Diprotic Acid Species and Charge

Introduction

A diprotic acid in water is never just "H₂A". Once it is dissolved, three related species share the acid: the fully protonated H₂A, the intermediate HA⁻ and the fully deprotonated A²⁻. To calculate anything reliably we need two bookkeeping tools met earlier in the unit, applied now to this three-species family: a mass balance , which tracks where the acid has gone, and a charge balance , which keeps the solution electrically neutral. Together with the two Ka expressions and Kw they give a complete description.

Core explanation

Mass balance. If a solution is made with an analytical (total) concentration C of a diprotic acid or any of its salts, every acid unit must be in one of three forms:

C = [H₂A] + [HA⁻] + [A²⁻]

This is true whether the solution was made from H₂A, from NaHA or from Na₂A. The starting compound only affects how C is distributed among the species, not the total.

Charge balance. Solutions are electrically neutral. Each ion's concentration is multiplied by the magnitude of its charge:

- Solution of H₂A only: [H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻] - Solution of NaHA: [Na⁺] + [H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻], with [Na⁺] = C - Solution of Na₂A: [Na⁺] + [H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻], with [Na⁺] = 2C

The factor of 2 in front of [A²⁻] is the most frequently forgotten detail. One mole of A²⁻ carries two moles of negative charge, so it must be balanced by two moles of positive charge.

Counting unknowns. For H₂A in water there are five unknowns: [H₂A], [HA⁻], [A²⁻], [H⁺] and [OH⁻]. The five equations are Ka1, Ka2, Kw, the mass balance and the charge balance. Five equations in five unknowns can be solved exactly; the rest of the unit is largely about sensible ways to simplify them.

Average charge and protons released. A useful single number is the average charge per acid unit:

average charge = −([HA⁻] + 2[A²⁻]) / C

It runs from 0 (all H₂A) to −2 (all A²⁻). Its magnitude equals the average number of protons each acid unit has lost. This quantity explains how the net charge of a molecule such as an amino acid or a protein changes with pH, and it is exactly what is measured, in effect, by a titration.

Proton condition. Subtracting the mass balance from the charge balance in a suitable way gives a proton condition: protons gained by some species equal protons lost by others. For a solution of H₂A alone it reads [H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻], which here is identical to the charge balance.

Step-by-step reasoning

1. List every species present, including H⁺, OH⁻ and spectator cations. 2. Write the mass balance for the acid, using its analytical concentration C. 3. Write the charge balance, multiplying each concentration by the size of its charge. 4. Express spectator ions (Na⁺, K⁺) in terms of C from the recipe. 5. Check that the number of independent equations equals the number of unknowns.

Visual explanation

Draw three boxes labelled H₂A (charge 0), HA⁻ (charge −1) and A²⁻ (charge −2). Share C tokens among them. Add up the charges carried by the tokens: the total negative charge must be matched by H⁺ and any Na⁺ on the other side of a balance scale.

Real-world analogy

Imagine C sets of twin coins. Each set can be in a purse with both coins, one coin or none. The mass balance counts purses; the charge balance counts coins given away. A purse with none contributes two missing coins, not one.

Real-world example

Oxalate in plant tissue and in urine exists as H₂C₂O₄, HC₂O₄⁻ and C₂O₄²⁻ depending on pH. Clinical chemists use mass and charge balances to predict how much free oxalate is available to form calcium oxalate kidney stones.

Why?

Why must mass and charge balances be written separately from the Ka expressions? Equilibrium constants only relate ratios of species. They say nothing about the total amount of acid added or the requirement of neutrality, so without the balances the problem has too many unknowns to solve.

Common misconception

"In a Na₂A solution the charge balance is [Na⁺] = [A²⁻]." This forgets both the factor of 2 and the other ions. Correctly, [Na⁺] + [H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻], with [Na⁺] = 2C.

Worked example

Question: A solution has total diprotic acid concentration C = 0.100 mol dm⁻³. At its pH, [HA⁻] = 0.060 mol dm⁻³ and [A²⁻] = 0.010 mol dm⁻³. Find [H₂A] and the average charge per acid unit.

Reasoning: Mass balance: [H₂A] = 0.100 − 0.060 − 0.010 = 0.030 mol dm⁻³. Average charge = −(0.060 + 2 × 0.010) / 0.100 = −0.080 / 0.100 = −0.80.

Answer: [H₂A] = 0.030 mol dm⁻³; the average charge is −0.80, meaning each acid unit has lost on average 0.80 protons.

Quick check

1. Write the charge balance for a solution containing only the diprotic acid H₂A in water. Answer: [H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻], with the doubly charged A²⁻ counted twice.

Exam focus

Examiners test the factor of 2 on [A²⁻] and the correct spectator term for salts: [Na⁺] = C for NaHA and 2C for Na₂A. Show both balances explicitly before simplifying, and state which terms you neglect and why.

Advanced insight

In precise work, charge balances are written in concentrations but equilibrium constants strictly apply to activities. At high ionic strength the doubly charged A²⁻ is affected most, because activity coefficients depend on the square of the ionic charge. Neutrality, however, is always exact in concentration terms.

Summary

A diprotic system contains H₂A, HA⁻ and A²⁻. The mass balance fixes their total at C; the charge balance equates positive and negative charge, counting A²⁻ twice. With Ka1, Ka2 and Kw these give enough equations to solve for every species. The average charge, −([HA⁻] + 2[A²⁻]) / C, summarises how deprotonated the acid is.

Practice questions

1. Write the mass balance for a 0.050 mol dm⁻³ solution of NaHA. Answer: 0.050 = [H₂A] + [HA⁻] + [A²⁻]. 2. Write the charge balance for a solution of Na₂A. Answer: [Na⁺] + [H⁺] = [HA⁻] + 2[A²⁻] + [OH⁻], where [Na⁺] = 2C. 3. At a certain pH, a diprotic acid is 50% HA⁻ and 50% A²⁻. What is its average charge? Answer: −(0.5 + 2 × 0.5) = −1.5. 4. How many unknowns and how many equations describe a solution of H₂A in water? Answer: Five unknowns ([H₂A], [HA⁻], [A²⁻], [H⁺], [OH⁻]) and five equations (Ka1, Ka2, Kw, mass balance and charge balance).