Strong Acids

Spec Mapping — OCR H432 Module 5.1.3 — Acids, bases and buffers, sub-topic on strong acids: the definition of a strong acid as one that undergoes essentially complete dissociation in water, calculation of pH from concentration for monobasic strong acids ($[\text{H}^+] = [\text{HA}]$[H+]=[HA]), pH calculation for dibasic strong acids treated as fully diprotic ($[\text{H}^+] = 2[\text{HA}]$[H+]=2[HA]), dilution and mixing of strong acid solutions, and reverse calculations from pH to concentration (refer to the official OCR H432 specification document for exact wording). This lesson is the simplest of all the pH-calculation lessons because the equilibrium lies so far to the right that we can treat dissociation as complete — no $K_a$Ka​ expression is needed.

Strong-acid pH calculation is the gateway routine for OCR Module 5.1.3 because it uses only one equation (pH = $-\log_{10}[\text{H}^+]$−log10​[H+]) and one piece of chemistry (full dissociation gives $[\text{H}^+] =$[H+]= basicity × $[\text{HA}]$[HA]). Students sometimes find this lesson deceptively easy and then make basicity errors on the dibasic case (H₂SO₄) or volume-conversion errors on the mixing case. The aim of this lesson is to embed the routine so deeply that you can do strong-acid pH in your head for any monobasic example, and reach for the factor-of-2 reflex any time you see H₂SO₄. The contrast with weak-acid pH (lessons 4–5) is fundamental: there you need $K_a$Ka​ and a square-root approximation; here you bypass equilibrium expressions because the equilibrium is fully on the dissociated side.

Key Equation: $\text{pH} = -\log_{10}[\text{H}^+] \qquad [\text{H}^+] = b \cdot c$pH=−log10​[H+][H+]=b⋅c where $b$b is basicity (1 for HCl/HNO₃, 2 for H₂SO₄) and $c$c is the analytical (formal) concentration of the acid. For dilutions: $c_2 = c_1 V_1 / V_2$c2​=c1​V1​/V2​. For mixtures: $[\text{H}^+] = \frac{\sum n(\text{H}^+)}{V_{\text{total}}}$[H+]=Vtotal​∑n(H+)​.

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What is a strong acid?

A strong acid is one that undergoes essentially complete dissociation in aqueous solution. The ionisation equilibrium

$\text{HA(aq)} \rightarrow \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)}$HA(aq)→H+(aq)+A−(aq)

lies so far to the right that we use a single forward arrow instead of an equilibrium arrow. For a 0.100 mol dm⁻³ HCl solution, every HCl molecule effectively gives up its proton — the residual concentration of undissociated HCl is below detection. The equilibrium constant $K_a$Ka​ for HCl is around $10^7$107 mol dm⁻³, so by the dissociation expression $K_a = [\text{H}^+][\text{A}^-] / [\text{HA}]$Ka​=[H+][A−]/[HA], with $K_a$Ka​ this large $[\text{HA}]$[HA] must be effectively zero. This is the operational meaning of "strong": $K_a \gg [\text{HA}]$Ka​≫[HA], ionisation > 99.9 %.

Three reminders before we calculate anything:

1. "Strong" is not "concentrated." Concentration is moles per dm³; strength is the dissociation fraction. A dilute (0.001 mol dm⁻³) HCl is still a strong acid — fully dissociated — but the pH is high (pH 3.00) because [H⁺] is low. A concentrated (17 mol dm⁻³) glacial ethanoic acid is still a weak acid — only ~0.3 % dissociated — yet has pH ≈ 2.4. The two axes are orthogonal.
2. Strong acids in water all have effectively the same operational strength. This is the levelling effect of water — water levels the strength of any acid stronger than H₃O⁺ down to the strength of H₃O⁺ itself. So HCl, HBr, HI, HNO₃ and HClO₄ all appear equally strong in dilute aqueous solution.
3. Conjugate base strength. A strong acid has a very weak conjugate base. Cl⁻, NO₃⁻, ClO₄⁻ are essentially inert as Brønsted bases in water (they do not accept H⁺ from H₃O⁺ to any measurable extent). This is the inverse-strength rule: strong acid ⇒ weak conjugate base.

