pH of Amphiprotic Salt Solutions

Introduction

An aqueous salt containing the intermediate ion HA⁻ of a diprotic acid H₂A has two competing reactions: HA⁻ can donate or accept a proton. Under common simplifying conditions, its pH lies near the average of the adjacent pKa values: pH ≈ ½(pKa1 + pKa2). This compact formula is useful, but it is an approximation with concentration, water-ionisation and chemical-system assumptions that must be checked.

Core explanation

Let H₂A ⇌ H⁺ + HA⁻ have Ka1 and HA⁻ ⇌ H⁺ + A²⁻ have Ka2. Suppose a salt such as NaHA supplies total acid-family concentration C. The middle ion can act as a base with Kb = Kw/Ka1 or as an acid with Ka2. Both processes occur, so setting pH from only Ka2 or only Kb neglects one side of its behaviour.

The familiar midpoint result comes from the complete mass-balance, charge-balance and equilibrium model when HA⁻ remains the dominant acid-family species and water's direct H⁺/OH⁻ contributions are small relative to the acid-family charge shifts. Then [H⁺] ≈ √(Ka1Ka2), giving pH ≈ (pKa1 + pKa2)/2. In this ideal regime, concentration C cancels from the leading approximation. That does not mean concentration never matters: it enters the exact balances and determines whether neglected terms are indeed small.

A more informative approximate expression for the same simple sodium salt is [H⁺]² ≈ (Ka2 C + Kw)/(1 + C/Ka1), under the usual concentration model and neglected higher-order corrections. If C is much larger than Ka1 and Ka2 C is much larger than Kw, this reduces to Ka1Ka2. If the solution is extremely dilute, water autoionisation is not negligible and pH tends toward the water-neutral value rather than a fixed pKa midpoint. If Ka1 and Ka2 are not well separated, HA⁻ may not dominate and a full solution is safer.

For the carbonate intermediate HCO₃⁻, an isolated closed-system calculation may give a useful estimate, but actual bicarbonate solutions can exchange CO₂ with air. That exchange changes the carbon balance, so a measured pH can depart from a sealed midpoint calculation. Similarly, phosphate salts at significant ionic strength call for activities rather than bare concentrations if precision is required.

The midpoint expression is not “average the pH of two acids.” The pKa values represent two adjacent dissociation constants in the same polyprotic family . Choosing unrelated pKa numbers from separate acids has no derivation. For a triprotic family, H₂A⁻ uses its adjacent pKa1 and pKa2, while HA²⁻ uses pKa2 and pKa3 in a comparable amphiprotic approximation. Select the pair that surrounds the actual ion.

A numerical example makes the assumptions visible. Let pKa1 = 4 and pKa2 = 10 for a hypothetical H₂A/HA⁻/A²⁻ family, with C = 0.010 M at 25 °C. Ka1 = 10⁻⁴, Ka2 = 10⁻¹⁰. Here C/Ka1 = 100 and Ka2 C = 10⁻¹², larger than Kw = 10⁻¹⁴, so the midpoint estimate is plausible: pH ≈ 7. The corrected expression gives [H⁺]² ≈ (1.0×10⁻¹² + 1.0×10⁻¹⁴)/(1+100) = 1.0×10⁻¹⁴ and hence pH approximately 7. The agreement is not accidental; the chosen conditions satisfy the simplifying inequalities reasonably well.

Step-by-step reasoning

1. Confirm the solute is an amphiprotic intermediate HA⁻, not H₂A or A²⁻. 2. Identify its neighbouring Ka1 and Ka2. 3. Test whether HA⁻ dominates and C is not extremely dilute. 4. Estimate pH as ½(pKa1 + pKa2). 5. If assumptions fail, solve mass and charge balances with water autoionisation.

Visual explanation

Draw pKa1 and pKa2 as markers on a pH line, with the amphiprotic ion HA⁻ dominant between them. Mark their arithmetic midpoint and label it an approximation, not a universal fixed value.

Real-world analogy

A balanced platform between two slopes often rests near the midpoint, but only if the platform is heavy compared with background disturbances. Dilution makes water's own ions a larger disturbance and shifts the result.

Real-world example

A sodium hydrogen phosphate solution is often first estimated from the two pKa values adjacent to HPO₄²⁻. A precise formulation then checks concentration, ionic strength and any CO₂ uptake or other competing equilibria.

Why?

Why does the simple pH estimate become unreliable at very low salt concentration? Water's own hydronium and hydroxide become comparable to the ions generated by amphiprotic reactions, so the neglected Kw terms matter.

Common misconception

“The midpoint formula works for every amphiprotic ion at every concentration.” It requires a consistent family model and suitable dominance and dilution conditions; otherwise solve the full coupled equilibria.

Worked example

For the hypothetical salt NaHA with pKa1 = 4.0 and pKa2 = 10.0 at 0.010 M, the midpoint estimate is pH = ½(4.0+10.0) = 7.0. Check C ≫ Ka1 (0.010 ≫ 0.00010) and Ka2 C ≫ Kw (10⁻¹² ≫ 10⁻¹⁴) at 25 °C. These checks support the estimate. If C were decreased drastically, recalculation including Kw would be needed rather than reusing 7.0 automatically.

Quick check

1. Which pKa values belong in the amphiprotic HA⁻ midpoint relation? Answer: The two stepwise pKa values adjacent to HA⁻ in the same H₂A/HA⁻/A²⁻ family: pKa1 and pKa2.

Exam focus

Show the chosen ion and adjacent equilibria before averaging pKa values. State at least one condition under which the approximation could fail.

Advanced insight

The midpoint is a leading approximation from mass, charge and equilibrium equations, not an independent law. A more accurate model may use activities and explicit counterions, particularly in concentrated salt solutions.

Summary

Suitable aqueous amphiprotic HA⁻ salts can have pH near the mean of their adjacent pKa values. The result assumes the intermediate dominates and water autoionisation is negligible relative to the salt's acid-base shifts. Very dilute or open gas-exchange systems require a fuller balance calculation.

Practice questions

1. Estimate pH when pKa1 = 5 and pKa2 = 9 under valid midpoint conditions. Answer: pH ≈ 7. 2. Which pKa pair surrounds HPO₄²⁻ in the phosphoric-acid family? Answer: pKa2 and pKa3, because HPO₄²⁻ lies between H₂PO₄⁻ and PO₄³⁻. 3. Does halving C always leave the exact pH unchanged? Answer: No. The midpoint estimate may be nearly unchanged in its valid regime, but the exact balance includes C and water ionisation. 4. Why can atmospheric CO₂ complicate a bicarbonate-salt calculation? Answer: CO₂ exchange changes total dissolved inorganic carbon and invalidates a fixed closed-system mass balance.