Y13 acids & bases lta 2017

1) Write down the formulae of the following acids or bases, if you can
also write the equation for their dissociation when dissolved in water;
a) Sulphuric acid
b) Nitric acid
c) Ethanoic acid
d) Potassium hydroxide
e) Calcium hydroxide
f) Ammonia
2) Define the following terms;
a) acid
b) base
c) alkali
3) Draw a mind map of everything you can remember from AS about ‘Acids
and Bases’
Starter Questions

Learning Objectives:
• Recall and state the formulae of common acids and bases &
state that an acid releases H+
ions in aqueous solution.
• State that an alkali is a soluble base that releases OH–
ions
in aqueous solution.
• Explain the difference between strong acids and bases and
concentrated vs. weak solutions of acids and bases.
• Describe and use the term conjugate acid-base pairs.
Key Words:
Acid, base, alkali, pH, neutral, neutralisation, conjugate pair
Acids and Bases

Acid: A species that is a proton (H+ ion) donor
Acids
The word Acid comes from the Latin
‘acidus’ – meaning sour.
Acids have been known for hundreds
of years but have only recently been
understood in detail.
The 3 common acids in A-Level
chemistry are
• Sulphuric acid –
• Hydrochloric acid -
• Nitric acid –
You need to know the name and
formula for these
H2SO4
HCl
HNO3
In water acids give a pH of less
than 7.0

Depending on their formulae and bonding, different acids can
release different numbers of protons.
Acids – mono, di and tri protic
HCl is a monoprotic acid as each molecule can release 1 proton.
HCl(aq) H+(aq) + Cl-(aq) (chloride)
H2SO4 is a diprotic acid - each molecule can release 2 protons.
H2SO4 (aq)H+(aq) + HSO4
-(aq) (hydrogensulphate)
HSO4
- (aq) ⇌ H+(aq) + SO4
2-(aq) (sulphate)
H3PO4 is a triprotic acid - each molecule can release 3 protons.
H3PO4 (aq)H+(aq) + H2PO4
-(aq) (dihydrogenphosphate)
H2PO4
-(aq) ⇌ H+(aq) + HPO4
2-(aq) (hydrogenphosphate)
HPO4
2-(aq) ⇌ H+(aq) + PO4
3-(aq) (phosphate)

Base: A species that is a proton (H+ ion) acceptor
Common bases are metal oxides and hydroxides:
Metal Oxides – MgO, CuO Metal Hydroxides – NaOH, Mg(OH)2
Ammonia – NH3 is also a base as are all amines e.g. CH3NH2
Bases
We can use bases in every day life:
For acid indigestion MgO is in milk of magnesia
Ca(OH)2 is used to treat acid soils

Alkali: a type of base that dissolves in water to form
hydroxide (OH-) ions.
A chemical that gives a solution with a pH greater than 7.0
when dissolved in water.
Common alkalis are:
• Sodium Hydroxide – NaOH
• Potassium hydroxide – KOH
• Ammonia – NH3
Alkalis
An alkali is a certain type of base that dissolves in water to
give aqueous hydroxide ions – OH-(aq)
E.g. NaOH + aq  Na+(aq) + OH-(aq)
Alkalis are very corrosive and
sometimes more dangerous
than acids!

When the protons (H+) from acids and the hydroxide ions
(OH-) from bases meet in solution. This is called a
neutralisation reaction. Water is formed:
H+(aq) + OH-(aq) H2O(l)
E.g. HCl + NaOH  NaCl + H2O
Acid + Base (or alkali)  Salt + Water
Acid + Carbonate  Salt + Water + carbon dioxide
Acid + metal  salt + hydrogen gas (redox reaction – not
strictly acid-base)
Neutralisation

Ammonia – NH3 is a gas that dissolves in water to form a
weak alkaline solution. It reacts with the water;
Ammonia as a Weak Base
Ammonia is a weak base because only a small proportion of the
dissolved NH3 actually reacts with the water.
As indicated by the equilibrium sign ⇋
NH3(aq) + H2O(l) ⇋ NH4
+(aq) + OH-(aq)

3) Write the equations for dissociation of the following acids
in water;
a) Nitric acid
b) Chromic acid H2CrO4
4) Write the full and ionic equations for the following acid-
base reactions;
a) sulphuric acid and solid magnesium carbonate
b) sulphuric acid and aqueous potassium carbonate
c) hydrochloric acid and solid calcium oxide
d) nitric acid and aqueous sodium hydroxide
e) hydrochloric acid and aluminium.
Questions – 4 minutes!

