Nitrogen and Phosphorus Oxoacids
Hold a bottle of concentrated phosphoric acid and a bottle of concentrated nitric acid side by side. You would never guess, from formula alone, that H₃PO₄ and HNO₃ belong to the same family of compounds. Nitric acid is a fuming, aggressive one-basic oxidiser. It reacts explosively with reducing agents. Phosphoric acid, by contrast, is a syrupy, non-oxidising triprotic acid used to acidify cola. The difference is not really about "N versus P" as elements. It is about what each element's orbitals allow it to build. Nitrogen, confined to period 2, cannot exceed four groups around itself. It leans on π-bonding to satisfy its valence. Phosphorus, one row down, routinely sits at the centre of a five-coordinate or distorted-tetrahedral σ-framework. This framework is stabilised by polarisable d-like or σ*-network orbitals. That single structural fact — traceable to The Inert-Pair Effect and to the hypervalent bonding described for Phosphorus Halides and Hypervalency — controls everything downstream: acid strength, oxidising power, and the ability to condense into chains and rings. This article works through the whole family systematically. It covers nitric acid (HNO₃) and nitrous acid (HNO₂) on the nitrogen side. It also covers phosphoric acid (H₃PO₄), phosphorous acid (H₃PO₃), and hypophosphorous acid (H₃PO₂) on the phosphorus side. It then covers the condensation chemistry — pyrophosphoric acid, triphosphoric acid, metaphosphates, and the cyclic and linear polyphosphates that underlie ATP and industrial water-softening. The oxidation states run from +5 (HNO₃, H₃PO₄) down through +3 (HNO₂, H₃PO₃) to +1 (H₃PO₂). At each step, the basicity — the number of ionisable O–H protons — is dictated not by the oxidation state alone. It depends on how many P–H or N-lone-pair "non-acidic" bonds are actually present in the structure. The big picture is a lesson in structure-before-formula reasoning. H₃PO₃ looks, from its formula, like a triprotic acid analogous to H₃PO₄. But it is diprotic, because one hydrogen is bonded directly to phosphorus (P–H). That hydrogen is not ionisable. H₃PO₂ looks triprotic but is monoprotic, because two of its three hydrogens are P–H. Nitrous acid, unlike phosphorous acid, keeps all its acidic character on oxygen. This is because nitrogen forms no analogous N–H tautomer under ordinary conditions. Meanwhile the condensation reactions — loss of water between two P–OH groups to form P–O–P linkages — turn simple monomeric acids into polyphosphate architectures. These structures store biological energy and sequester metal ions industrially. By the end you should be able to draw every acid from its central-atom valence electron count and VSEPR geometry. You should be able to predict the number of ionisable protons from the structure, not the formula. You should be able to rank acid strengths using Pauling's qualitative rules (foreshadowed here, formalised in Oxoacid Strength and Pauling's Rules). Finally, you should be able to explain why phosphorus — but not nitrogen — builds an entire condensed-phosphate chemistry.
- HNO₃ is trigonal planar at N (sp², three N–O σ-bonds, delocalised π over O–N–O with a formal N=O and two equivalent N–O⁻); it is a strong monoprotic acid and a strong oxidiser.
- HNO₂ is bent at N with a lone pair (sp² N, one N=O and one N–OH); it is a weak monoprotic acid (pKₐ ≈ 3.3) that exists as cis/trans forms and disproportionates readily.
- H₃PO₄ is tetrahedral at P (one P=O, three P–OH); triprotic acid, pKₐ₁,₂,₃ ≈ 2.1, 7.2, 12.4; non-oxidising because P(V) has no accessible lower-energy reduction pathway under mild conditions.
- H₃PO₃ is tetrahedral at P but with one P–H bond (P=O, one P–H, two P–OH); only diprotic (pKₐ₁ ≈ 1.3, pKₐ₂ ≈ 6.7); tautomerises slowly between P(III) trivalent and P(V)-with-P–H forms.
- H₃PO₂ is tetrahedral at P with two P–H bonds (P=O, two P–H, one P–OH); monoprotic (pKₐ ≈ 2.0) and a strong reducing agent because of those P–H bonds.
- Condensation of P–OH groups by loss of H₂O builds pyrophosphate, triphosphate, and cyclic/linear polymetaphosphates — the structural basis of ATP and industrial phosphate glasses.
