pH temperature
At what temperature is the pH of water exactly 7?
Problem
Water naturally dissociates into an equilibrium mixture of $H^+$ and $OH^-$ ions and $H_2O$ molecules
$$
H_2O \rightleftharpoons^{K_W} H^++OH^-\,,
$$
where the concentrations of $H^+$ and $OH^-$ ions, written as $[H^+]$ and $[OH^-]$ are related by the expression
$$
K_W = [H^+][OH^-].
$$
$K_W$ is called the dissociation constant, and depends on the temperature of the water.
The following table of data shows the dissociation constant for water at various temperatures and standard pressure.
Water temperature | $\quad K_W\times10^{14}\quad$ |
$0^\circ$ C | 0.1 |
$10^\circ$ C | 0.3 |
$18^\circ$ C | 0.7 |
$25^\circ$ C | 1.2 |
$30^\circ$ C | 1.8 |
$50^\circ$ C | 8.0 |
$60^\circ$ C | 13 |
$70^\circ$ C | 21 |
$80^\circ$ C | 35 |
$90^\circ$ C | 53 |
$100^\circ$ C | 73 |
From this table, work out an estimate for the temperature at which water has a $pH$ of exactly 7, 6.8 and 7.2. Recall that the $pH$ is defined as $pH=-\log_{10}([H^+])$
Getting Started
How is pH defined in terms of the concentration of $H^+$ ions?
Student Solutions
The initial key to this problem is to realise that for self-dissociating water, [H]$^+$ = [OH]$^-$.
Therefore, K$_W = [H^+]^2$
Since pH = $-log_{10}[H^+]$
$\mathbf{\Rightarrow K_W = 10^{-2pH}}$
When pH = 7; $K_W = 1 \times 10^{-14}$
pH =6.8; $K_W = 2.51 \times 10^{-14}$
pH = 7.2; $K_W = 0.398 \times 10^{-14}$
Plotting a graph of $K_W$ versus temperature gives:
By drawing a smooth curve of best fit through the points, the relevant temperatures can be read off the graph for the given values of $K_W$. If you draw the graph for the entire temperature range, it is difficult to read off accurate values in the required range, which are all smaller than $10\times 10^{14}$. Plotting the curve through the first three points allows us to read off accurate values.
Teachers' Resources
Why do this problem?
This problem involves the application of logarithms to pH. This interesting application, in which the critical value of pH 7 is familiar, will give a good motivation for engaging with the logarithm.
Possible approach
Key questions
Possible extension
Possible support