• Describe and use the term conjugate acid-base pairs.
Key Words:
Acid-base pair, alkali, pH, neutral, neutralisation
Conjugate Acid-Base pairs

Acid: A species that is a proton (H+ ion) donor
Acids are Proton Donors
A molecule of acid must contain a hydrogen that can be released as H+
However this does not just happen on its own. Acids only release protons if
there is something there to accept them.
Formation of the hydronium ion from HCl and H2O
Hydronium ions, H3O+ - water molecules accept protons.
HCl + H2O  H3O+ (aq) + Cl- (aq)

The equation for the process of HCl dissolving in water is
often simplified to: HCl + (aq)  H+ (aq) + Cl- (aq)
Using just H+ makes it easier to understand but remember
that H+(aq) and H3O+ both represent protons in aqueous
solution.
Acids are Proton Donors
Formation of the hydronium ion from HCl and H2O
Hydronium ions, H3O+ - water molecules accept protons.
HCl + H2O  H3O+ (aq) + Cl- (aq)

An acid-base pair: is a set of 2 species that transform into
each other by gain or loss a proton.
Below shows the acid-base dissociation of nitrous acid, HNO2.
It shows that the acid-base pair are linked by H+
Acid-Base Pairs
Conjugate acid–base pair

Acid-base equilibria involve 2 acid-base pairs. The equilibrium
below shows the dissociation of nitrous acid in water.
HNO2 (aq) + H2O(l) ⇌ H3O+ (aq) + NO2
- (aq)
In the forward direction:
• The acid HNO2 releases a proton to form its conjugate base NO2
-
• The base H2O accepts the proton to form its conjugate acid H3O+
In the reverse direction:
• The acid H3O+ releases a proton to form its conjugate base H2O -
• The base NO2
- accepts the proton to form its conjugate acid HNO2
HNO2 and NO2
- only differ by H+ and make up acid-base pair 1
H3O+ and H2O only differ by H+ and make up acid-base pair 2
Acid-Base Equilibria

1) Identify the acid-base pairs in the following acid-base
equilibria.
a) HIO3 + H2O ⇌ H3O+ + IO3
-
b) CH3COOH + H2O ⇌ H3O+ + CH3COO-
c) NH3 + H2O ⇌ NH4
+ + OH-
2) Write and equation for the formation of the hydronium ion
from hydrofluoric acid, HF, and water.
Checkpoint sheet
Questions

• Define pH as pH = –log[H+(aq)].
• Define [H+] = 10–pH.
• Convert between pH values and [H+(aq)].
Key Words:
pH, log, concentration
What is pH?

Aqueous solutions have concentrations of H+ (aq) ions in the
range of;
• From 10 mol dm-3 (101 mol dm-3)
• To 0.000,000,000,000,001 mol dm-3 (10-15 mol dm-3)
The pH Scale
This range is huge! It is therefore difficult to
work with such numbers.
So the Danish chemist Søren Sørensen devised
the pH scale as a more convenient way of
measuring the concentration of H+ ions.
This scale is still used by all scientists today!
pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
[H+(aq)]/
mol dm-3
1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-
11
10-
12
10-
13
10-
14

