Water Autoionisation in Dilute Solutions

Introduction

What is the pH of 1.0 × 10⁻⁸ mol dm⁻³ hydrochloric acid? The obvious answer, pH = 8, cannot be right: adding acid to pure water cannot make it alkaline. The paradox arises because the simple formula pH = −log c ignores the hydrogen ions supplied by water itself. In most solutions these are negligible, but once the acid is diluted to around 10⁻⁶ mol dm⁻³ or below, water's autoionisation becomes a major source of H⁺ and must be included.

Core explanation

Water's own ions. Pure water ionises slightly: H₂O ⇌ H⁺ + OH⁻, with Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. In pure water [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³.

Adding acid suppresses water's ionisation. When a strong acid is added, the extra H⁺ shifts the water equilibrium to the left (Le Chatelier). In 0.010 mol dm⁻³ HCl, [OH⁻] = 10⁻¹² mol dm⁻³, so water supplies only 10⁻¹² mol dm⁻³ of H⁺: completely negligible. The total [H⁺] is effectively just the acid concentration.

The exact treatment for a strong acid. For HCl at concentration c, the species are H⁺, OH⁻ and Cl⁻. The equations are:

Charge balance: [H⁺] = [Cl⁻] + [OH⁻] = c + [OH⁻] Kw: [OH⁻] = Kw/[H⁺]

Substituting: [H⁺] = c + Kw/[H⁺], which rearranges to the quadratic

[H⁺]² − c[H⁺] − Kw = 0, so [H⁺] = (c + √(c² + 4Kw)) / 2

Checking the limits. If c is large (say 10⁻³), c² dominates 4Kw and [H⁺] ≈ c, the familiar result. If c is tiny, 4Kw dominates and [H⁺] ≈ √Kw = 10⁻⁷, the pure-water result. The exact formula bridges both.

The answer to the paradox. For c = 1.0 × 10⁻⁸: c² = 1.0 × 10⁻¹⁶; 4Kw = 4.0 × 10⁻¹⁴; √(4.01 × 10⁻¹⁴) = 2.002 × 10⁻⁷. [H⁺] = (1.0 × 10⁻⁸ + 2.002 × 10⁻⁷)/2 = 1.05 × 10⁻⁷ mol dm⁻³, pH = 6.98. The solution is very slightly acidic, as it must be. Notice that water supplies over 90 % of the H⁺.

Strong bases. The same equation applies with [OH⁻] in place of [H⁺].

Weak acids in dilute solution. For a weak acid, water matters sooner, because the acid provides less H⁺. The full charge balance is [H⁺] = [A⁻] + Kw/[H⁺]. A useful approximate form when dissociation is small is [H⁺] ≈ √(Ka c + Kw). Water's contribution becomes significant when Ka c is no longer much larger than Kw, roughly when Ka c < 10⁻¹².

A working rule. If the calculated [H⁺] from the acid alone exceeds about 10⁻⁶ mol dm⁻³, water can be ignored (it then contributes less than 1 %). Between 10⁻⁶ and 10⁻⁸ mol dm⁻³ the exact treatment is needed. Below that, the pH is effectively 7.00 at 25 °C.

Formulae

Strong acid, exact: [H⁺] = (c + √(c² + 4Kw))/2. Weak acid, dilute (small dissociation): [H⁺] ≈ √(Ka c + Kw). Kw = 1.0 × 10⁻¹⁴ at 25 °C.

Step-by-step reasoning

1. Estimate [H⁺] from the acid alone. 2. If it is above about 10⁻⁶ mol dm⁻³, use the simple answer. 3. If not, write the charge balance including [OH⁻] = Kw/[H⁺]. 4. Solve the resulting quadratic and take the positive root. 5. Check that the pH is below 7 for an acid.

Visual explanation

Plot pH against −log c for HCl. At high concentrations the graph is a straight line of slope 1 (pH = −log c). As c falls below 10⁻⁶ the line curves and levels off, approaching pH 7 asymptotically but never crossing it.

