How to Calculate pH from Concentration

pH calculations look intimidating because of the logarithm, but almost every question at school level uses one of five patterns. Learn to recognise the pattern, and the maths is two or three calculator presses.

This guide assumes you know what pH measures. If not, start with the pH scale explained.

The formulas you need

Quantity	Formula
pH from hydrogen ion concentration	pH = −log₁₀[H⁺]
[H⁺] from pH	[H⁺] = 10⁻ᵖᴴ
pOH from hydroxide concentration	pOH = −log₁₀[OH⁻]
Link between them at 25 °C	pH + pOH = 14.00
Ionic product of water at 25 °C	[H⁺][OH⁻] = 1.0 × 10⁻¹⁴

Square brackets mean “concentration in mol/dm³” (the same as mol/L or M).

Pattern 1: pH of a strong acid

Strong acids ionise completely, so [H⁺] equals the acid concentration (times the number of acidic protons, for acids like H₂SO₄ where your course treats both as fully released).

Method

1. Write [H⁺] from the acid concentration.
2. Press −log on your calculator.

Example 1. What is the pH of 0.010 mol/dm³ hydrochloric acid?

• [H⁺] = 0.010
• pH = −log(0.010) = 2.00

Example 2. What is the pH of 0.025 mol/dm³ nitric acid?

• [H⁺] = 0.025
• pH = −log(0.025) = 1.60

Example 3. What is the pH of 0.0050 mol/dm³ sulfuric acid, assuming both protons are fully released?

• [H⁺] = 2 × 0.0050 = 0.010
• pH = 2.00

(At more advanced levels the second proton of sulfuric acid is treated as only partly released, so the true pH is slightly higher. Follow whatever assumption your course states.)

Significant figures tip: the number of decimal places in a pH should match the number of significant figures in the concentration. 0.025 has two significant figures, so the pH is given to two decimal places: 1.60.

Pattern 2: [H⁺] from pH

Going backwards uses the inverse function, 10ˣ (often SHIFT + log on a calculator).

Example 4. Rainwater has a pH of 5.6. What is [H⁺]?

• [H⁺] = 10⁻⁵·⁶ = 2.5 × 10⁻⁶ mol/dm³

Example 5. A sample of blood has a pH of 7.40. What is [H⁺]?

• [H⁺] = 10⁻⁷·⁴⁰ = 4.0 × 10⁻⁸ mol/dm³

A quick sense check: a pH between 5 and 6 must give [H⁺] between 10⁻⁵ and 10⁻⁶. If your answer isn’t in that range, you’ve pressed the wrong button.

Pattern 3: pH of a strong base

For a strong base you know [OH⁻], not [H⁺]. There are two equally good routes.

Route A (via pOH)

1. [OH⁻] from the base concentration.
2. pOH = −log[OH⁻]
3. pH = 14.00 − pOH

Route B (via Kw)

1. [OH⁻] from the base concentration.
2. [H⁺] = 1.0 × 10⁻¹⁴ ÷ [OH⁻]
3. pH = −log[H⁺]

Example 6. What is the pH of 0.050 mol/dm³ sodium hydroxide?

• [OH⁻] = 0.050
• pOH = −log(0.050) = 1.30
• pH = 14.00 − 1.30 = 12.70

Example 7. What is the pH of 0.020 mol/dm³ barium hydroxide, Ba(OH)₂?

• Each formula unit releases two OH⁻, so [OH⁻] = 0.040
• pOH = −log(0.040) = 1.40
• pH = 12.60

Forgetting the “×2” for Group 2 hydroxides is one of the most common exam errors. Our guide to pOH and pH goes through more base examples.

Pattern 4: dilution

Diluting doesn’t change the number of moles of H⁺, only the volume they’re spread through.

Method

1. Find the new concentration: c₂ = c₁V₁ ÷ V₂ (the same C₁V₁ = C₂V₂ you use for molarity).
2. Calculate pH as usual.

Example 8. 10.0 cm³ of 0.50 mol/dm³ HCl is diluted to 250 cm³. What is the new pH?

