Title: Lesson 6: pH, pOH and the Ionic
Product of Water
Learning Objectives:
– Understand the ionic product of water and use it calculate H+ and OHconcentrations
– Calculate pH
– Calculate pOH
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
Ethanoic acid, CH3COOH, is a weak acid.
a.
Define the term weak acid and state the equation for the
reaction of ethanoic acid with water.
b.
Vinegar, which contains ethanoic acid, can be used to clean
deposits of calcium carbonate from the elements of electric
kettles. State the equation for the reaction of ethanoic acid
with calcium carbonate.
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Relationship between strength and concentration



Acids and bases differ in strength according to the equilibrium position of
their ionisation reactions.
Different concentrations of aqueous solution according to the ratio of acid
or base to water used.
Both factors influence the pH of a solution.
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The ionic product of water, Kw


Water can be both an acid and a base, this leads to the following equilibrium:

2 H2O

H3O+ + OH-

H2O

H+ + OH-
OR MORE SIMPLY...
The equilibrium for this reaction is called the ionic product of water and has
the symbol Kw (at 298K):


KW  [ H ][OH ]  110




14
Dissociation of water is ENDOTHERMIC (bond breaking)
So increase in temperature will shift the equilibrium to the right and the Kw
will increase (Le Chatelier’s Principle)
This will cause an increase in the concentrations of H+(aq) and OH-(aq) and a
decrease in pH
A reduction in temperature will cause the opposite effect...
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Summary of effect of temperature on Kw

pH of water is only 7 at 298K (25oC)

Note that at temperatures above and below this, despite changes in pH, water is
still a neutral substance as its H+(aq) concentration is equal to its OH-(aq)
concentration. (Neither acidic or basic!)

Temperature should always be stated alongside pH measurements as Kw is
temperature dependent!
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Calculating [H+] and [OH-] of pure water:

Kw = [H+][OH-] = 1.00x10-14 mol2 dm-6

Since when pure, [H+] = [OH-]



[H+]2 = 1.00x10-14
[H+] = √1.00x10-14 = 1.00x10-7 mol dm-3
Kw varies with temperature (because it’s an equilibrium constant):
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

At 273 K, Kw = 1.14x10-15, calculate [H+]
At 373 K, Kw = 5.13x10-13, calculate [OH-]
Do these changes mean the self-dissociation of water is endothermic or exothermic?
Justify your answer.
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Calculating [H+] and [OH-] of strong acids

Must use the equilibrium

For example, what is the concentration of OH- in 2.00 mol dm-3 sulphuric acid
solution?

Calculate H+ from the data given in the question, there is a small effect from the equilibrium
but it can be ignored.

Since H2SO4 produces two protons:


[H+] = 2 x 2.00 = 4.00 mol dm-3
Now we use the ionic product of water:
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
[H+][OH-] = 1.00x10-14
[OH-]
= 1.00x10-14 / [H+] = 1.00x10-14 / 4.00
= 2.50x10-15 mol dm-3
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Calculating [H+] and [OH-] of strong bases

Must use the equilibrium

For example, what is the concentration of H+ in
hydroxide acid solution?
0.150 mol dm-3 sodium

Calculate OH- from the data given in the question, there is a small effect from the equilibrium
but it can be ignored.

Since NaOH only produces one hydroxide:


[OH-] = 0.150 mol dm-3
Now we use the ionic product of water:


[H+][OH-] = 1.00x10-14
[H+]
= 1.00x10-14 / [OH-] = 1.00x10-14 / 0.150
= 6.67x10-14 mol dm-3
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pH and pOH

You are familiar with the pH scale, based on [H+]:
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
0 (very strong acid)  7 (neutral)  14 (very strong alkali)
There is an analogue called pOH based on [OH-]:
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0 (very strong alkali)  7 (neutral)  14 (very strong acid)
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pH and pOH scales are inter-related
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pH scales allow for simplified expression of H+ concentration in a solution.
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Like H+ ions, OH- ions are often present in low concentrations so negative exponents can be
awkward to work with.
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The parallel scale known as pOH scale can describe the OH- content of solutions


pOH = -log10[OH-]
[OH-] = 10-pOH
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REMEMBER: Change in one unit of pH or pOH represents a 10x change in [H+] or [OH-]
pH = -log10[H+]
[H+] = 10-pH
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
pOH = -log10[OH-]
pH = -log10[H+]

[OH-] = 10-pOH
[H+] = 10-pH

[H+][OH-] = Kw = 1.00 x 10-14
(at 298K), it follows that:

10-ppH x 10-ppOH = 1.00 x 10-14
(at 298K)

By taking the negative logarithm to base 10 of both sides, we get

pH + pOH = 14.00
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Relationship between [H+], [OH-], pH, and pOH at 298K:
(at 298K)
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
In the same way as the negative logarithms to base 10 of H+ and OH- are known as
pH and pOH respectively, the same can be applied to Kw to derive pKw

pKw = -log10(Kw)

Kw = 10-pKw
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So we can rewrite the expression above in the form that will apply to all
temperatures:
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pH + pOH = pKw
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Converting H+ and OH- into pH and pOH
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Strong acids and bases: pH and pOH can be deduced from
their concentrations
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
We assume full dissociation for strong acids and bases.
We can deduce the ion concentrations and so calculate the pH or pOH directly from the initial
concentration of solution
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Calculating the pH of a strong base
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Calculations of strong base pH
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pH calculations summary
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pH calculations
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Calculating pH and pOH of strong acids
pH = -log10[H+]
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
AND
pOH = -log10[OH-]
For example, what is the concentration of pH and pOH of 2.00 mol dm-3 sulphuric acid
solution?

Step 1: pH = -log10[H+] = -log10(4.00) = -0.602
pH + pOH = 14.00

Step 2: pOH = -log10[OH-] = -log10(2.5x10-15)= 14.6
[OH-] = 10-14.6 = 2.5x10-15
Note: pH + pOH = 14*…..this allows us to take a short cut:

pOH = 14 – pH = 14 – (-0.602) = 14.6

pH = 14 – pOH = 14 – 14.6 = -0.602
*This ‘14’ is known as pKw,
i.e. –log10(1.00x10-14)
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Calculating pH and pOH of strong bases
pH = -log10[H+]


AND
pOH = -log10[OH-]
For example, what is the concentration of pH and pOH of 0.150 mol dm-3 sodium
hydroxide acid solution?

Step 1: pOH = -log10[OH-] = -log10(0.150)= 0.824
pH + pOH = 14.00

Step 2: pH = -log10[H+] = -log10(6.67x10-14) = 13.2
[H+] = 10-13.2 = 6.67x10-14
Note: pH + pOH = 14*…..this allows us to take a short cut:

pOH = 14 – pH = 14 – (13.2) = 0.824

pH = 14 – pOH = 14 – 0.824 = 13.2
*This ‘14’ is known as pKw,
i.e. –log10(1.00x10-14)
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Solutions
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Calculating pH and pOH

You will need to calculate the pH of a variety of mixtures of solutions and
then make those solutions to test them.
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Instructions can be found here
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Calculating [H+] and [OH-]
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Very simple:
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[H+] = 10-pH
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For example; solution of pH 6.2

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[H+] = 10-6.2 = 6.3x10-7 mol dm-3
[OH-] = 10-pOH

For example; solution of pH 6.2


pOH = 14 - 6.2 = 7.8
[OH-] = 10-7.8 = 1.6x10-8 mol dm-3
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Key Points

Kw = [H+][OH-] = 1.00x10-14

pH = -log10[H+] = 14 - pOH

pOH = -log10[OH-] = 14 - pH
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