- Key idea: the number of ionisable protons in an oxoacid is fixed by which hydrogens sit on oxygen versus directly on the central atom, and that partition is dictated by the central atom's period-dependent valence bonding, not by the molecular formula.
Read the structure, not the formula: only O–H hydrogens ionise, and phosphorus's ability to hold P–H bonds while still using a formal P=O (or its resonance-averaged equivalent) is what makes H₃PO₃ diprotic and H₃PO₂ monoprotic, despite formulas that look "triprotic."
Where this lives today
The distinction between phosphorus's P–H-bearing lower oxoacids and phosphoric acid itself is not a museum piece — it is industrial and biological chemistry running today. Hypophosphorous acid, H₃PO₂, and its sodium salt are used as reducing agents in electroless nickel plating, where the P–H bonds are the actual hydride-transfer sites that reduce Ni²⁺ to metallic nickel on a catalytic surface; this only works because H₃PO₂ genuinely has reactive P–H bonds, not because of its (single) acidic proton. Phosphorous acid derivatives and phosphite esters, built on the same P–H-retaining P(III)/P(V) tautomeric framework, are widely used as antioxidant stabilisers in polymer manufacturing and as key reagents in the Arbuzov and Michaelis–Becker reactions for constructing C–P bonds in organophosphorus synthesis, including some pesticide and flame-retardant chemistry. The condensation chemistry of phosphoric acid is even more consequential biologically: ATP is, structurally, a triphosphate ester built from three phosphate tetrahedra linked by P–O–P bridges exactly analogous to the pyrophosphate condensation shown above; the "high-energy" character of the terminal phosphoanhydride bond is a direct consequence of electrostatic repulsion between adjacent negatively charged tetrahedral phosphate groups and resonance stabilisation differences between the triphosphate and the hydrolysis products (ADP + inorganic phosphate). Nucleic acid backbones are themselves polymeric phosphate diesters, chemically continuous with the same P(V) tetrahedral chemistry described here. Industrially, sodium tripolyphosphate (built from the same triphosphate condensation) is used as a water softener and sequestrant in detergents, complexing Ca²⁺ and Mg²⁺ ions via its multiple terminal phosphate oxygens, while cyclic and linear polymetaphosphates form the basis of certain phosphate glasses used in specialty optical and dental materials. On the nitrogen side, nitric acid's dual identity as strong acid and strong oxidiser underlies its industrial role in the Ostwald process (catalytic oxidation of ammonia to make HNO₃ for fertiliser and explosives manufacture) and its use in metal etching and passivation, where its oxidising power — not its acidity — does the chemical work (for example, passivating stainless steel by oxidatively building a protective Cr₂O₃ layer). Nitrous acid's instability and rapid disproportionation (3 HNO₂ → HNO₃ + 2 NO + H₂O) is exploited in diazotisation reactions central to azo-dye synthesis, where HNO₂ generated in situ from NaNO₂ and mineral acid converts primary aromatic amines into diazonium salts.
Ask the tutor: Why does the tautomer of H₃PO₃ with a P–H bond win out thermodynamically over the P(OH)₃ form with a lone pair on phosphorus, in more quantitative bond-energy terms?
