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pH and the Acid Dissociation Constant
Learning Outcomes:
CHAPTER 4 - Acids and bases – Lesson 3
1. understand and use the terms strong acid and weak acid.
2. apply the knowledge of pH to calculate ion concentrations.
3. define mathematically the terms pH, Ka and Kw and use them in calculations (Kb and the equation Kw =
Ka × Kb will not be tested)
LESSON OBJECTIVE: Investigate pH, Kw, Ka and how to utilise them in
calculations.

Recap
Question
For the reaction below label the acid, base, conjugate
acid, conjugate base and identify the conjugate pairs:
NH3(aq) + H2O(l) NH
⇌ 4
+
(aq) + OH-
(aq)
Acid Conjugate
Base
Conjugate
Acid
Base
- Brønsted-Lowry theory of acids and bases: acids are proton donors and bases are
proton acceptors.
- Conjugate pairs: an acid and base on each side of an equilibrium equation that are
related to each other by the difference of a proton
- Equilibrium expression (Kc): A relationship that
links the equilibrium concentrations of
reactants/products to the stoichiometry that can be
used to calculate an equilibrium constant

- Water is amphoteric, it is able to
act as either an acid or a base:
H2O(l) + H2O(l) H
⇌ 3O+
(aq) + OH-
(aq)
- This is often simplified as:
H2O(l) H
⇌ +
(aq) + OH-
(aq)
- What would a Kc expression for this
reaction look like?
- Extent of ionisation is incredibly
low, therefore [H+
] and [OH-
] is too
- Thus can simplify further, assume
concentration of water is constant
Ionic Product of Water (Kw) is the
equilibrium constant for the
ionisation of water
Kw = [H+
][OH-
]
- Kw = 1.00 × 10-14
mol2
.dm-6
at 298K
- Pure water is neutral:
- [H+
] = [OH-
]
- Kw = [H+
]2
- [H+
] = 1.00 × 10-7
mol.dm-3
Ionic Product of Water, Kw
Kc = __________________
[H+
(aq)][OH-
(aq)]
[H2O(l)]

pH Scale
- What makes a solution acidic? The amount of H+
in solution
- pH is a measurement of the acidity of an aqueous solution:
pH = -log[H+
]
- The pH scale shows how acidic/basic a solution is on a scale from 1 to 14
- Invented by Danish chemist Søren Sørensen
- Uses a logarithmic scale to account for a very large range of hydrogen ion concentrations
- Negative sign results in (mostly) all positive values
- Pure water is neutral, calculate the pH of pure water ([H+
] = 1.00 × 10-7
mol.dm-3
)
neutral
7
basic
0 14
acidic
[H+
] > [OH-
] [H+
] < [OH-
]
Can rearrange
equation to get
[H+
] from pH:
[H+
] = 10-pH
Question
1) Calculate the pH of a solution with [H+
] = 1.6 × 10-4
mol.dm-3
2) Determine the [H+
] in a solution with pH = 10.5
pH = 3.80
[H+
] = 3.16 × 10-11
mol.dm-3

Acid-Base Strength and pH
Strong Acids:
- Strong monoprotic acids completely ionise in water: HCl(aq) H
→ +
(aq) + Cl-
(aq)
- Assume conc of strong acid is equal to conc of H+
and use that value in pH
equation
- Can ignore the negligent amount of H+
produced from ionisation of water
- Note: more steps required for polyprotic acids (e.g. H2SO4, H3PO4)
Strong Bases:
- Strong bases also completely ionise in water: NaOH(aq) Na
→ +
(aq) + OH-
(aq)
- Rearrange Kw expression: [H+
] = Kw
/[OH-] (Kw=1.00×10-14
mol2
.dm-6
at 298K)
- Can then use pH equation
Question
1) Determine the pH of a solution if [HCl] = 2.00×10-3
mol.dm-3
2) Determine the pH of a solution if [NaOH] = 0.0500 mol.dm-3
pH = 2.70
pH = 12.7

Ka =_________
[H+
][A-
]
[HA]
The Acid Dissociation Constant, Ka
Weak Acids
- Monoprotic weak acids do not fully dissociate:
HA(aq) H
⇌ +
(aq) + A-
(aq)
- Can apply equilibrium law to obtain an acid
dissociation constant, the equilibrium constant
for a weak acid:
The value of
Ka will imply
the degree of
ionisation
- For monoprotic acids [H+
] = [A-
] thus:
Ka = ______
[H+
]2
[HA]
Question
A weak acid has a Ka value of
6.3 × 10–6
mol.dm-3
, determine
the pH of a 0.15 mol.dm-3
solution of this weak acid.
pH = 3.0
[HA] = Total concentration – [H]

pKa
Very Weak Acids
- Ka values can be very small in which case we can give a pKa value, a value
of Ka expressed as a logarithm:
pKa = -logKa
Question
A 0.100 mol.dm-3
solution of ethanoic acid has a [H+
] of 1.32 × 10-3
mol.dm-3
, calculate the Ka and
pKa for this weak acid.
Ka = 1.74 × 10-5
mol.dm-3
pKa = 4.76

pH and the Acid Dissociation Constant
Learning Outcomes:
(taken from the Cambridge International AS and A Level Chemistry curriculum)
25.1 Acids and bases
1. understand and use the terms conjugate acid and conjugate base
2. define conjugate acid-base pairs, identifying such pairs in reactions
3. define mathematically the terms pH, Ka pKa and Kw and use them in calculations (Kb and the equation
Kw = Ka × Kb will not be tested)
LESSON OBJECTIVE: Investigate pH, Kw, Ka, pKa and how to utilise them
in calculations.

Work through each task, one column at a time. Start at the bottom and work your way up.
Explain how the ionic product of water, Kw,
can be simplified to Kw = [H+
]2
and why [H+
]
equals 1.00 × 10-7
mol.dm-3
at 298K.
Calculate the following:
1) The pH of a solution with a hydrogen
ion concentration of 2.7×10-2
mol.dm-3
2) The concentration of hydrogen ions
present in a solution with a pH equal to
9.3
A 0.010 mol.dm-3
solution of a weak acid
had a pH of 2.75, calculate its corresponding
Ka and pKa values.
Summarise the equation used to calculate Kw
from [H+
] and [OH-
] and state its value at
298K.
Summarise what information the pH of a
solution can provide and why it is necessary
to use a logarithmic scale to represent it.
Explain the concept of the acid dissociation
constant, Ka and how it can be used to
calculate a pKa value. Rationalise when it is
appropriate to use a pKa value.
Define the ionic product of water, Kw. Copy the equation to calculate pH from [H+
]
and then rearrange it to make [H+
] the
subject.
Summarise how to determine the pH for a
strong acid and a strong base.
START AT THE BOTTOM
WORK TOWARDS THE TOP

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