• If you compare the pH and [H+] below you will see a pattern.
• pH is the negative power of [H+(aq)]
• [H+(aq)] = 10-pH
The pH scale is logarithmic – the difference between each successive
number value is a factor of 10.
E.g. the difference between pH 3 and pH 5 is a factor of 10x10 = 100
• pH is equal to the negative logarithm to the base 10 of [H+(aq)]
• pH = -log[H+(aq)]
The pH Scale
pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
[H+(aq)]/
mol dm-3
1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-
11
10-
12
10-
13
10-
14

pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
[H+(aq)]/
mol dm-3
1 10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-10 10-
11
10-
12
10-
13
10-
14
What does a pH value mean?
The relationship between pH and [H+(aq)] is sometimes called a ‘see-saw’
As one goes up – the other goes down
• A low pH means a large [H+(aq)]
• A high pH means a small [H+(aq)]
• A pH change of 1 means [H+(aq)] changes by 10 times
• An acid with a pH 2 contains 1000x more [H+(aq)] of an acid with a pH 5

The table shows some examples of the relationship between pH & [H+(aq)]
pH 4.52 5.79 10.63 13.41
[H+(aq)]/
mol dm-3
10-4.52 10-5.79 10-10.63 10-13.41
Converting between pH and
[H+(aq)]
The [H+(aq)] values are not shown in conventional standard form. To get
them into standard form we can use our calculator. Input 10-pH
pH 4.52 5.79 10.63 13.41
[H+(aq)]/
mol dm-3
3.02 x 10 -5 1.62 x 10 -6 2.34 x 10 -11 3.89 x 10 -14

Converting between pH and
[H+(aq)]
Use this button on your calculator to convert between pH and [H+(aq)]
pH 4.52 5.79 10.63 13.41
[H+(aq)]/
mol dm-3
3.02 x 10 -5 1.62 x 10 -6 2.34 x 10 -11 3.89 x 10 -14

Hints for pH calculations
pH calculations are easy if you learn to use your
calculator!
Learn how to use your particular brand of calculator
and stick to it!
Once you have an answer, check if it looks sensible
Practical: Now carry out some practice pH readings
Ilpac 7.1 & 7.2

Questions
1. How many more times more hydrogen ions are in a solution with a pH 2
than one of pH 4?
2. Calculate the pH of solutions with the following hydrogen ion
concentrations. To 2 d.p.
a) 3.33 x 10-3 mol dm-3
b) 4.73 x 10-4 mol dm-3
c) 2.39 x 10-12 mol dm-3
d) 2.30 mol dm-3
3. Calculate [H+(aq)] of solutions with the following pH values. To 3 s.f
a) pH 6.53
b) pH 2.87
c) pH 9.58
d) pH -0.12
2.48
3.32
11.6
-0.36
2.95 x10 -7 mol dm-3
1.35 x10 -3 mol dm-3
2.63 x10 -10 mol dm-3
1.32 mol dm-3

Explain qualitatively the differences between strong and weak
acids.
Explain that the acid dissociation constant, Ka , shows the
extent of acid dissociation.
Deduce expressions for Ka and p Ka for weak acids.
Key Words:
Strong acid, weak acid, pH, Ka, pKa
Strong and Weak Acids

In aqueous solution acids dissociate and an equilibrium is set
up. The equilibrium below shows the dissociation of nitrous
acid in water.
HNO2 (aq) + H2O(l) ⇌ H3O+ (aq) + NO2
- (aq)
The strength of an acid is the extent to which this
dissociation occurs – or how far to the right the equilibrium is.
Strong acids: dissociate 100% - the equilibrium cam be
assumed to be completely over to the right.
There are comparatively few strong acids
Acid-Base Eqilibria

A weak acid: only partially dissociated in water.
Many naturally occurring acids are weak acids.
CH3COOH (aq) + H2O(l) ⇌ H3O+ (aq) + CH3COO- (aq)
When methanoic acid dissociates in water the equilibrium lies
well over to the left.
There is only a small concentration of the dissociated ion
compared to the un-dissociated methanoic acid molecule.
We can express the equilibrium constant for the dissociation
of an acid – this is called Ka – the acid dissociation constant
Weak Acids

• Calculate pH from [H+(aq)] and [H+(aq)] from pH for strong
and weak acids.
• Calculate Ka for a weak acid.
Key Words:
Strong acid, weak acid, pH, Ka, pKa
Calculating pH for Strong and
Weak Acids

