Approximations and Their Verification

Introduction

The complete equation set for even a simple weak acid leads to a cubic equation in [H⁺]. Nobody solves cubics by hand for every pH calculation. Instead, chemists make approximations : they drop terms that are expected to be small. Approximations are safe only if they are checked. This page shows the two standard approximations for weak acids and bases, a simple numerical test for each, and what to do when the test fails.

Core explanation

The two usual approximations. For a weak acid HA at analytical concentration c, the full treatment gives [H⁺] = [A⁻] + [OH⁻] (charge balance) and [HA] = c − [A⁻] (mass balance). Two simplifications follow:

1. Neglect water. If the solution is clearly acidic, [OH⁻] is tiny compared with [A⁻], so [H⁺] ≈ [A⁻]. Call this x. 2. Small x. If the acid is weak and not too dilute, only a small fraction dissociates, so [HA] = c − x ≈ c.

With both, Ka = x²/(c − x) ≈ x²/c, giving x = √(Ka × c).

The 5 % test. After calculating x, check that x/c is below 0.05 (5 %). If it is, the error in [H⁺] is at most about 2.5 %, which is smaller than the uncertainty in most tabulated Ka values. Also check that [OH⁻] = Kw/x is less than 5 % of x.

A quick predictor. The small-x approximation is usually safe when c/Ka is greater than about 400. For ethanoic acid (Ka = 1.8 × 10⁻⁵) at 0.10 mol dm⁻³, c/Ka ≈ 5600, well above 400. For the stronger chloroethanoic acid (Ka ≈ 1.4 × 10⁻³) at 0.010 mol dm⁻³, c/Ka ≈ 7, so the approximation will fail.

When the test fails: the quadratic. Keep the c − x term and solve exactly:

x² + Ka x − Ka c = 0, so x = (−Ka + √(Ka² + 4Ka c)) / 2

Only the positive root has physical meaning.

Successive approximations. An alternative to the quadratic formula is iteration. Start with x₁ = √(Ka c), then compute x₂ = √(Ka (c − x₁)), then x₃ = √(Ka (c − x₂)), and so on. The values converge within two or three cycles to the quadratic answer. This method extends naturally to more complicated equations where no formula exists.

When water cannot be neglected. If the acid is very dilute or very weak, [H⁺] from the acid approaches 10⁻⁷ mol dm⁻³, and water's own contribution matters. The approximation [H⁺] ≈ [A⁻] then fails; a separate treatment is needed.

Bases. The same logic applies to weak bases, with Kb and [OH⁻] replacing Ka and [H⁺].

Why checking matters. An unchecked approximation can give a pH that is plainly wrong. For example, applying √(Ka c) to a very dilute acid can predict a pH above 7 for an acid solution, which is impossible.

Formulae

Approximate: [H⁺] = √(Ka c). Exact (neglecting water): [H⁺] = (−Ka + √(Ka² + 4Ka c))/2. Test: x/c < 0.05. Predictor: c/Ka > about 400.

Step-by-step reasoning

1. Write the exact equation. 2. State the approximations you will make. 3. Solve the simplified equation. 4. Test each approximation numerically (5 % rule). 5. If a test fails, solve more exactly and check again.

Visual explanation

Plot [H⁺] against c on logarithmic axes for a weak acid. The approximate line √(Ka c) is straight. The exact curve follows it at high c, bends below it where the acid becomes substantially dissociated, and flattens at 10⁻⁷ mol dm⁻³ at very low c where water takes over.

Real-world analogy

When estimating a restaurant bill, you might ignore the price of a glass of tap water. That is fine if everything else costs pounds. It is not fine if you bought only a biscuit. An approximation is valid only relative to the size of what you keep.

Real-world example

Analytical chemists preparing calibration standards for pH meters must know the true pH, not an estimate. They use the exact equations, often with activity corrections, because a 0.05 pH-unit error in a standard would be carried into every subsequent measurement.

Why?

Why does a 5 % error in c − x give only about a 2.5 % error in [H⁺]? Because [H⁺] depends on the square root of the quantity containing the error, and taking a square root roughly halves a small percentage error.

Common misconception

"Weak acids are always less than 5 % dissociated, so the approximation always works." Percentage dissociation rises as the solution is diluted. A weak acid can be substantially dissociated in dilute solution.

Worked example

Question: Find the pH of 0.010 mol dm⁻³ chloroethanoic acid, Ka = 1.4 × 10⁻³.

Reasoning: Approximate: x = √(1.4 × 10⁻³ × 0.010) = 3.7 × 10⁻³; x/c = 37 %, which fails. Quadratic: x = (−1.4 × 10⁻³ + √(1.96 × 10⁻⁶ + 5.6 × 10⁻⁵))/2 = (−1.4 × 10⁻³ + 7.61 × 10⁻³)/2 = 3.1 × 10⁻³.

Answer: [H⁺] = 3.1 × 10⁻³ mol dm⁻³, pH = 2.51 (the approximation would have given 2.43).

Quick check

1. Is the small-x approximation safe for a weak acid with Ka = 1.0 × 10⁻⁴ at 0.20 mol dm⁻³? Answer: Yes: c/Ka = 2000, above 400, and x/c is about 2 %.

Exam focus

Always state the approximation and show a numerical check, for example "x/c = 1.3 % < 5 %, so the approximation is valid." If the check fails, use the quadratic; examiners expect you to reject an invalid approximation rather than ignore it.

Advanced insight

Iterative methods such as successive approximation or the Newton–Raphson method are how calculators and software solve equilibrium equations. They converge quickly when the starting guess is close, which is why the simple approximate answer remains useful even when it is not accurate enough on its own: it is an excellent first guess.

Summary

Weak acid calculations usually neglect water's contribution and assume c − x ≈ c, giving [H⁺] = √(Ka c). These approximations must be tested, commonly by the 5 % rule; c/Ka above about 400 predicts success. If a test fails, solve the quadratic or iterate by successive approximation. Checking by back-substitution guards against impossible answers.

Practice questions

1. State the two approximations used to derive [H⁺] = √(Ka c). Answer: Water's contribution to [H⁺] is negligible, and the amount dissociated is small compared with c, so c − x ≈ c. 2. Calculate the pH of 0.10 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵) and verify the approximation. Answer: x = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³, pH = 2.87; x/c = 1.3 %, below 5 %, so valid. 3. Perform one cycle of successive approximation for 0.010 mol dm⁻³ chloroethanoic acid starting from x₁ = 3.7 × 10⁻³. Answer: x₂ = √(1.4 × 10⁻³ × (0.010 − 0.0037)) = √(8.8 × 10⁻⁶) = 3.0 × 10⁻³, already close to the exact 3.1 × 10⁻³. 4. Explain why percentage dissociation of a weak acid increases on dilution. Answer: x/c ≈ √(Ka/c), which rises as c falls; dilution shifts the equilibrium towards more ions, the side with more dissolved particles.