Post on 01-Jan-2016
description
Finding the pH of weak acids
What is a “weak acid”?
What is a “weak base”?
Calculate the pH of a weak acid.
What is pH?What is pH?
pH = - log10 [H+(aq)]
where [H+] is the concentration of hydrogen ions in mol dm-3
to convert pH intohydrogen ion concentration [H+(aq)] = antilog (-pH)
IONIC PRODUCT OF WATER Kw = [H+(aq)] [OH¯(aq)] mol2 dm-6
= 1 x 10-14 mol2 dm-6 (at 25°C)
Calculating pH - Calculating pH - strong acids and alkalisstrong acids and alkalis
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.
Calculate the pH of 0.02M HCl
HCl completely dissociates in aqueous solution HCl H+ + Cl¯
One H+ is produced for each HCl dissociating so [H+] = 0.02M = 2 x 10-2 mol dm-3
pH = - log [H+] = 1.7
WORKEDEXAMPLEWORKEDEXAMPLE
Calculating pH - Calculating pH - strong acids and alkalisstrong acids and alkalis
Strong acids and alkalis completely dissociate in aqueous solution
It is easy to calculate the pH; you only need to know the concentrationonly need to know the concentration.
Calculate the pH of 0.02M HCl
HCl completely dissociates in aqueous solution HCl H+ + Cl¯
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
NaOH completely dissociates in aqueous solution NaOH Na+ + OH¯
One OH¯ is produced for each NaOH dissociating [OH¯] = 0.1M = 1 x 10-1 mol dm-3
The ionic product of water (at 25°C) Kw = [H+][OH¯] = 1 x 10-14 mol2 dm-6
therefore [H+] = Kw / [OH¯] = 1 x 10-13 mol dm-3
pH = - log [H+] = 13
WORKEDEXAMPLEWORKEDEXAMPLE
Calculating pH - Calculating pH - weak acidsweak acids
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
A weak acid is one which only partially dissociates in aqueous solution
Calculating pH - Calculating pH - weak acidsweak acids
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
A weak acid is one which only partially dissociates in aqueous solution
Calculating pH - Calculating pH - weak acidsweak acids
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
[HA(aq)]
A weak acid is one which only partially dissociates in aqueous solution
Calculating pH - Calculating pH - weak acidsweak acids
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
[HA(aq)]
Rearranging (3) gives [H+(aq)]2
= [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka
A weak acid is one which only partially dissociates in aqueous solution
Calculating pH - Calculating pH - weak acidsweak acids
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
[HA(aq)]
Rearranging (3) gives [H+(aq)]2
= [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka
pH = [H+(aq)]
A weak acid is one which only partially dissociates in aqueous solution
Calculating pH - Calculating pH - weak acidsweak acids
A weak acid, HA, dissociates as follows HA(aq) H+(aq) + A¯(aq)(1)
Applying the Equilibrium Law Ka = [H+(aq)] [A¯(aq)] mol dm-3 (2)
[HA(aq)]
The ions are formed in equal amounts, so [H+(aq)] = [A¯(aq)]
therefore Ka = [H+(aq)]2 (3)
[HA(aq)]
Rearranging (3) gives [H+(aq)]2
= [HA(aq)] Ka
therefore [H+(aq)] = [HA(aq)] Ka
pH = [H+(aq)]
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.
A weak acid is one which only partially dissociates in aqueous solution
Calculating pH - Calculating pH - weak acidsweak 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)
WORKEDEXAMPLEWORKEDEXAMPLE
Calculating pH - Calculating pH - weak acidsweak 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)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
WORKEDEXAMPLEWORKEDEXAMPLE
Calculating pH - Calculating pH - weak acidsweak 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)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and then rearrange equation
WORKEDEXAMPLEWORKEDEXAMPLE
Calculating pH - Calculating pH - weak acidsweak 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)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and the rearrange equation
ASSUMPTIONHA 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
WORKEDEXAMPLEWORKEDEXAMPLE
Calculating pH - Calculating pH - weak acidsweak 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)
Dissociation constant for a weak acid Ka = [H+(aq)] [X¯(aq)] mol dm-3
[HX(aq)]
Substitute for X¯ as ions are formed in [H+(aq)] = [HX(aq)] Ka mol dm-3
equal amounts and the rearrange equation
ASSUMPTIONHA 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)] = 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
ANSWER pH = - log [H+(aq)] = 2.699
WORKEDEXAMPLEWORKEDEXAMPLE
6. What is the pH of the following weak acids?a) Ka = 3.7 × 10-8 mol dm-3 [HA] = 0.01 mol dm-3
b) Ka = 5.8 × 10-10 mol dm-3 [HA] = 0.01 mol dm-3
c) Ka = 5.6 × 10-4 mol dm-3 [HA] = 0.01 mol dm-3
d) Ka = 1.74 × 10-5 mol dm-3 [HA] = 0.10 mol dm-3
e) Ka = 5.62 × 10-4 mol dm-3 [HA] = 0.20 mol dm-3
6. a) 4.72 b) 5.62 c) 2.68 d) 2.88 e) 1.97
0.1 mol dm-3 solution of HA has a pH of 5.10. What is its Ka value?
Exercise
Finding the pH of weak acids
What is a “weak acid”? What is a “weak base”? Calculate the pH of a weak acid.
CALCULATING THE pH OF MIXTURESCALCULATING 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 EXCESSSTRONG ACID and STRONG BASE - EITHER IN EXCESS
pH of mixturespH of mixtures
Strong acids and strong alkalis (either in excess)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 therefore
[H+] = Kw / [OH¯] then
pH = - log[H+]
Calculate the pH of a mixture of 25cm3 of 0.1M NaOH is added to 20cm3 of 0.1M HCl
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
WORKEDEXAMPLEWORKEDEXAMPLE
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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
25cm3 of
0.1M NaOH
20cm3 of
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
WORKEDEXAMPLEWORKEDEXAMPLE
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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
The reaction taking place is… HCl + NaOH NaCl + H2O
or in its ionic form H+ + OH¯ H2O (1:1 molar ratio)
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
WORKEDEXAMPLEWORKEDEXAMPLE
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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
The reaction taking place is… HCl + NaOH NaCl + H2O
or in its ionic form H+ + OH¯ 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
5.0 x 10-4
moles of OH¯
UNREACTED
25cm3 of
0.1M NaOH
20cm3 of
0.1M HCl
2.5 x 10-3 moles
2.0 x 10-3 moles
WORKEDEXAMPLEWORKEDEXAMPLE
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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
the volume of the solution is 25 + 20 = 45cm3
WORKEDEXAMPLEWORKEDEXAMPLE
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)
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
the volume of the solution is 25 + 20 = 45cm3
there are 1000 cm3 in 1 dm3
volume = 45/1000 = 0.045dm3
WORKEDEXAMPLEWORKEDEXAMPLE
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
pH of mixturespH of mixtures
Strong acids and alkalis (either in excess)Strong acids and alkalis (either in excess)
WORKEDEXAMPLEWORKEDEXAMPLE
[OH¯] = 5.0 x 10-4 / 0.045 = 1.11 x 10-2 mol dm-3