A strong monobasic acid HA has virtually complete dissociation
in water.:
HA (aq)  H+(aq) + A- (aq)
Meaning that the [H+(aq)] of a strong acid is equal to [HA(aq)]
[H+(aq)] = [HA(aq)]
So pH can be calculated using:
pH = -log[H+(aq)]
Calculating pH of Strong Acids

A sample of HCl has a concentration of 1.22 x 10-3 mol dm-3
What is the pH?
HCl (aq)  H+(aq) + Cl- (aq) – dissociates completely
Therefore; [H+(aq)] = [HCl(aq)] = 1.22 x 10-3 mol dm-3
And; pH = -log[H+(aq)]
So; -log(1.22 x 10-3 mol dm-3 ) = 2.91

A sample of HNO3 has a pH of 5.63
What is the concentration of t he HNO3?
HNO3 (aq)  H+(aq) + NO3
- (aq) – dissociates completely
Therefore; [H+(aq)] = [HNO3(aq)]
And; [H+(aq)] = 10-pH = 10-5.63 = 2.34 x 10-6 mol dm-3

In aqueous solution, a weak monobasic acid, HA, partially
dissociates, setting up the equilibrium;
HA (aq) ⇌ H+(aq) + A- (aq)
Ka =
𝐻
+
𝑎𝑞 [𝐴
−
𝑎𝑞 ]
[𝐻𝐴 𝑎𝑞 ]
To work out the pH we need to find [H+(aq)], this depends on;
• The concentration of the acid, [HA(aq)]
• The value of ka
Unlike strong acids [H+(aq)] is much less than [HA(aq)] is
dissociation not full.
Calculating pH of Weak Acids

HA (aq) ⇌ H+(aq) + A- (aq)
Ka =
𝐻
+
𝑎𝑞 [𝐴
−
𝑎𝑞 ]
However we do know that H+ and A- ions are formed in equal quantities.
[H+(aq)] = [A-(aq)]
So; [H+(aq)][A-(aq)] = [H+(aq)]2
As so few HA molecules dissociate [HA(aq)] will have reduced slightly.
The equilibrium, concentration of HA(aq) will be [HA(aq)] – [H+(aq)]
Therefore;
Ka =
𝐻+ 𝑎𝑞 2
𝐻𝐴 𝑎𝑞 −[𝐻+ 𝑎𝑞 ]

Making an Approximation:
We now make an approximation to simplify the pH calculation.
Only a very small proportion of the HA dissociates in weak
acids. So we can assume that the equilibrium concentration of
HA will be very nearly the same as the concentration od un-
dissociated HA.
So the expression approximates to;
Ka =
𝐻
+
𝑎𝑞 2
And therefore;
[H+(aq)]2 = Ka[HA(aq)]
and; [H+(aq)] = Ka[HA(aq)] and pH = -log[H+(aq)]

A sample of nitrous acid, HNO2 is 0.055 mol dm-3
Ka = 4.70 x 10-4 mol dm-3 at 25 ºC. Calculate pH
First find [H+(aq)]. HNO2(aq) ⇌ H+(aq) + NO2
-(aq)
Ka =
𝐻
+
𝑎𝑞 [𝑁𝑂2
−
(𝑎𝑞)]
[𝐻𝑁𝑂2
𝑎𝑞 ]
≈
𝐻
+
𝑎𝑞 2
[𝐻𝑁𝑂2
𝑎𝑞 ]
And therefore;
[H+(aq)]2 = Ka[HNO2(aq)]
and; [H+(aq)] = Ka[HA(aq)] = (4.70 x 10−4 x 0.055) = 5.08x10-13 mol
dm-3
and pH = -log[H+(aq)] = -log(5.08x10-13) = 2.29
Worked example of Calculating
pH of a Weak Acid

1. Find the pH of the following solutions of weak acids:
a) 0.65 mol dm-3 CH3COOH (Ka = 1.7x10-5 mol dm-3)
b) 4.4 x 10-2 mol dm-3 HClO (Ka = 3.7 x 10-8 mol dm-3)
2. Find the Ka and pKa values for the following weak acids.
a) 0.13 mol dm-3 solution with a pH of 3.52
b) 7.8 x 10-2 mol dm-3 solution with pH of 5.19
Questions

• State and use the expression for the ionic product of
water, Kw
• Understand the importance of Kw in controlling the
concentration of H+(aq) and OH-(aq) in aqueous solutions.
Key Words:
Ionic product of water, Kw , equilibrium, pH
The Ionisation of Water.