The physical picture
Start with the central-atom valence count. Neutral N has 5 valence electrons, neutral P has 5 valence electrons — same starting point. But nitrogen, restricted to the 2s/2p shell, can never exceed an octet: four σ-type electron domains (bonds + lone pairs) is the hard ceiling, and even that requires π-donation from oxygen lone pairs into an empty N-based orbital, or π back-donation into N=O to relieve formal charge. Phosphorus, with accessible 3s/3p and low-lying σ-antibonding networks (the modern replacement for "3d hybridisation," as clarified when discussing <a class="wiki-link" href="/wiki/chem-inorg-phosphorus-halides-hypervalency" data-track="wiki_article.body.wiki_link">Phosphorus Halides and Hypervalency</a>), routinely supports five or six electron domains and, more relevantly here, a P=O double bond that is properly described as a strongly polarised σ-bond reinforced by O(pπ)→P(σ) donation rather than a classical p(π)-p(π) or p(π)-d(π) double bond. The practical upshot: phosphorus in its +5 oxoacids is always four-coordinate and tetrahedral (one P=O, three other groups), while nitrogen in its +5 oxoacid is three-coordinate and trigonal planar (no room for a fourth σ-bond at all). For nitric acid, HNO₃, nitrogen forms three σ-bonds — to two equivalent terminal oxygens and to the OH oxygen — with a nominal N=O and N–O⁻ that delocalise into two equal 1.5-order N–O bonds (confirmed by two equal N–O bond lengths of 1.22 Å in the gas phase, distinct from the longer N–OH bond of 1.41 Å). Nitrogen is sp² hybridised, trigonal planar, no lone pair on N. One oxygen carries the ionisable proton. Nitrous acid, HNO₂, has nitrogen forming only two σ-bonds (to =O and to –OH) plus retaining one lone pair, giving a bent, sp²-ish geometry (O=N–OH angle ≈ 102–110° depending on the cis/trans isomer) — the extra electron pair on N that HNO₃ does not have. This is the structural signature of the drop from N(V) to N(III): losing one N–O bond frees up a lone pair rather than removing an oxygen atom outright, because nitrogen must keep an octet. For the phosphorus series the coordination number at P stays at four all the way from P(V) down to P(I), because phosphorus resolves the electron-count problem by adding P–H bonds instead of removing σ-domains. H₃PO₄: P(V), one P=O, three P–OH, tetrahedral, no P–H, no lone pair on P — three ionisable protons. H₃PO₃: P(III) and P(V) tautomers exist, but the dominant, thermodynamically stable tautomer is the tetrahedral P(V) form with one P=O, one P–H, two P–OH — two ionisable protons. H₃PO₂: tetrahedral P(V) form, one P=O, two P–H, one P–OH — one ionisable proton. In every phosphorus case P stays tetracoordinate; what changes is the O/H partition around it, which is exactly why "counting oxygens" or "counting hydrogens" from the formula alone is a trap — you must know how many hydrogens are P–H.
The condensation chemistry follows directly from the presence of multiple P–OH groups on adjacent tetrahedra. Two H₃PO₄ molecules can eliminate one water across a shared P–O–P bridge to give pyrophosphoric acid, H₄P₂O₇; repeating this gives triphosphoric acid (H₅P₃O₁₀), long linear polyphosphate chains, and, if the chain closes on itself, cyclic metaphosphates like the trimetaphosphate ion (P₃O₉)³⁻. Nitrogen has no comparable chemistry at ordinary conditions: N(V) has no N–OH group left over once it forms NO₃⁻'s three-fold symmetric anion (loss of the proton leaves no second OH to condense with), and nitrogen's lack of accessible higher coordination number blocks the analogous N–O–N bridging seen in phosphates (there exist some N–N-bonded oxoacids, e.g. hyponitrous acid H₂N₂O₂, but that is a different structural motif built on an N–N bond, not O–bridged condensation).
The core ideas
The central relationship threading this whole family is Pauling's empirical rule for oxoacid strength, which will be derived formally elsewhere but is used here descriptively: writing an oxoacid as XOₘ(OH)ₙ, with m the number of non-hydroxyl (formal X=O or X–O⁻) oxygens,
where:
- = number of terminal, non-protonated oxygens on the central atom (the "extra" oxo groups beyond the OH groups)
- the formula is calibrated against measured pKₐ₁ values and is only a rough (±1–2 unit) guide, not a rigorous thermodynamic law
Reading: each additional strongly electron-withdrawing terminal oxygen delocalises negative charge in the conjugate base and inductively stabilises the O–H bond's departing proton, dropping pKₐ₁ by roughly 5 units per oxo group. HNO₃ has m = 2 (two non-OH oxygens), predicting pKₐ₁ ≈ 8 − 10 = −2, consistent with it being a strong acid (measured pKₐ ≈ −1.4). HNO₂ has m = 1, predicting pKₐ₁ ≈ 3, matching the measured 3.3 almost exactly. H₃PO₄ has m = 1 (one P=O, three P–OH), predicting pKₐ₁ ≈ 3 — close to, though a bit higher than, the measured 2.15 (phosphorus's larger size and different inductive behaviour relative to nitrogen account for the residual mismatch, a limitation discussed further under Oxoacid Strength and Pauling's Rules). H₃PO₃ (in its dominant tautomer) also has m = 1 relative to its two ionisable OH groups, and indeed pKₐ₁ ≈ 1.3–2 is in the same ballpark as H₃PO₄'s first ionisation. H₃PO₂, with m = 1 and one OH, has pKₐ ≈ 2.0, again consistent. The second core relationship is the successive-ionisation spacing rule, again empirical but structurally grounded:
where:
- indexes successive ionisation steps of a polyprotic acid (removing a second, third proton from an already-anionic species)
- the ≈5 unit gap reflects the extra electrostatic work needed to remove a proton from a species that already carries negative charge
Reading: each successive deprotonation is roughly 10⁵ times weaker because you are pulling H⁺ away from an increasingly negative conjugate base — pure Coulombic penalty, essentially independent of which oxoacid family you are in. H₃PO₄'s pKₐ₁, pKₐ₂, pKₐ₃ ≈ 2.15, 7.2, 12.4 show gaps of 5.05 and 5.2 — textbook agreement. H₃PO₃'s two pKₐs, ≈1.3 and ≈6.7, show a gap of 5.4, again consistent, and confirming (independently of the tautomer argument) that only two protons are really ionisable — if a third existed it should appear near pKₐ₃ ≈ 11–12, and no such ionisation is observed.
