Second Dissociation and Its Contribution

Introduction

When we calculate the pH of a diprotic acid from Ka1 alone, we quietly throw away the second dissociation. Good chemistry does not leave such decisions unexamined. This page shows how to estimate the amount of fully deprotonated A²⁻ produced by the second step, how to test whether the protons it releases really are negligible, and how to recognise the cases, such as the hydrogensulfate ion, where they are not.

Core explanation

Estimating A²⁻. Once the first-dissociation calculation has given [H⁺] and [HA⁻], rearrange Ka2:

[A²⁻] = Ka2 × [HA⁻] / [H⁺]

For a solution of H₂A alone, [HA⁻] ≈ [H⁺], so [A²⁻] ≈ Ka2. The concentration of A²⁻ is then independent of how much acid was dissolved — a useful and slightly surprising result.

Testing the neglected protons. Every A²⁻ formed releases one extra H⁺ and removes one HA⁻. The proton condition for H₂A in water is

[H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻]

so the first-step picture ignores [A²⁻] (the extra proton). The test is the ratio [A²⁻] / [H⁺]. If it is below about 5%, the neglect is justified; if it is below 1%, it is negligible at any normal precision.

When the second step matters. Three situations make the second dissociation important:

- Close Ka values. If Ka1/Ka2 is below about 10³, both steps release comparable amounts of H⁺ in dilute solution. - Dilute solutions. As C falls, [H⁺] from the first step falls, but [A²⁻] ≈ Ka2 does not, so the ratio [A²⁻]/[H⁺] grows. - Strong first step. For sulfuric acid the first dissociation is complete and the second, HSO₄⁻ ⇌ H⁺ + SO₄²⁻, has Ka2 ≈ 1.0 × 10⁻². In 0.010 mol dm⁻³ H₂SO₄ the second step adds roughly 40% more H⁺ to the 0.010 mol dm⁻³ from the first, so it cannot be ignored.

Refining the answer. When the second step is significant, include it: write [H⁺] = x + y, [HA⁻] = x − y and [A²⁻] = y, and solve Ka2 = (x + y)y / (x − y). Alternatively iterate: estimate [H⁺], use it to find [A²⁻], correct [H⁺], and repeat until the value is stable.

Step-by-step reasoning

1. Solve the first dissociation to obtain [H⁺] and [HA⁻]. 2. Calculate [A²⁻] = Ka2[HA⁻]/[H⁺]. 3. Compare [A²⁻] with [H⁺] as a percentage. 4. If the percentage is small, report the first-step pH; if not, solve the two coupled equilibria or iterate. 5. State the result of the test explicitly in your answer.

Visual explanation

Plot [H⁺] and [A²⁻] against log C for a diprotic acid. The [H⁺] line slopes downwards as the acid is diluted, while the [A²⁻] line stays flat at Ka2. Where the lines approach each other, the second step can no longer be neglected.

Real-world analogy

A shop's main cash register takes most of the money, and a small tip jar adds a little. On a busy day the tips are trivial compared with the takings; on a quiet day the tip jar stays about the same while takings fall, so the tips become a noticeable share.

Real-world example

In acid rain, sulfur dioxide forms sulfurous and sulfuric acids at low concentration. Because the solutions are dilute and hydrogensulfate is a fairly strong second acid, environmental chemists must include the second dissociation when relating measured sulfate to rainwater pH.

Why?

Why does [A²⁻] stay near Ka2 when the solution is diluted? Dilution lowers [H⁺] and [HA⁻] together, and they appear as a ratio in [A²⁻] = Ka2[HA⁻]/[H⁺]. Since the ratio stays close to 1, [A²⁻] barely changes.

Common misconception

"If Ka2 is small, the second step never matters." Its importance depends on Ka2 relative to [H⁺], not on Ka2 alone. A small Ka2 can still matter in a very dilute solution or when Ka1 and Ka2 are close.

Worked example

Question: For 0.0100 mol dm⁻³ oxalic acid (Ka1 = 5.9 × 10⁻², Ka2 = 6.4 × 10⁻⁵), the first step gives [H⁺] = [HC₂O₄⁻] = 8.7 × 10⁻³ mol dm⁻³. Test the neglect of the second step.

Reasoning: [C₂O₄²⁻] = Ka2 × [HC₂O₄⁻] / [H⁺] ≈ 6.4 × 10⁻⁵ mol dm⁻³. Ratio = 6.4 × 10⁻⁵ / 8.7 × 10⁻³ = 0.0074, or 0.74%.

Answer: The second dissociation adds less than 1% to [H⁺], so pH = −log(8.7 × 10⁻³) = 2.06 stands. The ratio is larger than for 0.100 mol dm⁻³ oxalic acid (about 0.1%), illustrating the effect of dilution.

Quick check

1. In a solution of a weak diprotic acid alone, what is the approximate concentration of A²⁻? Answer: Approximately equal to Ka2, because [HA⁻] and [H⁺] are nearly equal and cancel in the Ka2 expression.

Exam focus

Examiners like the result [A²⁻] ≈ Ka2 and the percentage test that follows. Always quote the percentage and conclude clearly. For sulfuric acid questions, remember that the first dissociation is complete and the second is only partial.

Advanced insight

The approximation [A²⁻] ≈ Ka2 is used in classical sulfide chemistry: in a solution saturated with H₂S alone, [S²⁻] is fixed near Ka2 regardless of concentration. Adding a strong acid breaks the equality [HA⁻] ≈ [H⁺] and lowers [A²⁻] sharply, which is the principle behind controlling A²⁻ in selective precipitation later in this unit.

Summary

The second dissociation produces A²⁻ at a concentration Ka2[HA⁻]/[H⁺], approximately Ka2 for the acid alone. Its extra protons are tested by comparing [A²⁻] with [H⁺]. The contribution grows on dilution and when Ka values are close, and it is large for hydrogensulfate; in such cases solve the coupled equilibria or iterate.

Practice questions

1. For 0.100 mol dm⁻³ ascorbic acid, [H⁺] = 2.8 × 10⁻³ mol dm⁻³ and Ka2 = 1.6 × 10⁻¹². Estimate [A²⁻] and the percentage it adds to [H⁺]. Answer: [A²⁻] ≈ 1.6 × 10⁻¹² mol dm⁻³, about 6 × 10⁻⁸ %, which is completely negligible. 2. Explain why diluting a diprotic acid makes the second dissociation more significant. Answer: [H⁺] falls roughly as √C, but [A²⁻] stays near Ka2, so the ratio [A²⁻]/[H⁺] increases. 3. Why must the second dissociation be included for 0.010 mol dm⁻³ sulfuric acid? Answer: Ka2 for HSO₄⁻ (about 1.0 × 10⁻²) is similar in size to [H⁺], so the second step adds a large fraction of extra H⁺. 4. Write the expressions for [H⁺], [HA⁻] and [A²⁻] used when both steps are included, with x from step one and y from step two. Answer: [H⁺] = x + y, [HA⁻] = x − y and [A²⁻] = y, substituted into Ka1 and Ka2.