Pure water ionises into H+(aq) and OH-(aq) ions but the extent
of the ionisation is very very small – only 1 H2O molecule out of
every 500 000 000 dissociates.
H2O (l) ⇌ H+(aq) + OH- (aq)
The position of equilibrium is well to the left.
The Ionisation of Water.
Kc =
[H
+
(aq)][OH−(aq)]
[H2O(l)]
so Kc x [H2O(l)] = [H+(aq)][OH- (aq)]
[H2O(l)] =
1000
18
= 55.6 mol dm-3

Kc and [H2O(l)] are both constants – combined together they
give a new constant, the ionic product of water, Kw.
Kw = Kc x [H2O(l)] = [H+(aq)][OH- (aq)]
constant, Kw ionic product
The Ionic Product of Water, Kw
At 25℃ the pH of water is 7, so [H+(aq)] = 10-7 mol dm-3
• as [H+(aq)] = [OH-(aq)], so [OH-(aq)] = 10-7 mol dm-3
• And 10-7 mol dm-3 x 10-7 mol dm-3 = 10-14 mol2 dm-6
• At 25℃ , Kw = 1.00 x 10-14 mol2 dm-6

• Kw controls the balance between [H+(aq)] and [OH-(aq)] in all
aqueous solutions.
• At 25℃ the pH of 7 is the neutral point where the [H+(aq)]
= [OH-(aq)] = 10-7 mol dm-3
The Significance of Kw
All aqueous solutions contain H+ and OH- ions.
• In water and neutral solutions [H+(aq)] = [OH-(aq)]
• In acidic solutions [H+(aq)] > [OH-(aq)]
• In alkaline solutions [H+(aq)] < [OH-(aq)]
The relative concentrations [H+(aq)] and [OH-(aq)] are
determined by Kw
At 25℃ Kw = [H+(aq)][OH-(aq)] must always equal 1 x 10-14 mol2
dm-6

It is easy to find [H+(aq)] & [OH-(aq)] if you know the pH.
Because [H+(aq)][OH-(aq)] = 1 x 10-14 mol2 dm-6
Finding the values of [H+] &
[OH-]
Therefore, [H+(aq)] = 1 x 10-14 mol2 dm-6/[OH-(aq)]
And, [OH-(aq)] = 1 x 10-14 mol2 dm-6/[H+(aq)]

Questions
1. Define the term ionic product of water.
2. The following solutions have the shown [H+(aq)]. Calculate [OH-(aq)].
a) 10-6 mol dm-3
b) 10-2 mol dm-3
c) 10-11 mol dm-3
3. The following solutions have the shown [OH-(aq)]. Calculate [H+(aq)].
a) 10-13 mol dm-3
b) 10-4 mol dm-3
c) 10-10 mol dm-3

• Calculate pH from [H+(aq)] for strong bases using Kw.
• Calculate [H+(aq)] from pH for strong bases using Kw.
Key Words:
Ionic product of water, Kw , base, pH, alkali
pH values of bases.

Bases: a substance that can accept protons, reacts
with an acid to form a salt and water.
Alkali: a soluble base that releases OH- ions.
Base Strength:
The strength of a base is a measure of it’s dissociation in
water to release OH- ions.
NaOH is a strong base which dissociates completely in water.
NH3 is a weak base as it only partially dissociates water in an
equilibrium that is well to the left.
NH3(aq) + H2O(l) ⇌ NH4
+ (aq) + OH-(aq)
pH values of bases.