• Moles of H⁺ = 0.50 × 0.0100 = 0.0050 mol
• New concentration = 0.0050 ÷ 0.250 = 0.020 mol/dm³
• pH = −log(0.020) = 1.70

Useful shortcut: diluting a strong acid by a factor of 10 raises its pH by exactly 1. Diluting a strong base by 10 lowers its pH by 1. This works until you get close to pH 7.

Pattern 5: mixing an acid with a base

When a strong acid and a strong base are mixed, they neutralise each other. Whatever is left in excess decides the pH.

Method

1. Moles of H⁺ = concentration × volume (in dm³).
2. Moles of OH⁻ = concentration × volume.
3. Subtract: the larger amount minus the smaller is the excess.
4. Divide the excess by the total volume.
5. Calculate pH (or pOH, if OH⁻ is in excess).

Example 9. 25.0 cm³ of 0.100 mol/dm³ HCl is mixed with 20.0 cm³ of 0.100 mol/dm³ NaOH. Find the pH.

• H⁺: 0.100 × 0.0250 = 0.00250 mol
• OH⁻: 0.100 × 0.0200 = 0.00200 mol
• Excess H⁺ = 0.00050 mol
• Total volume = 45.0 cm³ = 0.0450 dm³
• [H⁺] = 0.00050 ÷ 0.0450 = 0.0111 mol/dm³
• pH = 1.95

Example 10. 20.0 cm³ of 0.100 mol/dm³ HCl is mixed with 25.0 cm³ of 0.100 mol/dm³ NaOH.

• H⁺: 0.00200 mol; OH⁻: 0.00250 mol
• Excess OH⁻ = 0.00050 mol in 0.0450 dm³ → [OH⁻] = 0.0111 mol/dm³
• pOH = 1.95, so pH = 14.00 − 1.95 = 12.05

The most common mistake here is dividing by the volume of only one solution. After mixing, the ions are spread through the combined volume.

The trap: very dilute acids

What’s the pH of 1.0 × 10⁻⁸ mol/dm³ HCl? The obvious answer, 8, is wrong: an acid can’t make water alkaline.

At this concentration, the H⁺ from the acid is smaller than the H⁺ that water itself produces (1.0 × 10⁻⁷ mol/dm³). You have to include both. Solving the full equation gives [H⁺] ≈ 1.05 × 10⁻⁷ mol/dm³, so the pH is about 6.98, just slightly acidic, as common sense suggests.

Rule of thumb: if a strong acid’s concentration is below about 10⁻⁶ mol/dm³, water’s own contribution matters and the simple method fails.

What about weak acids?

Weak acids don’t ionise completely, so [H⁺] is much smaller than the acid concentration. You need the acid dissociation constant, Ka. For a weak acid HA at concentration c, when only a small fraction ionises:

[H⁺] ≈ √(Ka × c)

Example 11. 0.10 mol/dm³ ethanoic acid, Ka = 1.8 × 10⁻⁵.

• [H⁺] ≈ √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol/dm³
• pH ≈ 2.87

Compare that with 0.10 mol/dm³ HCl, which has a pH of 1.00. Same concentration, but about 75 times less H⁺. That’s the practical meaning of “weak”. See strong vs weak acids for more.

Checking your answer

Mistakes in pH questions are almost always caught by a ten-second check at the end, so it is worth building the habit.

Before you move on, run three quick checks:

1. Direction. Acids must give pH below 7, alkalis above 7.
2. Size. A concentration of 0.01–0.1 mol/dm³ should give a pH between 1 and 2 for a strong acid.
3. Log sense. Every factor of 10 in concentration is one pH unit, never more.

Key takeaways

• Strong acid: pH = −log[acid] (times the number of protons released).
• Strong base: find pOH = −log[OH⁻], then pH = 14 − pOH at 25 °C.
• Dilution and mixing: work in moles first, then divide by the final total volume.
• Very dilute acids never go above pH 7; include water’s own H⁺.
• Weak acids need Ka: [H⁺] ≈ √(Ka × c).
• Check your working with our pH calculator, which handles strong and weak acids and bases exactly.