**Contents**

## Calculating pH of strong acids

A **strong acid** is a substance that completely dissociates in aqueous solution to produce H^{+} ions.

The fact that a strong acid completely dissociates in solution means that the concentration of hydrogen ions in solution is the same as the concentration of the acid, assuming that the concentration of H^{+} ions arising from the ionisation of water molecules is negligible.

HCI is an example of a strong acid because it completely dissociates in aqueous solution to produce H^{+}(aq) ions and Cl^{–}(aq) ions as follows:

HCl(aq) →

H^{+}(aq) + Cl^{–}(aq)

So, for example, if we have 1 mol/dm^{3} HCl it follows that [H^{+}(aq)] = 1 mol/dm^{3} too.

pH = -log[H+(aq)]

- = -log 1
- = 0

Therefore pH of 1 mol/dm^{3} HCl = 0.

### Example 1 – How to calculate pH of a strong acid given [H^{+}]

Calculate the pH of a 0.16 mol/dm^{3} solution of HCl.

### Solution

We have 0.16 mol/dm^{3} HCl it follows that [H^{+}(aq)] = 0.16 mol/dm^{3} too.

pH = -log[H+(aq)]

- = -log 0.16

**pH = 0.80** to 2 d.p.

## Calculating the pH of strong bases

A **strong base** is a substance that completely dissociates in aqueous solution to produce OH^{–} ions.

The fact that a strong base completely dissociates in solution means that the concentration of OH^{–} ions in solution is the same as the concentration of the acid, assuming that the concentration of OH^{–} ions arising from the ionisation of water molecules is negligible.

To calculate the pH of a solution of strong base we need to know the:

- concentration of H
^{+}or OH^{–}ions in the solution. - equilibrium expression for the ionisation of water K
_{w}= [H^{+}][OH^{–}], to be calculate to deduce [H^{+}] if we are given [OH^{–}]. - the value of K
_{w}for water.

### Example 2 – How to calculate the pH of a strong base given [OH^{–}]

Calculate the pH of a solution of sodium hydroxide of concentration 0.0500mol/dm^{3}. (K_{w} = 1.00 × 10^{–14}mol^{2}dm^{–6} at 298K).

### Solution – using the first method

First determine [H^{+}] using K_{w} = [H^{+}][OH^{–}]

$$[H^+] = \frac{K_w}{[OH^–]}$$

$$ = \frac{1.00 × 10^{–14}}{0.0500}$$

$$ = 2.00 × 10^{–13} mol/dm^{3}$$

Then calculate pH

pH = –log10 [H^{+}]

= –log10 (2.00 × 10–13)

= 12.7

### Solution – using the second method

An alternative method of calculating pH is by first calculating pOH and then subtracting it from 14, since pH + pOH = 14.

pOH = –log_{10} [OH^{–}]

= –log_{10}(0.0500)

= 1.3

pH = 14 – pOH

= 14 – 1.3

= 12.7