# Extreme dissociation

In this question we push the pH formula to its theoretical limits.

The pH of a solution is defined using logarithms as

$$

pH = -\log_{10}[H^+],

$$

where $[H^+]$ is the concentration of $H^+$ ions in mol/l of the solution.

Imagine that this equation works for any concentration of $H^+$, and imagine that scientists have created conditions in a 100ml sample of pure water in which only one molecule of water has dissociated into $H^+$ and $OH^{-}$ (such conditions are, or course, science fiction!)

What would the pH of this fluid be?

Next imagine that conditions arise in which every single molecule of water dissociates.

What would the pH be now? What assumptions do you make in the calculation?

Under real-world situations, what volume of water would we expect to contain exactly one $H^+$ ion?

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NOTES AND BACKGROUND

This problem uses the concept of dissociation in which molecules of water $H_2O$ dissociate into $H^+$ and $OH^{-}$ ions. In reality, the pH formula given here fails to hold for weak acids or bases because not all of the water in the solution splits into $H^+$ and $OH^-$ ions. Only some of the water will dissociate; the dissociation constant tells you how many molecules split in this way.

Whilst the pH scale is usually quoted as running from 1-14, very strong acids can have zero or even negative pH values, running as low as -20 for superacids .

This problem uses the concept of dissociation in which molecules of water $H_2O$ dissociate into $H^+$ and $OH^{-}$ ions. In reality, the pH formula given here fails to hold for weak acids or bases because not all of the water in the solution splits into $H^+$ and $OH^-$ ions. Only some of the water will dissociate; the dissociation constant tells you how many molecules split in this way.

Whilst the pH scale is usually quoted as running from 1-14, very strong acids can have zero or even negative pH values, running as low as -20 for superacids .

Remember that the pH of water under standard conditions is 7.

1 molecule of water dissociated into $H^+$ and $OH^-$ in 100ml of water.

Thus in 1l of water, there will be 10 molecules of $H^+$.

$\therefore [H^+] = 10\ molecules/l$

By dividing by Avogadro's number, this gives:

$ [H^+] = 1.66 \times 10^{-23}\ mol/l$

$$\therefore pH = -log_{10}[H^+] = 22.78\ (2\ d.p.)$$

Under the condition that every molecule of water dissociates, we need to make an assumption that 100ml of water weighs approximately 100g.

Therefore 100ml of water is $\frac{100}{18} = 5.56$ moles

Each mole of water gives 1 mole of $[H^+]$.

Therefore $$[H^+] = \frac{5.56}{0.1} = 55.6 mol/l$$

$$\therefore pH = -log_{10}[H^+] = -1.74$$

Under real world situations, the pH of water is 7.

Therefore $[H^+] = 10^{-7} mol/l = 6.02 \times 10^{16} molecules /l$

Therefore there is a single $H^+$ in $\frac{1}{6.02 \times 10^{16}} = 1.66 \times 10^{-17} l$

### Why do this problem?

This problem provides a good exercise in linking two areas of mathematics: concentration and logarithms. Students will need to flip between the two descriptions of acidity which will require careful, clear thinking. The context is fascinating and this will encourage a 'concrete' and practical engagement with logarithms. It will also help to give some numerical intuition as to the concentrations involved in pH calculations.### Possible approach

This question uses the advanced concept of logarithms but will
also use basic ideas about concentrations and moles. Students will
need to remember basic ideas about the meaning of a 'concentration'
in order to work out the extreme concentrations given in the
problem.

### Key questions

What equations do you need to flip between the pH and the
concentration of $H^+$ ions?

What knowledge of physical chemistry is required to do this
problem?

### Possible extension

Try the final question in the sequence:
Blood Buffers

### Possible support

You might try and get started using the pH formula with the
problem Temperature
pH