Acid base equilibria
Starter
44.0g of ethyl ethanoate was mixed with 36.0g of
water containing HCl as a catalyst and allowed
to reach equilibrium over several days. The
equilibrium mixture was then made up to 250cm3
with pure water. A 25.0 cm3 sample of diluted
mixture was titrated with 1.00mol dm-3 NaOH.
After allowing for the acid catalyst present, the
ethanoic acid in the equilibrium mixture was
found to require 29.5 cm3 of NaOH for
neutralisation. Find the Kc for the reaction:
CH3COOC2H5(l)+ H2O(l) CH3COOH(l) + C2H5OH(l)
The equation of for the neutralisation reaction is:
CH3COOH + NaOH  CH3COONa + H2O
Starter
44.0g of ethyl ethanoate was mixed with 36.0g of
water containing HCl as a catalyst and allowed
to reach equilibrium over several days. The
equilibrium mixture was then made up to 250cm3
with pure water. A 25.0 cm3 sample of diluted
mixture was titrated with 1.00mol dm-3 NaOH.
After allowing for the acid catalyst present, the
ethanoic acid in the equilibrium mixture was
found to require 29.5 cm3 of NaOH for
neutralisation. Find the Kc for the reaction:
CH3COOC2H5(l)+ H2O(l) CH3COOH(l) + C2H5OH(l)
The equation of for the neutralisation reaction is:
CH3COOH + NaOH  CH3COONa + H2O
= 0.249 (no units)
Learning Objectives
• Define pH
• Calculate pH and H+ ion concentration for
strong acids
• Define the ionic product for water
• Calculate the pH and the OH- ion
concentration for strong bases
Defining pH
pH = -log10[H+]
100 is 102 the logarithm to base 10 of 100 is 2
1000 is 103 the logarithm to base 10 of 1000 is 3
2 is 100.3010 the logarithm to base 10 of 2 is 0.3010
The logarithm to base 10 of a number is the power
you have to raise 10 to in order to equal that
number
Using your calculator
• To find log102:
[2] [log][=] [0.301029995]
OR
[log] [2][=] [0.301029995]
To unlog use the 10x often found above the
log button and using the shift key.
Calculating pH from hydrogen ion
concentration
What is the pH if [H+] = 0.1 mol dm-3
Find log10(0.1) = -1
NB: pH = -log10[H+]
“-(-1) is 1”
pH = 1
Calculating hydrogen ion
concentration from pH
NB: hydrogen ion concentrations will almost
always be less than 1 mol dm-3
pH is 2.80 what is the hydrogen ion concentration?
pH = -log10[H+]
Rearranging this: log10[H+] = -pH
log10[H+] = -2.80
[shift][log][-][2.8][=][1.58 x 10-3]
= 1.58 x 10-3 mol dm-3
Questions
Convert the following hydrogen ion
concentrations (all in mol dm-3) into pHs
a) 0.0100
b) 0.0250
c) 3.00 x 10-4
d) 1.00 x 10-7
e) 7.50 x 10-10
Convert the following pHs into hydrogen ion
concentrations in mol dm-3
a) 3.42
b) 1.20
c) 5.65
d) 8.40
e) 13.0
The pH of strong acids
• What is the pH of 0.1 mol dm-3 HCl?
• Because HCl is a strong acid every 1 mole of
HCl is entirely split up into 1 mole of H+(aq) and 1
mole of Cl-(aq). The concentration of hydrogen
ions is therefore exactly the same as the
concentration of the acid
• [H+] = 0.1
• pH = log10[H+]
• pH = -log10(0.1)
• pH = 1
What is the pH of 0.00100mol dm-3
H2SO4?
•
•
•
•
H2SO4  2H+(aq) + SO42-(aq)
[H+] = 2 x 0.0100
pH = – log10(0.0200)
pH = 1.70
Finding the concentration of a strong acid
from its pH
Simply the reverse the process
• Convert pH into [H+]
• Use [H+] to find the concentration of the
acid
• What is the concentration of HCl whose
pH is 1.60?
• “Unlogging” the pH gives [H+] = 0.0251
mol dm-3
• If you got an answer of 39.8107 you forgot
to make the pH ‘-’ before you unlogged it.
• As HCl is a monoprotic acid the
concentration of the acid is the same as
the concentration of the H+ ions.
What is the concentration of H2SO4, if its pH is
1.00?
•
•
•
•
•
Sulfuric acid is diprotic.
Unlogging th pH gives [H+] = 0.1 mol dm-3
Now stop and think:
H2SO4(aq)  2H+(aq) + SO42-(aq)
Each mole af acid gives 2 moles of H+(aq).
There are only half the number of moles of
acid as of H+ ions.
• The concentration of the acid is therefore
0.0500mol dm-3.
