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A strong acid is one that is essentially fully dissociated in aqueous solution. The equilibrium:
HA(aq) -> H+(aq) + A-(aq)
lies so far to the right that we write it with a single arrow. For every mole of HA dissolved, very nearly one mole of H+ and one mole of A- are produced.
Contrast this with a weak acid, where only a few percent (or less) of molecules dissociate and the equilibrium arrow is <=>.
Being "strong" is not the same as being "concentrated". Concentration refers to how much acid is present per unit volume; strength refers to the fraction that dissociates. A dilute strong acid and a concentrated weak acid can have similar pH.
| Name | Formula | Basicity | Conjugate base |
|---|---|---|---|
| Hydrochloric acid | HCl | 1 | Cl- |
| Hydrobromic acid | HBr | 1 | Br- |
| Hydroiodic acid | HI | 1 | I- |
| Nitric acid | HNO3 | 1 | NO3- |
| Sulfuric acid | H2SO4 | 2 | HSO4- / SO4^2- |
| Perchloric acid | HClO4 | 1 | ClO4- |
At OCR A-Level you are expected to recognise HCl, HNO3 and H2SO4 as strong, and treat H2SO4 as dibasic (donating two protons per molecule, fully for the first step and essentially fully for the second at A-Level precision).
Note that HF is a weak acid despite being a halogen hydride. Do not assume all hydrogen halides are strong.
For a strong monobasic acid of concentration c:
[H+] = c (since dissociation is complete and each HA gives one H+) pH = -log10(c)
Worked Example 1: Calculate the pH of 0.100 mol dm-3 HCl.
[H+] = 0.100 mol dm-3 pH = -log10(0.100) = 1.00
Worked Example 2: Calculate the pH of 2.35 x 10^-2 mol dm-3 HNO3.
[H+] = 2.35 x 10^-2 mol dm-3 pH = -log10(2.35 x 10^-2) = 1.63 (2 dp)
Worked Example 3: Calculate the pH of 4.50 x 10^-4 mol dm-3 HCl.
[H+] = 4.50 x 10^-4 mol dm-3 pH = -log10(4.50 x 10^-4) = 3.35 (2 dp)
The relationship is beautifully simple: for a strong monobasic acid, pH depends only on the initial concentration.
A strong dibasic acid such as H2SO4 donates two protons per molecule. At A-Level we assume both steps are essentially complete:
H2SO4 -> 2H+ + SO4^2-
So [H+] = 2 x c.
Worked Example 4: Calculate the pH of 0.0500 mol dm-3 H2SO4.
[H+] = 2 x 0.0500 = 0.100 mol dm-3 pH = -log10(0.100) = 1.00
Note that 0.0500 mol dm-3 H2SO4 has the same pH as 0.100 mol dm-3 HCl because each H2SO4 delivers twice as many protons.
Real-world caveat: the second dissociation of H2SO4 has Ka2 around 10^-2, so at very low concentrations it is not fully ionised. OCR A-Level ignores this subtlety and treats H2SO4 as fully dibasic.
When a strong acid is diluted, [H+] decreases in proportion to the dilution factor. Recall from Year 1:
c1V1 = c2V2
so c2 = c1 x V1 / V2.
Worked Example 5: 25.0 cm3 of 0.200 mol dm-3 HCl is diluted to 250.0 cm3 with distilled water. Calculate the pH of the diluted solution.
c2 = (0.200 x 25.0) / 250.0 = 0.0200 mol dm-3 [H+] = 0.0200 mol dm-3 pH = -log10(0.0200) = 1.70 (2 dp)
The pH has risen by 1.00 (from 0.70 to 1.70) for a 10-fold dilution. Every 10-fold dilution raises the pH of a strong acid by 1.
Worked Example 6: 50.0 cm3 of 0.100 mol dm-3 HNO3 is mixed with 450 cm3 of water. Find the pH.
Total volume = 500 cm3. Moles of H+ = 0.100 x 0.0500 = 5.00 x 10^-3 mol. [H+] after dilution = 5.00 x 10^-3 / 0.500 = 0.0100 mol dm-3. pH = -log10(0.0100) = 2.00.
If two strong acids are mixed, add the moles of H+ and divide by the total volume.
Worked Example 7: 25.0 cm3 of 0.100 mol dm-3 HCl is mixed with 75.0 cm3 of 0.0400 mol dm-3 H2SO4. Find the pH of the mixture.
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