0.1M NaOH
0.1M HCl
2.5 x 10-3
moles
5.0 x 10-4
moles of OH¯
2.0 x 10-3
moles
The reaction taking place is…
HCl + NaOH
H+ +
or in its ionic form
UNREACTED
OH¯
NaCl
+
H2O
H2O
(1:1 molar ratio)
2.0 x 10-3 moles of H+ will react with the same number of moles of OH¯
this leaves
2.5 x 10-3
-
2.0 x 10-3 = 5.0 x 10-4 moles of OH¯ in excess
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
the volume of the solution is 25 + 20 = 45cm3
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
the volume of the solution is 25 + 20 = 45cm3
there are 1000 cm3 in 1 dm3
volume = 45/1000 = 0.045dm3
pH of mixtures
WORKED
EXAMPLE
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
[OH¯]
= 5.0 x 10-4 / 0.045
= 1.11 x 10-2 mol dm-3
WORKED
EXAMPLE
pH of mixtures
Strong acids and alkalis (either in excess)
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
1. Calculate the number of moles of H+ and OH¯ ions present
2. As the ions react in a 1:1 ratio, calculate the unreacted moles of the excess species
3. Calculate the volume of the solution by adding the two individual volumes
4. Convert volume to dm3 (divide cm3 by 1000)
5. Divide moles by volume to find concentration of excess ion in mol dm-3
6. As the excess is OH¯ use
or
[OH¯]
[H+]
pH
= 5.0 x 10-4 / 0.045
= Kw / [OH¯]
= - log[H+]
pOH = - log[OH¯]
Kw
= [H+][OH¯]
then
pH + pOH = 14
so [H+] = Kw / [OH¯]
= 1.11 x 10-2 mol dm-3
= 9.00 x 10-13 mol dm-3
= 12.05
Kw = 1 x 10-14
mol2 dm-6 (at 25°C)
REVISION CHECK
What should you be able to do?
Calculate pH from hydrogen ion concentration
Calculate hydrogen ion concentration from pH
Write equations to show the ionisation in strong and weak acids
Calculate the pH of strong acids and bases knowing their molar concentration
Calculate the pH of weak acids knowing their Ka and molar concentration
Calculate the pH of mixtures of strong acids and strong bases
CAN YOU DO ALL OF THESE?
YES
NO
You need to go over the
relevant topic(s) again
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pH calculations
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