To calculate pH you need to know [H+(aq)]. This depends on;
• The concentration of base
• The ionic product of water, Kw = 1.00 x 10-14 mol2 dm-6
NaOH is a strong monobasic alkali. Meaning is the same as the
concentration of the NaOH.
• NaOH(aq) ⟶ Na+(aq) + OH-(aq)
• So [NaOH(aq)] = [OH-(aq)]
Calculating the pH value of
Strong Bases.
We can find [H+(aq)] from Kw and [OH-(aq)]
• Kw = [H+(aq)][OH-(aq)]
• So [H+(aq)] = Kw/ [OH-(aq)]
• And pH = -log [H+(aq)]

KOH(aq) has a concentration of 0.050 mol dm-3. What is the
pH?
KOH is a strong base; [KOH(aq)] = [OH-(aq)] = 0.050 mol dm-3
Examples
Method 1:
Find [H+(aq)] from [OH-(aq)]
• Kw = [H+(aq)][OH-(aq)] = 1.00 x 10-14 mol2 dm-6
• So, [H+(aq)] = Kw/ [OH-(aq)] = 1.00 x 10-14 / 0.050
= 2.0 x 10-13 mol dm-3
• Now find pH = -log [H+(aq)] = -log(2.0 x 10-13 ) = 12.70

KOH(aq) has a concentration of 0.050 mol dm-3. What is the
pH?
KOH is a strong base; [KOH(aq)] = [OH-(aq)] = 0.050 mol dm-3
Examples
Method 2:
pOH is defined as; pOH = -log[OH-(aq)]
• Now find pOH = -log [OH-(aq)] = -0.050) = 1.30
• To get pH subtract the pOH value from 14
pH= 14 – 1.30 = 12.70

All aqueous solutions contain both H+(aq) and OH-(aq).
We know that Kw is important in controlling the balance
between [H+] and [OH-] in all aqueous solutions.
This relationship is critical:
• At 25℃ Kw = [H+(aq)][OH-(aq)] = 1.00 x 10-14 mol2 dm-6
Back to Kw
So using Kw:
• We can find [OH-(aq)] if we know [H+(aq)]
• We can find [H+(aq)] if we know [OH-(aq)]

Questions
1. Find the [H+(aq)] and pH of the following alkalis at 25℃
a) 2 x 10-3 mol dm-3 of OH-(aq)
b) 5.7 x 10-1 mol dm-3 of OH-(aq)
3. Find the pH of the following solutions at 25℃
a) 0.0050 mol dm-3 KOH(aq)
b) 3.56 x 10-2 mol dm-3 NaOH(aq)
3. Find the concentration in mol dm-3 of OH-(aq) at 25℃
a) pH = 12.43
b) pH = 13.82

• Describe what is meant by a buffer solution.
• State that a buffer solution can be made from the weak
acid and a salt of a weak acid.
• Explain the role of the conjugate acid-base pair in an acid
buffer solution.
Key Words:
Buffer solution, acid, base, pH
Buffer Solutions

Buffer Solution: a mixture that minimises pH changes on
addition of small amount of acid or base.
A buffer cannot prevent the pH from changing completely, but
pH changes are minimised. At least for as long as some of the
buffer solution remains.
A buffer solution is a mixture of:
• A weak acid, HA
• Its conjugate base, A-
It can be made from a weak acid and a salt of that weak acid.
e.g. ethanoic acid, CH3COOH and sodium ethanoate CH3COONa
Buffer Solutions

In the ethanoic acid, CH3COOH and sodium ethanoate CH3COONa buffer
system;
• the weak acid dissociates partially;
CH3COOH (aq) ⇌ H+(aq) CH3COO-(aq)
• The salt dissociates completely, generating the conjugate base;
CH3COO-Na+ (aq)  CH3COO-(aq) + Na+(aq)
The mixture formed contains a lot of the un-dissociated weak acid
CH3COOH and its conjugate base CH3COO-.
The high concentration of the conjugate base pushes the equilibrium to the
left so [H+] is very small.
Buffer Solutions

The resulting buffer solution contains large ‘reservoirs’ of the weak acid
and the conjugate base;
Buffer Solutions