Goal. Show, from phosphorus's valence electron count and octet/bonding rules, why phosphorous acid settles into a structure with one non-ionisable P–H bond rather than a P(OH)₃ structure with a lone pair on P (which would make it look like a triprotic, "pyramidal" acid analogous to the way sulfite SO₃²⁻ retains a lone pair on S).
Step 1 — count valence electrons available to phosphorus. Neutral P contributes 5 valence electrons. In the acid H₃PO₃, the empirical formula tells us there are 3 H, 1 P, 3 O. If we tried a trivalent, lone-pair-retaining structure P(OH)₃ — three P–O single bonds plus a lone pair on P, directly analogous to P(OH)₃ being the tautomer of "phosphorous acid" written as a phosphine-oxide-free trivalent P(III) species — phosphorus would have 3 bonding pairs + 1 lone pair = four electron domains, pyramidal, all three O's bearing OH, and all three H's would sit on oxygen. This tautomer indeed exists as a minor, kinetically accessible form (the P(III) tautomer), and it is this form that is the immediate product of some syntheses.
Step 2 — compare energetics: P=O bond formation versus a lone pair on P. The alternative, thermodynamically dominant tautomer converts one P–OH into a P–H bond plus a P=O bond: HP(=O)(OH)₂. This costs breaking one O–H bond and forming one P–H bond, but gains a strong P=O σ-bond heavily reinforced by O(2p)→P(σ*) donation — the same type of strong, short (≈1.50 Å) P=O bond found in H₃PO₄, phosphine oxides, and phosphonates. Because this P=O bond is worth roughly 540–580 kJ mol⁻¹ of bond energy (much stronger than a typical P–O single bond, ≈335 kJ mol⁻¹), the tautomerisation P(OH)₃ ⇌ HP(=O)(OH)₂ is strongly exothermic, and equilibrium lies essentially completely (>1000:1) on the P=O side at room temperature.
Step 3 — locate the resulting hydrogens. In HP(=O)(OH)₂, phosphorus has four σ-bonds: one to H (P–H, non-polar, non-ionisable — this hydrogen never sees water's base character because the P–H bond has negligible bond dipole toward H and no resonance stabilisation of P⁻ + H⁺ dissociation), one P=O, and two P–OH. Only the two O–H hydrogens are acidic.
Step 4 — confirm by physical measurement. ³¹P NMR of H₃PO₃ shows a large one-bond P–H coupling (¹J(P,H) ≈ 630–700 Hz) characteristic of a direct P–H σ-bond, and IR shows a P–H stretch near 2400 cm⁻¹ absent in H₃PO₄. Potentiometric titration shows exactly two equivalence points (pKₐ₁ ≈ 1.3, pKₐ₂ ≈ 6.7), not three. Both independent experiments confirm the diprotic, P–H-containing structure over the trivalent P(OH)₃ tautomer.
Conclusion. The formula H₃PO₃ hides a structural fact: phosphorus's ability to form a strong P=O bond makes the P–H tautomer thermodynamically preferred, and the resulting molecule is diprotic, not triprotic — a direct consequence of period-3 phosphorus's bonding flexibility relative to nitrogen.
Go deeper · Derivation B · 2 minBond-order and resonance structure of HNO₃: why N-O bonds average to 1.33 and only one proton ionises
Goal. Derive the bonding picture of nitric acid from nitrogen's electron count and resonance, and show why exactly one hydrogen is acidic while the two "non-OH" oxygens become fully equivalent in the nitrate ion.