Real-world analogy

Adding a teaspoon of dye to an ocean does not make the ocean the colour of the dye. The background dominates. A trace of acid in water is like that teaspoon: the water's own H⁺, small as it is, becomes the background that controls the result.

Real-world example

Ultrapure water used in semiconductor manufacture has a conductivity close to the theoretical value for water's own ions alone. Engineers monitor it continuously; any rise signals contamination at parts-per-billion levels, a regime where water autoionisation dominates the ionic content.

Why?

Why does water supply a large share of H⁺ in the dilute limit, even though acid suppresses its ionisation? When the added acid is tiny, the suppression is small too. At c = 10⁻⁸, [OH⁻] is still about 9.5 × 10⁻⁸, so water still ionises almost as much as in pure water.

Common misconception

"Diluting an acid enough makes it alkaline." Dilution can only bring the pH closer to 7. The simple formula pH = −log c fails in the dilute limit because it omits water's H⁺.

Worked example

Question: Find the pH of 5.0 × 10⁻⁷ mol dm⁻³ NaOH at 25 °C.

Reasoning: By symmetry, [OH⁻] = (c + √(c² + 4Kw))/2 = (5.0 × 10⁻⁷ + √(2.5 × 10⁻¹³ + 4.0 × 10⁻¹⁴))/2 = (5.0 × 10⁻⁷ + 5.39 × 10⁻⁷)/2 = 5.19 × 10⁻⁷. pOH = 6.28.

Answer: pH = 14.00 − 6.28 = 7.72 (the simple formula gives 7.70).

Quick check

1. What is the approximate pH of 1.0 × 10⁻¹⁰ mol dm⁻³ HCl at 25 °C, and why? Answer: About 7.00, because water's own H⁺ (10⁻⁷ mol dm⁻³) vastly exceeds the acid's contribution.

Exam focus

Expect "trap" questions on very dilute acids or bases. Show the charge balance including OH⁻, solve the quadratic and comment that the answer is correctly just below (acid) or above (base) pH 7.

Advanced insight

At concentrations where water autoionisation matters, dissolved atmospheric CO₂ also matters. Water exposed to air absorbs CO₂ and reaches a pH of about 5.6, so a real "10⁻⁸ mol dm⁻³ HCl" solution open to air would be controlled by carbonic acid, not by the added HCl.

Summary

Water autoionisation supplies 10⁻⁷ mol dm⁻³ of H⁺ and OH⁻ at 25 °C. Added acid or base suppresses it, so in ordinary solutions water's contribution is negligible. When the solute provides less than about 10⁻⁶ mol dm⁻³ of H⁺ or OH⁻, the charge balance with Kw must be solved exactly, giving [H⁺] = (c + √(c² + 4Kw))/2 for a strong acid. Dilution can never push an acid past neutrality.

Practice questions

1. Explain why pH = −log c fails for 1.0 × 10⁻⁸ mol dm⁻³ HCl. Answer: It ignores the H⁺ from water, which is larger than the acid's contribution, and so wrongly predicts an alkaline pH of 8. 2. Calculate the pH of 2.0 × 10⁻⁷ mol dm⁻³ HCl at 25 °C. Answer: [H⁺] = (2.0 × 10⁻⁷ + √(4.0 × 10⁻¹⁴ + 4.0 × 10⁻¹⁴))/2 = (2.0 × 10⁻⁷ + 2.83 × 10⁻⁷)/2 = 2.41 × 10⁻⁷; pH = 6.62. 3. For a weak acid with Ka = 1.0 × 10⁻⁸ at 1.0 × 10⁻⁴ mol dm⁻³, estimate [H⁺] and state whether water matters. Answer: Ka c = 1.0 × 10⁻¹², comparable to Kw; [H⁺] ≈ √(1.0 × 10⁻¹² + 1.0 × 10⁻¹⁴) = 1.005 × 10⁻⁶, so water contributes about 1 % and is only just negligible. 4. Why does adding HCl reduce the amount of H⁺ that water itself supplies? Answer: Extra H⁺ shifts the equilibrium H₂O ⇌ H⁺ + OH⁻ to the left, lowering [OH⁻] and hence the amount of water that has ionised.