The strength of an acid (HA) is determined by its ability to donate protons or H+ ions to water.
The [H+(aq)] varies from 1 x 10-1 mol dm-3 to 1 x 10-10 mol dm-3 for acids so it is more convenient to use the pH scale. Since this scale is logarithmic, moving 1 point on the pH scale is equivalent to a tenfold change in [H+(aq)].
pH = – log10 [H+(aq)] and [H+(aq)] = 10 -pH
Example: calculate the pH of a solution with a [H+(aq)] of 2.63 x 10-3 mol dm-3
pH = – log[H+(aq)]
= – log (2.63 x 10-3) = 2.58
Example: what is the [H+(aq)] of a solution that has a pH of 3.70
[H+(aq)] = 10-pH
= 10-3.70 = 2.00 x 10-4 mol dm-3
Strong acids
The concentration of a strong acid = [H+(aq)] because we are assuming that the acid, HA, is fully dissociated in solution, which makes finding the pH of a strong acid very straightforward 😊.
Example: find the pH of 0.01 mol dm-3 HCl
HCl(aq) ⇾ H+(aq) + Cl–(aq)
1 : 1
0.01 : 0.01
pH = – log [H+(aq)] = – log (0.01) = 2
Weak acids
Since weak acids are only partially dissociated, we cannot assume that the concentration of the acid = [H+(aq)]. We need to know where the equilibrium position lies in order to determine [H+(aq)].
In using this expression, either to determine Ka for an acid (given the pH and concentration of the solution) or to determine the pH of a weak acid (given the value for Ka and the concentration of the acid), we always make TWO assumptions.
- When the acid, HA, dissociates, the concentration of H+ ions equals the concentration of the anion A–(aq) and that the dissociation of water producing H+ ions is negligible.
- For a weak acid the actual number of acid molecules, HA, that dissociate is so small (the Ka for ethanoic acid is 1.7 x 10-5 mol dm-3 , for example) that we can assume that the concentration of HA at equilibrium is unchanged from the initial HA concentration. Clearly this assumption holds less true for moderately weak acids than for very weak acids.
Ka values are influenced by temperature but not concentration of the acid so Ka is an accurate measure of the extent of dissociation of an acid and therefore the strength of that acid.
Concentration and strength are NOT the same thing! Be careful with your language.
The larger the Ka value for an acid, the greater the [H+(aq)] and the stronger the acid (position of equilibrium lies further t the right). Once again, the range in Ka values for acids is huge, so a logarithmic scale is often used …
pKa = – log10 Ka
(a low Ka will correspond to a high pKa and vice versa).
| Ka / mol dm-3 | pKa | ||
| chloric(III) acid | HClO2 | 1.1 x 10-2 | 1.94 |
| methanoic acid | HCO2H | 1.8 x 10-4 | 3.75 |
| phenol | C6H5OH | 1.1 x 10-10 | 9.99 |
Example: calculate the Ka for weak acid solution given that the concentration is 0.100 mol dm-3 and the pH is 5.10 at 25°C.
The assumption that the concentration of the acid at equilibrium is the same as it was initially holds less true for stronger weak acids because the stronger the weak acid, the greater the dissociation.
Example: Determine the pH of 0.100 mol dm-3 ethanoic acid at 25°C, given that its Ka is 1.7 x 10-5 mol dm-3.
Practice questions
- (a) Write an equation to show the reaction of ethanoic acid with water.
(b) Calculate the pH of a 0.03 mol dm-3 ethanoic acid solution, given that the Ka = 1.70 x 10-5 mol dm-3
- In aqueous solution, boric acid dissociates into ions in three steps. The equation below shows the first:
H3BO3 (aq) ⇌ H+ (aq) + H2BO3– (aq)
(i) Write an equation to show the second step.
(ii) The pKa for step 1 is 9.24. Calculate the pH of a 0.075 mol dm-3 solution of boric acid.
(iii) State any assumptions you made in carrying out the calculation in (b).
Answers
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