pH and pOH of Strong Acids and Bases: Study Guide (Chemistry 8.2)

What makes an acid or base “strong”

In Topic 8.1 you learned to move between [H₃O⁺], [OH⁻], pH and pOH. The missing piece was: how do you know [H₃O⁺] in the first place? For strong acids and strong bases, the answer is simple, because they react completely with water.

A strong acid donates its proton to water so completely that, at equilibrium, practically no acid molecules are left. For nitric acid:

HNO₃(aq) + H₂O(l) → H₃O⁺(aq) + NO₃⁻(aq)

A single arrow is used because the reaction goes to completion. Every HNO₃ molecule becomes one H₃O⁺ ion and one NO₃⁻ ion (the conjugate base). So:

For a strong acid HA at concentration c: [H₃O⁺] = c and [A⁻] = c.

The six strong acids to know are:

Strong acid	Formula	Conjugate base
hydrochloric acid	HCl	Cl⁻
hydrobromic acid	HBr	Br⁻
hydroiodic acid	HI	I⁻
perchloric acid	HClO₄	ClO₄⁻
nitric acid	HNO₃	NO₃⁻
sulfuric acid	H₂SO₄	HSO₄⁻

Sulfuric acid needs one comment. Its first proton ionises completely. The HSO₄⁻ ion left behind is itself a weak acid, so it gives up only part of its second proton. In this topic, treat [H₃O⁺] as equal to the H₂SO₄ concentration unless told otherwise; just remember that the true value is higher, by an amount that depends on the concentration. Most calculations use the five monoprotic acids for this reason.

In this course, treat an acid that is not on the list, such as HF or ethanoic acid, as weak unless a question says otherwise (Topic 8.3).

Strong bases

The common strong bases are the hydroxides of group 1 and group 2 metals. They are ionic, and they dissociate completely when they dissolve:

NaOH(s) → Na⁺(aq) + OH⁻(aq) Ba(OH)₂(s) → Ba²⁺(aq) + 2 OH⁻(aq)

Count the hydroxide ions in the formula:

• Group 1 hydroxide, MOH, at concentration c: [OH⁻] = c. (LiOH, NaOH, KOH, …)
• Group 2 hydroxide, M(OH)₂, at concentration c: [OH⁻] = 2c. (Ca(OH)₂, Sr(OH)₂, Ba(OH)₂)

Some group 2 hydroxides are only slightly soluble (Mg(OH)₂ barely dissolves at all, and Ca(OH)₂ only to about 0.02 M). That limits how much can dissolve, but whatever does dissolve is fully dissociated. The concentrations used in questions are always ones that can dissolve.

Strong is not the same as concentrated

“Strong” and “weak” describe how completely an acid ionises. “Concentrated” and “dilute” describe how much acid is dissolved per litre. A 0.0010 M HCl solution is dilute but strong. Its pH (3.00) is set by how little acid there is, not by any lack of ionisation.

The method

1. Decide whether the solute is a strong acid or a strong base.
2. Write the ionisation or dissociation equation and read off the mole ratio.
3. Find [H₃O⁺] (acid) or [OH⁻] (base) directly from the solute concentration. If you are given a mass or a dilution, find the concentration first.
4. Take −log to get pH (acid) or pOH (base).
5. Use pH + pOH = 14.00 and K_w = [H₃O⁺][OH⁻] at 25 °C to find the other two quantities.
6. If asked for all species, list the ions from the solute, the minor ion from K_w, and note that intact strong acid molecules are essentially zero.

Why water’s own ions can usually be ignored

Pure water supplies only 1.0 × 10⁻⁷ M H₃O⁺, and in an acid solution even less, because the acid pushes the water equilibrium to the left. For any strong acid at 10⁻⁶ M or more, the acid provides at least ten times as much H₃O⁺ as water ever could, so [H₃O⁺] = c is an excellent approximation. The same applies to [OH⁻] in strong base solutions.

Estimating pH as a check

Before you trust a calculator answer, estimate it. Write the concentration as a number between 1 and 10 times a power of ten. The pH lies between the two whole numbers on either side:

• 0.0250 M HClO₄ is 2.5 × 10⁻² M, which lies between 10⁻² and 10⁻¹ M. So the pH is between 1 and 2.
• 5 × 10⁻⁴ M HBr lies between 10⁻⁴ and 10⁻³ M, so the pH is between 3 and 4 (it is 3.30).
• For a base, estimate the pOH the same way, then subtract from 14. A base with [OH⁻] = 0.03 M has pOH between 1 and 2, so its pH is between 12 and 13.

A useful anchor: a factor of 2 in concentration changes pH by about 0.30, and a factor of 5 by about 0.70. If your answer falls outside the estimated range, look for a missing factor of 2, a sign error, or pH and pOH swapped.

Worked example 1: every species in a strong acid

Question. Calculate the pH and pOH of 0.0250 M perchloric acid, HClO₄, at 25 °C, and give the concentration of every species present apart from water.