Common strong acids

Name	Formula	Basicity	Conjugate base	OCR-examined?
Hydrochloric acid	HCl	1	Cl⁻	Yes
Hydrobromic acid	HBr	1	Br⁻	Background only
Hydroiodic acid	HI	1	I⁻	Background only
Nitric acid	HNO₃	1	NO₃⁻	Yes
Sulfuric acid	H₂SO₄	2	HSO₄⁻ → SO₄²⁻	Yes (treat as fully dibasic)
Perchloric acid	HClO₄	1	ClO₄⁻	Background only

At OCR A-Level you must recognise HCl, HNO₃ and H₂SO₄ as strong, and treat H₂SO₄ as dibasic with both ionisation steps essentially complete. The second dissociation of sulfuric acid is in reality only partially complete (HSO₄⁻ has $K_a \approx 10^{-2}$Ka​≈10−2, marginally weak), but OCR explicitly tells you to assume it is fully diprotic for pH calculations.

Warning — hydrogen fluoride is weak. Do not assume all hydrogen halides are strong. HF has $K_a \approx 6 \times 10^{-4}$Ka​≈6×10−4 mol dm⁻³, making it a weak acid. The unusual O–H...F hydrogen-bonding network and the high bond enthalpy of H–F (565 kJ mol⁻¹) keep HF molecular at moderate concentrations.

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Strong monobasic acid — the master routine

For a strong monobasic acid HA at concentration $c$c:

$[\text{H}^+] = c \quad \Rightarrow \quad \text{pH} = -\log_{10}(c)$[H+]=c⇒pH=−log10​(c)

Worked example 1 — 0.100 mol dm⁻³ HCl

$[\text{H}^+] = 0.100 \text{ mol dm}^{-3}$[H+]=0.100 mol dm−3 $\text{pH} = -\log_{10}(0.100) = 1.00$pH=−log10​(0.100)=1.00

Worked example 2 — 2.35 × 10⁻² mol dm⁻³ HNO₃

$[\text{H}^+] = 2.35 \times 10^{-2} \text{ mol dm}^{-3}$[H+]=2.35×10−2 mol dm−3 $\text{pH} = -\log_{10}(2.35 \times 10^{-2}) = 1.63 \quad (2 \text{ dp})$pH=−log10​(2.35×10−2)=1.63(2 dp)

Worked example 3 — 4.50 × 10⁻⁴ mol dm⁻³ HCl

$[\text{H}^+] = 4.50 \times 10^{-4} \text{ mol dm}^{-3}$[H+]=4.50×10−4 mol dm−3 $\text{pH} = -\log_{10}(4.50 \times 10^{-4}) = 3.35 \quad (2 \text{ dp})$pH=−log10​(4.50×10−4)=3.35(2 dp)

The relationship is mechanical: for any strong monobasic acid, pH depends only on the analytical concentration of the acid, via a single logarithm. Once you have done a handful of examples the calculator routine becomes automatic.

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Strong dibasic acid — the factor-of-2 reflex

A strong dibasic acid such as H₂SO₄ donates two protons per molecule. Treating both dissociation steps as complete:

$\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}$H2​SO4​→2H++SO42−​

so $[\text{H}^+] = 2c$[H+]=2c, where $c$c is the concentration of H₂SO₄.

Worked example 4 — 0.0500 mol dm⁻³ H₂SO₄

$[\text{H}^+] = 2 \times 0.0500 = 0.100 \text{ mol dm}^{-3}$[H+]=2×0.0500=0.100 mol dm−3 $\text{pH} = -\log_{10}(0.100) = 1.00$pH=−log10​(0.100)=1.00

Note: 0.0500 mol dm⁻³ H₂SO₄ has the same pH as 0.100 mol dm⁻³ HCl, because each H₂SO₄ delivers two protons.

Subtlety beyond OCR: the rigorous pH of dilute H₂SO₄ is not exactly $-\log_{10}(2c)$−log10​(2c) at very low concentrations because the second ionisation (HSO₄⁻ → H⁺ + SO₄²⁻) is only partially complete. For OCR A-Level questions the "assume fully dibasic" instruction is given explicitly; for true precision below 0.001 mol dm⁻³ you would solve the second-stage equilibrium too.

flowchart TD
    A["Strong acid concentration c"] --> B{"Monobasic or dibasic?"}
    B -- "Monobasic HCl HNO3" --> C["[H+] = c"]
    B -- "Dibasic H2SO4" --> D["[H+] = 2 x c"]
    C --> E["pH = -log10[H+]"]
    D --> E
    E --> F["Quote pH to 2 dp"]

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Dilution and pH

When a strong acid is diluted, [H⁺] decreases in direct proportion to the dilution factor:

$c_1 V_1 = c_2 V_2 \quad \Rightarrow \quad c_2 = \frac{c_1 V_1}{V_2}$c1​V1​=c2​V2​⇒c2​=V2​c1​V1​​

Worked example 5 — 10× dilution of HCl

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.