Step 1 — build the Lewis structure and count electron domains at N. Neutral N contributes 5 valence electrons; in HNO₃, N is bonded to three oxygens (one bears H) and has no lone pair (if it did, it would exceed the observed trigonal-planar, not pyramidal, geometry, and would leave N electron-deficient relative to an octet unless a lone pair is retained — check by formal charge). Assign one N=O double bond, one N–O(⁻) single bond bearing a full negative formal charge, and one N–OH single bond. Total bonds at N: one double + two single = 4 bond pairs, consistent with an octet (2 electrons from the double bond's π-component plus 6 from three σ-bonds = 8, no lone pair) — nitrogen is sp² hybridised, 3 σ-domains, trigonal planar, exactly as VSEPR predicts for AX₃ with no lone pairs.
Step 2 — resonance-average the two non-OH oxygens. Two resonance structures place the formal N=O double bond on either of the two non-hydroxyl oxygens, with the negative formal charge migrating to whichever oxygen is left as N–O⁻. Because these two structures are degenerate (identical energy, related by simple relabelling of equivalent atoms), the true structure is an equal-weighted resonance hybrid: each of those two N–O bonds carries bond order
Step 3 — compare to the N–OH bond, which does not participate in this resonance. The N–OH bond is a genuine single bond throughout (it is never doubly bonded in any low-energy resonance structure, because doing so would require moving the ionisable proton or breaking the O–H bond) — bond order 1, and correspondingly longer (≈1.41 Å, vs ≈1.22 Å for the resonance-averaged 1.5-order bonds). This asymmetry is exactly what is observed by microwave spectroscopy and X-ray/neutron diffraction of nitric acid, both in the gas phase and in the crystalline hydrate.
Step 4 — identify the acidic proton. Only the oxygen engaged in the N–OH single bond carries a proton; the other two oxygens have no attached hydrogen at all. Removing that one proton (loss of H⁺ from O–H) gives NO₃⁻, in which all three N–O bonds become equivalent (bond order 4/3 = 1.33, the classic D₃ₕ nitrate ion, confirmed by three equal N–O bond lengths of 1.24 Å and a single sharp ¹⁵N NMR signal / single symmetric IR stretching pattern). There is no second or third proton to remove — the acid is unambiguously monoprotic.
Conclusion. Nitric acid's resonance structure explains simultaneously why it has only one ionisable proton (only one oxygen ever carries H), why its two "double-bond" oxygens are experimentally indistinguishable (resonance averaging), and why the nitrate ion formed on deprotonation becomes fully symmetric (bond order drops uniformly from an average of 1.5 in the neutral acid's non-OH oxygens to 1.33 across all three oxygens in the anion, since the negative charge is now shared three ways instead of two).
Worked examples
Example 1 (foundational) — Assign the P oxidation state and count ionisable protons in H₃PO₂. Formula: H₃PO₂, i.e., 3 H, 1 P, 2 O. First find the oxidation state of P using the standard rule that H bonded to a nonmetal counts as +1 and O counts as −2, with the overall molecule neutral:
So phosphorus is formally P(+1) by the naive oxidation-state bookkeeping rule that assumes all H's are equivalent — but this naive number is misleading, because it silently assumes all three hydrogens are on oxygen. We must check the actual structure. Phosphorus needs four σ-bonds to be tetrahedral (its universal geometry in oxoacids): the Lewis structure that satisfies phosphorus's tetravalent, four-domain preference is H–P(=O)(H)–OH, i.e., one P=O, two P–H, one P–OH. Redo the oxidation state bookkeeping correctly, now distinguishing H bonded to P (which, being more electronegative than H, makes P the negative partner in that bond, so that H is assigned +1 and P effectively "gains" from those bonds) from H bonded to O: Using the correct structural formula: P is bonded to =O (contributes −2 formally, standard for a double bond to O), two H atoms directly on P (each H is +1, but since P is more electronegative than H, in oxidation-state counting each P–H bond assigns the shared electron pair to P, so P effectively contributes −1 per P–H bond to its own oxidation state — equivalently, each P–H bond lowers P's oxidation number by 1 relative to treating that position as a bond to a +1 partner not assigned to P), and one P–OH. The standard result, cross-checked against tabulated values, is P(+1) — so the naive bookkeeping above (treating all H as +1 and assigning them away from P) happens to give the right oxidation number in this case, but the reasoning matters for identifying acidity, not oxidation state: oxidation state is a bookkeeping tool, whereas ionisable-proton count depends on which atom the H is physically bonded to. The structural formula HP(H)(=O)(OH) shows only ONE oxygen-bound hydrogen. Hypophosphorous acid is therefore monoprotic, with pKₐ ≈ 2.0, even though its formula "H₃PO₂" naively suggests three acidic protons. Limiting-case check: if instead all three H were O-bound (a hypothetical, non-existent tautomer P(OH)₃ minus one O, which is not geometrically sensible for a 2-oxygen formula), the compound would need to be triprotic; the fact that experiment shows only one titratable proton (single equivalence point in titration with NaOH, 1:1 stoichiometry) directly falsifies that hypothetical structure and confirms the two P–H bonds.
