## 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/dm3 HCl it follows that [H+(aq)] = 1 mol/dm3 too.

pH = -log[H+(aq)]

• = -log 1
• = 0

Therefore pH of 1 mol/dm3 HCl = 0.

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

Calculate the pH of a 0.16 mol/dm3 solution of HCl.

### Solution

We have 0.16 mol/dm3 HCl it follows that [H+(aq)] = 0.16 mol/dm3 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 Kw = [H+][OH], to be calculate to deduce [H+] if we are given [OH].
• the value of Kw 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/dm3. (Kw = 1.00 × 10–14mol2dm–6 at 298K).

### Solution – using the first method

First determine [H+] using Kw = [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 = –log10 [OH]

= –log10(0.0500)

= 1.3

pH = 14 – pOH

= 14 – 1.3

= 12.7