$c_2 = \frac{0.200 \times 25.0}{250.0} = 0.0200 \text{ mol dm}^{-3}$c2​=250.00.200×25.0​=0.0200 mol dm−3 $\text{pH} = -\log_{10}(0.0200) = 1.70 \quad (2 \text{ dp})$pH=−log10​(0.0200)=1.70(2 dp)

The pH has risen by 1.00 (from 0.70 to 1.70) for a 10× dilution. Every 10-fold dilution of a strong acid raises the pH by exactly 1.00 unit, as long as we stay well above the $\sim 10^{-6}$∼10−6 mol dm⁻³ regime where water self-ionisation contributes.

Worked example 6 — mixing acid with water

50.0 cm³ of 0.100 mol dm⁻³ HNO₃ is mixed with 450 cm³ of water. Find the pH.

$n(\text{H}^+) = 0.100 \times 0.0500 = 5.00 \times 10^{-3} \text{ mol}$n(H+)=0.100×0.0500=5.00×10−3 mol $[\text{H}^+] = \frac{5.00 \times 10^{-3}}{0.500} = 0.0100 \text{ mol dm}^{-3}$[H+]=0.5005.00×10−3​=0.0100 mol dm−3 $\text{pH} = -\log_{10}(0.0100) = 2.00$pH=−log10​(0.0100)=2.00

The pH rose from 1.00 to 2.00 — exactly $\log_{10}(10) = 1.00$log10​(10)=1.00 unit for a 10× total-volume dilution.

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Mixing strong acids

If two strong-acid solutions are mixed, add the moles of H⁺ from each and divide by the total volume:

$[\text{H}^+]_{\text{mix}} = \frac{n_1(\text{H}^+) + n_2(\text{H}^+)}{V_1 + V_2}$[H+]mix​=V1​+V2​n1​(H+)+n2​(H+)​

Worked example 7 — HCl mixed with H₂SO₄

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.

$n(\text{H}^+)_{\text{HCl}} = 0.100 \times 0.0250 = 2.50 \times 10^{-3} \text{ mol}$n(H+)HCl​=0.100×0.0250=2.50×10−3 mol $n(\text{H}^+)_{\text{H}_2\text{SO}_4} = 2 \times 0.0400 \times 0.0750 = 6.00 \times 10^{-3} \text{ mol}$n(H+)H2​SO4​​=2×0.0400×0.0750=6.00×10−3 mol $n(\text{H}^+)_{\text{total}} = 8.50 \times 10^{-3} \text{ mol}$n(H+)total​=8.50×10−3 mol $V_{\text{total}} = 0.100 \text{ dm}^3$Vtotal​=0.100 dm3 $[\text{H}^+] = \frac{8.50 \times 10^{-3}}{0.100} = 0.0850 \text{ mol dm}^{-3}$[H+]=0.1008.50×10−3​=0.0850 mol dm−3 $\text{pH} = -\log_{10}(0.0850) = 1.07 \quad (2 \text{ dp})$pH=−log10​(0.0850)=1.07(2 dp)

Note the factor of 2 on the H₂SO₄ contribution — the most common mistake in this class of problem.

Worked example 8 — equal volumes

25.0 cm³ of 0.200 mol dm⁻³ HCl is mixed with 25.0 cm³ of 0.0800 mol dm⁻³ H₂SO₄. Find the pH.

$n(\text{H}^+)_{\text{HCl}} = 0.200 \times 0.0250 = 5.00 \times 10^{-3} \text{ mol}$n(H+)HCl​=0.200×0.0250=5.00×10−3 mol $n(\text{H}^+)_{\text{H}_2\text{SO}_4} = 2 \times 0.0800 \times 0.0250 = 4.00 \times 10^{-3} \text{ mol}$n(H+)H2​SO4​​=2×0.0800×0.0250=4.00×10−3 mol $n(\text{H}^+)_{\text{total}} = 9.00 \times 10^{-3} \text{ mol}$n(H+)total​=9.00×10−3 mol $V_{\text{total}} = 0.0500 \text{ dm}^3$Vtotal​=0.0500 dm3 $[\text{H}^+] = \frac{9.00 \times 10^{-3}}{0.0500} = 0.180 \text{ mol dm}^{-3}$[H+]=0.05009.00×10−3​=0.180 mol dm−3 $\text{pH} = -\log_{10}(0.180) = 0.74 \quad (2 \text{ dp})$pH=−log10​(0.180)=0.74(2 dp)

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Reverse calculations — pH to concentration

If a question gives pH and asks for acid concentration, invert via $[\text{H}^+] = 10^{-\text{pH}}$[H+]=10−pH, then relate $[\text{H}^+]$[H+] to acid concentration by dividing by basicity.

Worked example 9 — monobasic acid

A strong monobasic acid has pH = 2.45. Calculate its concentration.