Example 2 (applied) — Rank the acid strengths of HNO₃, HNO₂, H₃PO₄, and H₃PO₃ using Pauling's m-rule, then compare to real pKₐ values. Tabulate m (number of non-hydroxyl oxygens on the central atom) for each acid using its correct structure:
| Acid | Structure (central atom) | m | Predicted pKₐ₁ = 8 − 5m | Measured pKₐ₁ |
|---|---|---|---|---|
| HNO₃ | trigonal N, 2×(N–O, resonance) + 1×N–OH | 2 | −2 | ≈ −1.4 |
| H₃PO₄ | tetrahedral P, 1×P=O + 3×P–OH | 1 | 3 | 2.15 |
| H₃PO₃ | tetrahedral P, 1×P=O, 1×P–H, 2×P–OH | 1 | 3 | 1.3 |
| HNO₂ | bent N, 1×N=O + 1×N–OH | 1 | 3 | 3.3 |
Reading the table: the m-rule correctly places HNO₃ far below the others (strong acid, m = 2) and correctly groups H₃PO₄, H₃PO₃, and HNO₂ together in the "m = 1, moderate acid, pKₐ ≈ 1–4" bracket. It does not perfectly rank the fine order within that bracket (H₃PO₃ is actually somewhat stronger than H₃PO₄ despite equal m, because the electron-withdrawing P–H bond and the acid's smaller effective molecular size shift the balance slightly) — a known limitation of this zeroth-order rule, refined by more detailed electronegativity/charge-distribution treatments elsewhere. Limiting-case check: setting m = 0 (a hypothetical acid with no oxo groups at all, e.g. a simple alcohol-like X–OH with no other X=O) predicts pKₐ₁ ≈ 8, matching the observed weakly acidic behaviour of simple alcohols and boric acid's B(OH)₃-type first ionisation region (pKₐ ≈ 9.2, in the right ballpark for m ≈ 0), which is a sensible physical limit: no adjacent electron-withdrawing oxo group means minimal charge delocalisation assistance for O–H ionisation.
Example 3 (exam/competition level) — Given 5.00 g of white phosphorus P₄ burned completely in a limited supply of oxygen to give P₄O₆ (not P₄O₁₀), and the P₄O₆ then hydrolysed completely in excess cold water, calculate the mass of phosphorous acid H₃PO₃ produced, and state whether the resulting solution can be exactly neutralised by NaOH in a 1:2 mole ratio of acid:base.
Step 1 — moles of P₄. Molar mass of P₄ = 4 × 30.97 = 123.9 g mol⁻¹.
Step 2 — stoichiometry of combustion to P₄O₆. The reaction is P₄ + 3 O₂ → P₄O₆ (limited oxygen; each P ends at P(+3), consistent with phosphorus trioxide's structure, an adamantane-like cage of P₄ with six bridging oxygens, no terminal P=O). One mole of P₄ gives one mole of P₄O₆:
Step 3 — hydrolysis of P₄O₆. Each P₄O₆ hydrolyses with six molecules of water to give four molecules of H₃PO₃ (this stoichiometry follows from P₄O₆ having 4 P atoms, each needing to pick up one P–H is not what happens here — rather, each bridging P–O–P is cleaved by water addition, each P ending with one terminal P=O, one P–H is incorrect for this route; the correct hydrolysis is P₄O₆ + 6 H₂O → 4 H₃PO₃, where the P–H bond is generated internally by disproportionation-free tautomerisation of the initially formed P(OH)₃ units into HP(=O)(OH)₂). Balance check: left side has 4 P, 6 + 6 = 12 O total (6 from P₄O₆ + 6 from 6 H₂O), 12 H; right side, 4 × H₃PO₃ = 4 P, 4×3 = 12 H, 4×3 = 12 O. Balanced.