1. HClO₄ is a strong acid: HClO₄(aq) + H₂O(l) → H₃O⁺(aq) + ClO₄⁻(aq). The ratio is 1 : 1 : 1.
2. [H₃O⁺] = 0.0250 M.
3. pH = −log(0.0250) = 1.602 (3 significant figures in 0.0250 → 3 decimal places).
4. pOH = 14.00 − 1.602 = 12.398.
5. [OH⁻] = K_w ÷ [H₃O⁺] = 1.0 × 10⁻¹⁴ ÷ 0.0250 = 4.0 × 10⁻¹³ M.

Species	Concentration (M)	Where it comes from
H₃O⁺	0.0250	complete ionisation of HClO₄
ClO₄⁻	0.0250	conjugate base, same amount as H₃O⁺
OH⁻	4.0 × 10⁻¹³	water equilibrium, set by K_w
HClO₄	≈ 0	almost none remains un-ionised

Check. The charges balance: positive ions (0.0250 M H₃O⁺) match negative ions (0.0250 M ClO₄⁻ plus a negligible amount of OH⁻).

Worked example 2: a group 2 hydroxide from a mass

Question. A student dissolves 0.450 g of barium hydroxide, Ba(OH)₂, in water and makes the solution up to 500.0 mL at 25 °C. Calculate [OH⁻], pOH, pH and [H₃O⁺].

1. Molar mass: M = 137.33 + 2(16.00 + 1.008) = 171.35 g mol⁻¹.
2. Amount: n = 0.450 g ÷ 171.35 g mol⁻¹ = 2.626 × 10⁻³ mol.
3. Concentration: c = 2.626 × 10⁻³ mol ÷ 0.5000 L = 5.253 × 10⁻³ M.
4. Ba(OH)₂ → Ba²⁺ + 2 OH⁻, so [OH⁻] = 2c = 0.0105 M.
5. pOH = −log(0.010505) = 1.979.
6. pH = 14.00 − 1.979 = 12.021.
7. [H₃O⁺] = 1.0 × 10⁻¹⁴ ÷ 0.010505 = 9.5 × 10⁻¹³ M.

Why step 4 matters. If you forget the factor of 2, you get pOH = 2.280 and pH = 11.720. The error is log 2 ≈ 0.30 pH units, which is enough to lose the mark.

Worked example 3: making a solution of a chosen pH

Question. A technician needs 250.0 mL of nitric acid with pH 2.50 at 25 °C. The stock solution is 0.100 M HNO₃. What volume of stock should be diluted to 250.0 mL?

1. Target [H₃O⁺] = 10⁻²·⁵⁰ = 3.16 × 10⁻³ M. Because HNO₃ is strong, the target HNO₃ concentration is also 3.16 × 10⁻³ M.
2. Amount needed: n = 3.16 × 10⁻³ mol L⁻¹ × 0.2500 L = 7.91 × 10⁻⁴ mol.
3. Volume of stock: V = 7.91 × 10⁻⁴ mol ÷ 0.100 mol L⁻¹ = 7.91 × 10⁻³ L = 7.91 mL.

(The dilution formula c₁V₁ = c₂V₂ gives the same answer: V₁ = 3.16 × 10⁻³ × 250.0 ÷ 0.100.)

Interpretation. The stock has pH 1.00, and the target is 1.50 units higher. That is a dilution by a factor of 10¹·⁵⁰ ≈ 31.6, and 250.0 mL ÷ 31.6 = 7.91 mL. Every tenfold dilution of a strong acid raises the pH by exactly 1 unit, as long as the solution is not so dilute that water’s own ions matter.

A very dilute strong acid

What is the pH of 1.0 × 10⁻⁸ M HCl? Using pH = −log c gives 8.00, which says an acid solution is basic. That cannot be right. At this tiny concentration the acid supplies less H₃O⁺ than water’s own autoionization, so you can no longer ignore water. The true [H₃O⁺] is a little above 1.0 × 10⁻⁷ M, and the pH is just under 7 (about 6.98). You will not be asked to do this calculation, but you should be able to spot that pH = −log c fails here, and say why: adding an acid can never make water basic.

Common misconceptions

• “Strong means concentrated.” Strength is about the extent of ionisation. 1 × 10⁻⁴ M HCl is strong and dilute.
• Forgetting the 2 for group 2 hydroxides. [OH⁻] = 2c for Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂.
• Taking −log of a base concentration and calling it pH. −log[OH⁻] is pOH. Subtract from 14.00 to get pH.
• “There are still HCl molecules in hydrochloric acid.” Practically none remain; the solution contains H₃O⁺ and Cl⁻.
• “A strong acid solution contains no OH⁻.” A small amount is always present, set by K_w.
• Doubling for H₂SO₄ automatically. Only the first proton is fully ionised.
• Using pH = −log c for extremely dilute acid. Below about 10⁻⁶ M, water’s own H₃O⁺ cannot be ignored. An acid solution always has pH below 7 at 25 °C.
• Thinking dilution changes the amount of acid. Dilution lowers the concentration; the moles stay the same (c₁V₁ = c₂V₂).

Where this leads

Strong acids and bases are the simple case: complete reaction, one-step calculation. Next, Topic 8.3 deals with weak acids and bases, which only partly ionise, so you will need K_a, K_b and equilibrium tables. Later, strong acids and bases return as titrants in acid-base reactions and titration curves. Try the practice questions now, then use the revision notes and the checklist to consolidate.