First-Dissociation Approximation for Diprotic Acids

Introduction

The complete description of a diprotic acid needs five equations, but in most real problems we do not need to solve them all. For a solution of H₂A on its own, the first dissociation usually provides almost all of the hydrogen ions, and the second step is a tiny correction. This page explains when we may treat a diprotic acid as if it were monoprotic, using Ka1 only, and how to check that the shortcut is honest rather than merely convenient.

Core explanation

The approximation. For H₂A at analytical concentration C, suppose only the first step matters:

H₂A ⇌ H⁺ + HA⁻ Ka1 = x² / (C − x), where x = [H⁺] = [HA⁻]

This is exactly the weak monoprotic acid calculation met earlier. If x is small compared with C, then x ≈ √(Ka1C).

Condition 1: the steps are well separated. The second step releases extra H⁺ equal to [A²⁻]. From the Ka2 expression, [A²⁻] = Ka2[HA⁻] / [H⁺]. Under the first-dissociation picture [HA⁻] ≈ [H⁺], so [A²⁻] ≈ Ka2. The extra H⁺ is negligible if Ka2 is much smaller than x. A practical guide is Ka1/Ka2 ≥ 10³ and Ka2 < 0.05x. For most diprotic acids Ka2 is thousands to millions of times smaller than Ka1, so this condition is easily met.

Condition 2: water is not significant. The acid must produce far more H⁺ than water's autoionisation, about 10⁻⁷ mol dm⁻³. In practice x should exceed about 10⁻⁶ mol dm⁻³, which fails only for extremely dilute or extremely weak acids.

Condition 3: the square-root shortcut. Replacing C − x by C needs x < 5% of C. If Ka1 is large (for example oxalic acid, Ka1 = 5.9 × 10⁻²) or the solution is dilute, this fails and the quadratic x² + Ka1x − Ka1C = 0 must be solved. Note that this condition concerns the first step only; failing it does not mean the second step matters.

What the approximation gives. Once x is known we have [H⁺], [HA⁻] ≈ x, [H₂A] ≈ C − x and, as a bonus, [A²⁻] ≈ Ka2. All four species follow from one simple calculation.

Step-by-step reasoning

1. Compare Ka1 and Ka2; if the ratio is at least about 10³, proceed with Ka1 alone. 2. Estimate x = √(Ka1C). 3. Check x / C: if below 5%, accept; otherwise solve the quadratic. 4. Check that Ka2 is small compared with x. 5. Check that x is well above 10⁻⁶ mol dm⁻³, then calculate pH = −log x.

Visual explanation

Picture two taps filling a basin with H⁺. The first tap (Ka1) runs strongly; the second (Ka2) drips. Once the basin is full enough, the drip changes the level so little that you can read the level from the first tap alone.

Real-world analogy

A household's income might include a salary and a tiny interest payment on savings. To estimate monthly spending power you use the salary alone; the interest is real but too small to change the answer at the precision you care about.

Real-world example

The acidity of a vitamin C tablet dissolved in water is set by ascorbic acid's first dissociation. Its second proton, with Ka2 ≈ 1.6 × 10⁻¹², is so weakly acidic that it plays no role in the taste or the pH of the drink.

Why?

Why does the second step contribute so little? The H⁺ produced by the first step pushes the second equilibrium to the left, by Le Chatelier's principle, and Ka2 is already small. Both effects keep [A²⁻] near the tiny value Ka2.

Common misconception

"Because the acid is diprotic, [H⁺] = 2C" or "[H⁺] = 2√(Ka1C)". Neither is true for a weak diprotic acid: the second proton is barely released, so [H⁺] ≈ √(Ka1C).

Worked example

Question: Find the pH of 0.100 mol dm⁻³ ascorbic acid (Ka1 = 8.0 × 10⁻⁵, Ka2 = 1.6 × 10⁻¹²).

Reasoning: Ka1/Ka2 = 5 × 10⁷, so the steps are well separated. x = √(8.0 × 10⁻⁵ × 0.100) = √(8.0 × 10⁻⁶) = 2.83 × 10⁻³ mol dm⁻³. x / C = 2.8%, below 5%. Ka2 is negligible compared with x, and x is far above 10⁻⁶.

Answer: pH = −log(2.83 × 10⁻³) = 2.55. The approximation is fully justified.

Quick check

1. Why can the pH of 0.10 mol dm⁻³ carbonic acid be found from Ka1 alone? Answer: Ka2 is about ten thousand times smaller than Ka1, so the second dissociation adds a negligible amount of H⁺.

Exam focus

Show your checks: state the Ka ratio, the percentage dissociation and the size of Ka2 relative to [H⁺]. Examiners often give an acid with a large Ka1, such as oxalic acid, precisely so that you must spot the need for the quadratic even though the second step is still negligible.

Advanced insight

The approximation [A²⁻] ≈ Ka2 is a striking result: in a solution of pure H₂A, the concentration of the fully deprotonated ion is roughly independent of the acid concentration. It holds only while [H⁺] ≈ [HA⁻], which fails when strong acid or a salt such as NaHA is added.

Summary

For a diprotic acid alone in water, the first dissociation usually controls the pH. The approximation needs well-separated steps (Ka1/Ka2 ≥ about 10³), Ka2 much smaller than [H⁺], and negligible water contribution. Use √(Ka1C) if dissociation is under 5%; otherwise solve the quadratic. The second step then gives [A²⁻] ≈ Ka2.

Practice questions

1. Calculate the pH of 0.100 mol dm⁻³ oxalic acid (Ka1 = 5.9 × 10⁻², Ka2 = 6.4 × 10⁻⁵). Answer: The shortcut fails (x/C is large), so solve x² + 0.059x − 0.0059 = 0, giving x = 0.053 mol dm⁻³ and pH = 1.28. 2. For the oxalic acid solution above, estimate [C₂O₄²⁻] and comment on the approximation. Answer: [C₂O₄²⁻] ≈ Ka2 = 6.4 × 10⁻⁵ mol dm⁻³, about 0.1% of [H⁺], so neglecting the second step is justified. 3. State three checks needed before using the first-dissociation approximation. Answer: Ka1/Ka2 is at least about 10³; Ka2 is small compared with [H⁺]; the acid's [H⁺] is far above the 10⁻⁷ mol dm⁻³ from water. 4. Calculate the pH of 0.0100 mol dm⁻³ ascorbic acid. Answer: x = √(8.0 × 10⁻⁷) = 8.9 × 10⁻⁴ mol dm⁻³ (8.9%, so the quadratic gives 8.6 × 10⁻⁴); pH ≈ 3.07.