Step 4 — mass of H₃PO₃. Molar mass of H₃PO₃ = 3(1.008) + 30.97 + 3(16.00) = 3.024 + 30.97 + 48.00 = 81.99 g mol⁻¹.
Step 5 — neutralisation stoichiometry. Because H₃PO₃ is diprotic (one P–H bond, two ionisable O–H, as established structurally above), complete neutralisation requires exactly 2 mol NaOH per mole of acid:
so yes, the 1:2 acid:base mole ratio given in the problem is exactly the stoichiometric ratio required — the solution is neutralised precisely, with no excess acid or base, producing 0.3228 mol NaOH-equivalent consumed, i.e.
A student who assumed H₃PO₃ was triprotic (naively reading "H₃PO₃" as three acidic H's) would have wrongly demanded a 1:3 ratio and miscalculated the required NaOH by 50%; the structural analysis above is what prevents that error.
Limiting cases, scope & common traps
| Situation | What actually happens | Why |
|---|---|---|
| Reading "H₃PO₃" as triprotic from the formula | It is diprotic | One H is P–H, non-ionisable; confirmed by ¹J(P–H) NMR coupling and 2-step titration |
| Reading "H₃PO₂" as triprotic | It is monoprotic | Two H's are P–H; only one O–H |
| Assuming HNO₃'s two "non-OH" oxygens are chemically distinct | They are resonance-equivalent | Two degenerate Lewis structures average to 1.5 bond order each |
| Assuming HNO₂ has no lone pair on N (by analogy to HNO₃) | HNO₂'s N carries one lone pair | Dropping from N(V) to N(III) removes an oxygen without removing an electron domain — freeing a lone pair, giving bent geometry |
| Assuming nitrogen condenses into polymeric oxoacids like phosphorus does | It does not, under ordinary conditions | N(V) has no second OH to condense; N's small size and lack of low-lying acceptor orbitals block extended O-bridged polymerisation |
| Assuming Pauling's m-rule gives exact pKₐ | It is only accurate to ±1–2 units | It ignores solvent structure, exact bond polarisation, and central-atom-specific effects (formalised further elsewhere) |
| Treating H₃PO₄ as an oxidising acid like HNO₃ | H₃PO₄ is essentially non-oxidising | P(V)→P(III) reduction has unfavourable kinetics/thermodynamics under normal aqueous conditions, unlike N(V)→ lower oxides, which are kinetically accessible via nitrogen's smaller reorganisation energy and multiple accessible oxidation states |
The chemist behind it
Origins · 1 minLinus Pauling and how this was discovered
Pauling's 1930s–1940s work on the nature of the chemical bond included an empirical scheme — refined across successive editions of The Nature of the Chemical Bond — for estimating oxoacid pKₐ values directly from the number of non-hydroxyl oxygens on the central atom. It was one of the first genuinely predictive, structure-based rules in inorganic acid–base chemistry, bridging simple Lewis-structure counting and quantitative thermodynamic behaviour, decades before high-level computational methods could confirm the underlying charge-delocalisation argument numerically.
Pauling built his career on turning qualitative bonding pictures into testable numerical rules — electronegativity, resonance energy, and the oxoacid strength rule discussed here are all products of the same instinct: that a good structural diagram, drawn correctly, should let you predict a number. He was equally at home reasoning about the ionic/covalent character of a bond and about the folding of a protein helix, and his insistence that oxidation states and formal charges are bookkeeping devices — useful, but subordinate to the actual electron distribution — is precisely the discipline this article has applied throughout: never trust a formula's apparent proton count without first drawing the structure.
Origins · 1 minhow this was discovered
Why is HNO3 such a strong oxidiser while H3PO4 is essentially not?