The weak acid and the conjugate base are both responsible for controlling
pH. The buffer minimises the pH changes by using the equilibrium;
HA(aq) ⇌ H+(aq) + A-(aq)
• The weak acid, HA, removes added alkali.
• The conjugate base, A-, removes added acid.
How do Buffers Act?
When an acid (H+) is added to a buffer solution;
• [H+(aq)] increases
• The conjugate base, A-, reacts with the H+ ions
• The equilibrium shifts to the left, removing most of the added H+ ions

The weak acid and the conjugate base are both responsible for controlling
pH. The buffer minimises the pH changes by using the equilibrium;
• The weak acid, HA, removed added alkali.
• The conjugate base, A-, removes added acid.
How do Buffers Act?
When an alkali (OH-) is added to a buffer solution;
• [OH-(aq)] increases
• The small concentration of H+ ions react with the OH- ions
H+ + OH-  H2O
• HA dissociates shifting the equilibrium to the right to restore most of
the H+ ions that have reacted

How do Buffers Act?
Shifting the buffer equilibrium

Questions
1. What is meant by the term buffer solution.
2. A buffer is made from propanoic acid and sodium propanoate.
Explain using equations the role of the conjugate acid-base pair in this
buffer solution.

• Calculate the pH of a buffer solution from the Ka value of a
weak acid and the equilibrium concentrations of the
conjugate acid–base pair.
• Explain the role of the carbonic acid–hydrogencarbonate ion
system as a buffer in the control of blood pH.
Key Words:
Buffer solution, acid, base, pH
pH values of Buffer Solutions

The pH of a buffer solution depends on;
• The acid dissociation constant, Ka, of the buffer system.
• The concentration ratio of the weak acid to its conjugate base.
There are two methods for doing this.
Method 1:
Ka =
𝐻
+
𝑎𝑞 [𝐴
−
𝑎𝑞 ]
So, [H+(aq)] = Ka x
[𝐴− 𝑎𝑞 ]
Only a very small proportion of HA dissociates so we assume;
[HA(aq)]equilibrium = [HA(aq)]undissociated
and once [H+(aq)] is known we find pH from -log[H+(aq)]
Calculations involving Buffer
Solutions

The pH of a buffer solution depends on;
• The acid dissociation constant, Ka, of the buffer system.
• The concentration ratio of the weak acid to its conjugate base.
There are two methods for doing this.
Method 2: the Henderson-Hasselbalch method
pH= pKa + log
[A− aq ]
[HA aq ]
Where pKa = -logKa
Calculations involving Buffer
Solutions

Calculate the pH at 25℃ of a buffer containing 0.050 mol dm-3
CH3COOH(aq) and 0.10 mol dm-3 CH3COO-Na+(aq).
For CH3COOH(aq) Ka = 1.7 x 10-5 mol dm-3
Worked example
Using Method 1:
[H+(aq)] = Ka x
[𝐇𝐀 𝐚𝐪 ]
[𝐀− 𝐚𝐪 ]
= 1.7 x 10-5 x
𝟎.𝟎𝟓𝟎
𝟎.𝟏𝟎
= 8.5 x 10-6 mol dm-3
pH = -log(8.5 x 10-6)
= 5.07

Calculate the pH at 25℃ of a buffer containing 0.050 mol dm-3
CH3COOH(aq) and 0.10 mol dm-3 CH3COO-Na+(aq).
For CH3COOH(aq) Ka = 1.7 x 10-5 mol dm-3
Worked example
Using Method 2:
pH= pKa + log
[𝐀− 𝒂𝒒 ]
[𝐇𝐀 𝒂𝒒 ]
=-logKa + log
[𝐇𝐂𝐎𝐎− 𝐚𝐪 ]
[𝐇𝐂𝐎𝐎𝐇 𝐚𝐪 ]
= 4.77 + 0.30
= 5.07

