chemistry 5632650 640 Calculating pH of strong acids and strong bases (A level Chemistry)

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 Clโ€“(aq) ions as follows:

HCl(aq) โ†’
H+(aq) + Clโ€“(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 โ€“ 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 OHโ€“ ions.

The fact that a strong base completely dissociates in solution means that the concentration of OHโ€“ ions in solution is the same as the concentration of the acid, assuming that the concentration of OHโ€“ 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 OHโ€“ ions in the solution.
  • equilibrium expression for the ionisation of water Kw = [H+][OHโ€“], to be calculate to deduce [H+] if we are given [OHโ€“].
  • the value of Kw for water.

Example 2 โ€“ How to calculate the pH of a strong base given [OHโ€“]

Calculate the pH of a solution of sodium hydroxide of concentration 0.0500mol/dm3. (Kw = 1.00 ร— 10โ€“14mol2dmโ€“6 at 298K).

Solution โ€“ using the first method

First determine [H+] using Kw = [H+][OHโ€“]

$$[H^+] = \frac{K_w}{[OH^โ€“]}$$

$$ = \frac{1.00 ร— 10^{โ€“14}}{0.0500}$$

$$ = 2.00 ร— 10^{โ€“13} mol/dm^{3}$$

Then calculate pH

pH = โ€“log10 [H+]

= โ€“log10 (2.00 ร— 10โ€“13)

= 12.7

Solution โ€“ 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 = โ€“log10 [OHโ€“]

= โ€“log10(0.0500)

= 1.3

pH = 14 โ€“ pOH

= 14 โ€“ 1.3

= 12.7