# pH temperature

At what temperature is the pH of water exactly 7?

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 $pH=-\log_{10}([H^+])$

How is pH defined in terms of the concentration of $H^+$
ions?

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:

Image

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.

### 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

To do this question, students will need to understand the
precise mathematical meaning of both the dissociation constant and
the pH (both are clearly described in the question, so no
particular knowledge of chemistry is required). Initially students
might need to think on how they can connect the various pieces of
information.

### Key questions

What equations will you need to use to solve the
problem?

How can you use the table of data in the question?

Why is an approximation in the question needed? At what point
in a question will an approximation be made?

### Possible extension

Students will need to interpolate the data in the table to be
able to provide an approximation. Can they suggest
differentpossibilities for this interpolation? Which one is
best?

The next question in this series is
extreme dissociation

### Possible support

Ask the students to work out the pH at the different
temperatures given in the table.