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PROPERTIES OF LOWRY–BRØNSTED ACIDS AND BASES 249

Worked Example 6.5 What is the pH of a solution of sodium hydroxide of concentra-
−3
tion 0.02 mol dm ? Assume the temperature is 298 K.

At first sight, this problem appears to be identical to those in previ-
ous Worked Examples, but we soon appreciate how it is complicated
because we need first to calculate the concentration of the free protons pK w =− log 10 K w
before we can convert to a pH. However, if we know the concentration pH =− log [H 3 O + ]
of the alkali, we can calculate the pH thus: 10 (aq)
−
pOH =− log [OH (aq) ]
10
pK w = pH + pOH (6.22)
where pK w and pH have their usual definitions, and we define pOH as
−
pOH =− log [OH ] (6.23) We must look up the
10
value of K w from Table
Inserting the concentration [NaOH] = 0.02 mol dm −3 into Equation 6.2 if the temperature
(6.22) yields the value of pOH = 2. The value of pK w at 298 K is 14 differs from 298 K,
(see Table 6.2). Therefore: and then calculate a
different value using
pH = pK w − pOH
Equation (6.26).
pH = 14 − 2
pH = 12

SAQ 6.6 What is the pH of a solution of potassium hydroxide of concen-
tration 6 × 10 −3 mol dm −3 . Again, assume the temperature is 298 K.

Justification Box 6.2

Water dissociates to form ions according to Equation (6.2). The ionic product of the
concentrations is the autoprotolysis constant K w , according to Equation (6.4).
Taking logarithms of Equation (6.4) yields

−
+
log 10 K w = log [H 3 O ] + log [OH ] (6.24)
10
10
Next, we multiply each term by ‘−1’ to yield
+
−
− log 10 K w =− log [H 3 O ] − log [OH ] (6.25)
10
10
+
The term ‘− log [H 3 O ]’ is the solution pH and the term ‘− log K w ’is defined
10 10
according to
pK w =− log (6.26)
10 K w
−
We give the name pOH to the third term ‘− log [OH ]’. We thereby obtain Equa-
10
tion (6.22).