$[\text{H}^+] = 10^{-2.45} = 3.55 \times 10^{-3} \text{ mol dm}^{-3}$[H+]=10−2.45=3.55×10−3 mol dm−3

Since the acid is monobasic and strong, $[\text{HA}] = [\text{H}^+] = 3.55 \times 10^{-3}$[HA]=[H+]=3.55×10−3 mol dm⁻³.

Worked example 10 — sulfuric acid from pH

A solution of H₂SO₄ has pH = 1.20. Calculate $[\text{H}_2\text{SO}_4]$[H2​SO4​].

$[\text{H}^+] = 10^{-1.20} = 6.31 \times 10^{-2} \text{ mol dm}^{-3}$[H+]=10−1.20=6.31×10−2 mol dm−3 $[\text{H}_2\text{SO}_4] = \frac{[\text{H}^+]}{2} = 3.16 \times 10^{-2} \text{ mol dm}^{-3}$[H2​SO4​]=2[H+]​=3.16×10−2 mol dm−3

Worked example 11 — preparing a target-pH solution

A chemist wants 250 cm³ of HCl(aq) at pH = 1.30. What mass of HCl ($M_r = 36.5$Mr​=36.5) is required?

$[\text{H}^+] = 10^{-1.30} = 5.01 \times 10^{-2} \text{ mol dm}^{-3}$[H+]=10−1.30=5.01×10−2 mol dm−3 $n(\text{HCl}) = 5.01 \times 10^{-2} \times 0.250 = 1.25 \times 10^{-2} \text{ mol}$n(HCl)=5.01×10−2×0.250=1.25×10−2 mol $m(\text{HCl}) = 1.25 \times 10^{-2} \times 36.5 = 0.457 \text{ g}$m(HCl)=1.25×10−2×36.5=0.457 g

In practice HCl is supplied as a concentrated aqueous solution; the chemist would dilute the stock rather than weigh out anhydrous HCl gas.

Worked example 12 — concentrated H₂SO₄: where the "fully dibasic" assumption strains

A 2.00 mol dm⁻³ solution of H₂SO₄ — applying the OCR "fully dibasic" rule blindly gives $[\text{H}^+] = 2 \times 2.00 = 4.00$[H+]=2×2.00=4.00 mol dm⁻³ and pH = $-\log_{10}(4.00) = -0.60$−log10​(4.00)=−0.60. In reality the second-stage dissociation HSO₄⁻ $\rightleftharpoons$⇌ H⁺ + SO₄²⁻ ($K_{a2} \approx 1.2 \times 10^{-2}$Ka2​≈1.2×10−2 mol dm⁻³) is suppressed by the very high [H⁺] already present (Le Chatelier — common-ion effect by the first-stage H⁺), so the actual $[\text{H}^+]$[H+] is closer to 2.4 mol dm⁻³ (pH $\approx -0.38$≈−0.38). For OCR pH calculations you still apply the "treat as 2c" rule, but you should be able to recognise the limitation when comparing to measured pH values from textbook tables. The lesson: the "fully dibasic" simplification is fine at typical lab concentrations (~0.01–0.5 mol dm⁻³), strains above 1 mol dm⁻³, and breaks completely below $10^{-3}$10−3 mol dm⁻³ where the second-stage equilibrium has plenty of room to lie further to the right.

Worked example 12b — practical implications of the H₂SO₄ "fully dibasic" rule

A textbook problem gives 0.0250 mol dm⁻³ H₂SO₄ and asks for the pH. Three different approaches:

Approach	$[\text{H}^+]$[H+] / mol dm⁻³	pH	Notes
OCR "fully dibasic" rule	$2 \times 0.0250 = 0.0500$2×0.0250=0.0500	1.30	Mark-scheme expected answer
Only first stage ionises	0.0250	1.60	Wrong — ignores significant second stage
Rigorous (with $K_{a2}$Ka2​ = 0.012)	0.0455 (numerically solved)	1.34	Most accurate; the OCR answer is within 0.04 of this

The lesson: at A-Level concentrations (typically 0.01–0.5 mol dm⁻³), the "fully dibasic" OCR rule gives an answer within 0.05 pH units of the rigorous value — well within experimental precision. The rule is therefore practically exact for the concentrations OCR examines, even though it is formally an approximation. Mark schemes credit the OCR rule explicitly; do not "improve" on it by attempting a rigorous treatment unless the question signals it.

Worked example 13 — logarithmic shifts under serial dilution

A student starts with 1.00 mol dm⁻³ HCl and performs five successive 10-fold dilutions (each time taking 10.0 cm³ and diluting to 100.0 cm³). Tabulate the pH at each stage and comment on the trend.