Questions
1. Two different buffer solutions, A and B, were made using CH3COOH
and CH3COONa. Ethanoic acid had a Ka of 1.7 x 10-5 mol dm-3 at 25
degrees.
Calculate the pH of each buffer solution if;
A has the same concentrations of the acid and the salt
B has the concentrations 0.75 mol dm-1 of acid and 0.25 mol dm-3 of the
salt
2. What ratio of [CH3COOH ]:[CH3COONa] Would be needed to make a solution
that buffers at a pH of 5.47

Human blood plasma needs to have a pH of between 7.35 and 7.45. If it
falls below a condition called acidosis is experienced. If it goes above it is
called alkalosis
The carbonic acid-
hydrogencarbonate buffer
Blood pH is controlled by a mix of buffers. This one is the most important.
• Carbonic acid, H2CO3 is the weak acid
• Hydrogencarbonate, HCO3
- is the conjugate base
H2CO3(aq) ⇌ HCO3
-(aq) + H+(aq)
Any increase in H+ is removed by the base and the equilibrium shifts to the left.
Any increase in OH- is removed by the acid ions reacting with the OH- to form
water H+ + OH-  H2O
More H2CO3 dissociates and equilibrium moves to the right to restore H+

The carbonic acid-
H2CO3(aq) ⇌ HCO3
-(aq) + H+(aq)
Ka for this equilibrium is 4.3 x 10-7 mol dm-3
Most materials released into the blood are acidic. The hydrogencarbonate
ions remove these by being converted to H2CO3.
The carbonic acid is converted to dissolved CO2 through the action of an
enzyme and it is excreted ay the lungs as CO2 gas.
The CO2 content of the blood can be changed by changing breathing rate.
Faster and heavier breathing removes more CO2
Calmer, slower breathing removes less CO2

The carbonic acid-
H2CO3(aq) ⇌ HCO3
-(aq) + H+(aq)
The blood contains about 10 times more HCO3
- (aq) than H2CO3(aq).
The pH of healthy blood is 7.4
Therefore [H+(aq)] = 10-7.4
= 3.98 x 10-8 mol dm-3
[H+(aq)] = Ka x
[𝐇𝟐𝐂𝐎𝟑 𝐚𝐪 ]
[𝐇𝐂𝐎 𝟑
− 𝐚𝐪 ]
[𝐇𝐂𝐎 𝟑
−
𝐚𝐪 ]
[𝐇𝟐𝐂𝐎𝟑 𝐚𝐪 ]
=
Ka
[𝐇+ 𝐚𝐪 ]
= 4.3 x 10-7 / 3.98 x 10-8 = 10.8 / 1

Questions
1. Calculate the pH of a buffer solution containing 0.15 mol dm-3 methanoic
acid HCOOH(aq), and 0.065 mol dm-3 sodium methanoate, HCOONa. For
HCOOH Ka = 1.6 x 10-4 mol dm-3
2. If extra acid is produced in the blood how do buffers prevent the pH
from falling.

• Interpret and sketch acid-base titration pH curves for
strong and weak acids and bases.
• Explain the choice of suitable indicators for acid-base
reactions.
Key Words:
, acid, base, pH, Equivalence point, end point,
Neutralisation – Titration Curves

Neutralisation – Titration Curves
In a titration, a solution of known concentration is titrated
with a solution of unknown concentration.
An acid and a base can be titrated to find the concentration
of one of them.
During an acid-base titration, an indicator is often used to
show when the solution has been neutralized. This point is
called the equivalence point.
If a pH meter is used, a pH
curve can be plotted. This is
a graph showing how the pH
of a solution changes as base
(or acid) is added.

colourless pink
yellow blue
red yellow
yellow purple
red yellow
red
yellow
Selecting an Appropriate
Indicator
Around the equivalence point of a titration, the pH changes very rapidly.
Indicators change colour over a narrow pH range approximately centred
around the pKa of the indicator. An indicator will be appropriate for a
titration if the pH range of the indicator falls within the rapid pH change
for that titration.
bromophenol blue
methyl orange
methyl red
phenolphthalein
bromothymol blue
thymol blue
Indicator Colour in acid pH range Colour in alkali
1.2–2.8
3.1–4.4
4.4–6.2
6.0–7.6
8.3–10.0
3.0–4.6
red

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