Michael S. answered • 08/05/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
The key idea is that HPO42− is amphiprotic — it can donate a proton or accept one. Which role it plays is decided by whatever you put it with. That is exactly why this question gives you an acid and its conjugate base as the two partners.
1. With the acid HNO — HPO42− accepts a proton (acts as a base)
HPO42−(aq) + HNO(aq) ⇌ H2PO4−(aq) + NO−(aq)
2. With the base NO− — HPO42− donates a proton (acts as an acid)
HPO42−(aq) + NO−(aq) ⇌ PO43−(aq) + HNO(aq)
How to build these without memorising anything
Move exactly one H+ in the direction the partner demands, then follow the charges:
• Gaining H+ makes the charge one unit more positive: HPO42− → H2PO4−
• Losing H+ makes it one unit more negative: HPO42− → PO43−
Then check both sides. Equation 1: left is (−2) + 0 = −2, right is (−1) + (−1) = −2 ✓. Equation 2: left is (−2) + (−1) = −3, right is (−3) + 0 = −3 ✓. Charge must balance in every equation, and it catches most errors instantly.
A note on the formulas as written
"HNO" and "NO1−" are an unusual pair to see in an intro problem — HNO (nitroxyl) and NO− are genuinely a conjugate acid/base pair, so the equations above are correct as asked. But it is worth checking whether your problem set actually means HNO2 (nitrous acid) and NO2− (nitrite), which is the far more common textbook pair.
The logic is identical either way — just substitute:
HPO42− + HNO2 ⇌ H2PO4− + NO2−
HPO42− + NO2− ⇌ PO43− + HNO2
Which way does each one lie?
An acid–base equilibrium always favours the side holding the weaker acid, so compare pKa values. For phosphoric acid: pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.35.
Taking the nitrous acid version: HNO2 has pKa = 3.40, making it much stronger than H2PO4− (7.20). So the first equilibrium lies far to the right — roughly 103.8 in favour of products. The second lies far to the left, since HNO2 is also much stronger than HPO42− (12.35).
That asymmetry is the real lesson: being amphiprotic means HPO42− can go either way, but it does not do both equally well. Against a reasonably strong acid it behaves as a base almost exclusively.