Calculate the pHs of the following strong
acids
1. 0.0300 mol dm-3 HCl
2. 0.00500 mol dm-3 H2SO4
3. 0.120 mol dm-3 HNO3
Calculate the pHs of the following strong
acids
1. 0.0300 mol dm-3 HCl 1.52
2. 0.00500 mol dm-3 H2SO4 2.00
3. 0.120 mol dm-3 HNO3 0.92
Calculate the concentrations of the following
strong acids from their pHs
1. HCl of pH 0.70
2. H2SO4 of pH 1.5
3. HNO3 of pH 2.0
Calculate the concentrations of the following
strong acids from their pHs
1. HCl of pH 0.70 0.2 mol dm-3
2. H2SO4 of pH 1.5 0.0016 mol dm-3
3. HNO3 of pH 2.0 0.010 mol dm-3
Ionic product of water
• Whenever liquid water is present, this
equilibrium is established
• H2O(l) H+(aq)+ OH-(aq)
• Just like any other equilibrium, you can write an
expression for the equilibrium constant Kc. You
might expect it to look like this:
Kc = [H+] [OH-]
[H2O]
But it doesn’t! such a tiny amount of water
ionises that the concentration of the water
is effectively constant. You need to go to 9
significant figures before you noticed any
change in the concentration of the wtare
following ionisation!
Ionic product of water
Kw = [H+][OH-]
Kw is usually 1.00 x 10-14
The units are [concentration]2
Calculating the pH of pure water
• pH varies with temperature,
• The ionisation process is endothermic
• If the water is pure the concentrations of
[H+] and [OH-] will be equal
• H2O(l) H+(aq) + OH-(aq)
If Kw is 1.00 x 10-14 mol2dm-6 (at 24oC)
• Kw = [H+][OH-]
• And [OH-] = [H+] because the water is pure
• So Kw = [H+]2
• [H+]2 = 1.00x 10-14
• [H+] = 1.00 x 10-7
• pH = -log10 [H+]
• pH = 7.00
• If Kw is 5.3 x 10-13 mol2dm-6 (at 100oC)
• pH = 6.14
• The neutral point has moved from the
more familiar value.
Calculate the pH of pure water at:
1. 15oC Kw = 4.52 x 10-15 mol2dm-6
2. 50oC Kw = 5.48 x 10-14 mol2dm-6
Calculate the pH of pure water at:
1. 15oC Kw = 4.52 x 10-15 mol2dm-6 7.17
2. 50oC Kw = 5.48 x 10-14 mol2dm-6 6.63
The pH of strong bases
• When most bases dissolve in water, the
solution consist of free hydrated metal ions
and hydroxide ions
• E.g NaOH(s) + aq  Na+(aq) + OH- (aq)
• Ca(OH)2(s) + aq  Ca2+(aq) + 2OH-(aq)
• Typical pHs are 13 or 14
• This means [H+] is 10-13 or 10-14
Calculating the pH of a strong base
• Use the concentration of the base to find
[OH-]
• Use Kw to find [H+]
• Convert [H+] into pH
• In the next examples I will take Kw to be
1.00 x 10-14 mol2dm-6
What is the pH of 0.10mol dm-3 NaOH?
Each mole of NaOH gives 1 mole of OH- in
solution, so the concentration of OH- is also 0.1
mol dm-3
Now use Kw and substitute in the value for [OH-]
[H+][OH-] = 1.00 x 10-14
[H+] x 0.1 = 1.00 x 10-14
[H+] = 1.00 x 10-14/0.10
[H+]= 1.0 x 10-13
Convert [H+] into pH = 13
Finding the concentration of a strong base
from its pH
Again you reverse the process
• Convert pH into [H+]
• Use Kw to find [OH-]
• Use the [OH-] to find the concentration of
the base.
What is the concentration of KOH solution if
its pH is 12.8?
• “Unlog” the pH to give the [H+]
pH =-log10[H+]
12.8 = -log10[H+]
Log10[H+] = -12.8
[H+] = 1.585 x 10-13
Now use Kw to find [OH-]
• [H+][OH-] = 1.00 x 10-14
• 1.585 x 10-13 x [OH-] = 1.00 x 10-14
• [OH-] = 1.00 x 10-14/1.585 x 10-13
• [OH-] = 0.0631 mol dm-3
• Because each mole of KOH produces 1
mole of OH-, the concentration of KOH =
0.0631 mol dm-3
What is the concentration of Ba(OH)2 if its
pH is 12.0?
“Unlogging” the pH gives
[H+] = 1.00 x 10-12
Now use Kw to find [OH-]:
[H+][OH-] = 1.00 x 10-14
1.00 x 10-12 x [OH-] = 1.00 x 10-14
[OH-] = 1.00 x 10-14/ 1.00 x 10 -12
[OH-] = 0.0100 mol dm-3
Each mole of Ba(OH)2 gives 2 moles of OH- this
means the concentration will only be half
= 0.00500mol dm-3
Calculate the pHs of the following strong
bases
1. 0.250 mol dm-3 NaOH
2. 0.100 mol dm-3 Ba(OH)2
3. 0.00500 mol dm-3 KOH
Calculate the pHs of the following strong
bases
1. 0.250 mol dm-3 NaOH 13.4
2. 0.100 mol dm-3 Ba(OH)2 13.3
3. 0.00500 mol dm-3 KOH 11.7
Calculate the concentrations of the following
strong bases from their pHs
1. NaOH of pH 13.2
2. Sr(OH)2 of pH 11.3
Calculate the concentrations of the following
strong bases from their pHs
1. NaOH of pH 13.2 0.158 mol dm-3
2. Sr(OH)2 of pH 11.3 1.00 x 10-3 mol dm-3
Summarise
pH [OH-]
log10
diprotic acids
10x
[H+]
strong acids
monoprotic acids
Strong bases
Kw
Apply to exam questions