Both are +5 oxoacids of their respective central atoms, but oxidising power depends on the thermodynamics and kinetics of reducing the central atom, not on acid strength. Nitrogen has many accessible, closely spaced oxidation states (+5, +4, +2, 0, −3) with relatively small reorganisation energies between them, so N(V) in NO₃⁻ can be reduced to NO₂, NO, or lower species with modest activation barriers, especially in the presence of reducing agents and acidic conditions. Phosphorus's reduction from P(V) to P(III) or lower requires breaking strong P=O bonds and reorganising a rigid tetrahedral framework, which is kinetically sluggish and thermodynamically less favourable under ordinary aqueous conditions — so H₃PO₄ survives essentially unreduced even in contact with many reducing agents that HNO₃ would readily oxidise.
Is the P=O bond in phosphoric acid a true double bond involving phosphorus 3d orbitals?
Not in the classical sense once attributed to it. Early treatments invoked p(π)–d(π) back-bonding using empty phosphorus 3d orbitals, but modern computational and spectroscopic work shows phosphorus 3d orbitals are too high in energy and too diffuse to contribute significantly. The better description, consistent with treatments of Phosphorus Halides and Hypervalency, is a strongly polarised P–O σ-bond reinforced by oxygen lone-pair donation into a low-lying P–O σ* orbital (negative hyperconjugation), which shortens the bond and raises its formal bond order without requiring literal d-orbital participation.
Why does nitrous acid exist as separate cis and trans isomers but nitric acid does not?
HNO₂ is bent at nitrogen (sp²-like, with one lone pair on N), and the O=N–O–H dihedral arrangement allows the terminal OH to sit either on the same side as the lone pair (cis, more common in the gas phase, roughly 80:20 over trans at room temperature) or the opposite side (trans), because rotation about the N–OH bond is restricted by partial double-bond character from delocalisation of the lone pair. HNO₃, in contrast, is trigonal planar with full three-fold-related resonance among all three oxygens (no restricted single-bond rotation issue of the same kind, and no lone pair on N), so there is no analogous cis/trans isomerism to speak of — the molecule already has a fixed, symmetric planar shape.
Why does phosphorous acid H3PO3 slowly disproportionate on heating?
On heating, H₃PO₃ undergoes an internal redox disproportionation: 4 H₃PO₃ → 3 H₃PO₄ + PH₃. Phosphorus, already sitting at the P(III)/P(V) tautomeric borderline (with one P–H bond already present), can be pushed by thermal energy to redistribute — some P centres are driven to full oxidation (P(V) in H₃PO₄) while others are driven to the fully reduced hydride PH₃ (P(−III)). This reflects the general instability of intermediate oxidation states relative to their neighbouring extremes when a low-energy reaction pathway exists, a pattern also seen in the disproportionation of HNO₂ into HNO₃ and NO.
How do you predict the number of ionisable protons in an unfamiliar oxoacid you have never seen before?
Draw the Lewis structure of the central atom explicitly, applying the correct coordination number for that element in that oxidation state (nitrogen: three σ-domains maximum with an octet; phosphorus in oxoacids: essentially always four σ-domains, tetrahedral). Identify every hydrogen and determine whether it is bonded to oxygen (acidic, ionisable in water) or directly to the central atom (non-acidic, essentially non-ionisable, since X–H bonds where X is not more electronegative than H, or where no resonance stabilises the resulting anion, do not behave as Brønsted acids under normal aqueous conditions). Count only the O–H hydrogens as ionisable protons — never read the number of acidic protons directly off the molecular formula.
Why does the pKa gap between successive ionisations of a polyprotic acid tend to be about 5 units, and does this rule ever fail?
The roughly 5-unit gap is essentially electrostatic: removing a second proton from an already-negatively-charged conjugate base costs extra Coulombic work because you are separating a positive H⁺ from a more negatively charged, more strongly attracting anion than in the neutral acid's first ionisation. This rule is fairly reliable for simple oxoacids like phosphoric or phosphorous acid where the successive anions are structurally similar and differ mainly in charge. It becomes less reliable when a structural change accompanies ionisation (for example, a conformational shift, intramolecular hydrogen bonding stabilising one particular anion unusually well, or a change in coordination number), so it should be treated as a useful estimate rather than an exact law.
Does the number of P–H bonds in these acids relate to their use as reducing agents?
Yes, directly. A P–H bond is a hydridic (relatively electron-rich toward H) bond compared to an O–H bond, and it can act as a two-electron reductant when phosphorus is oxidised further (P(III)→P(V) or P(I)→P(III)/P(V)) while transferring hydride-like character to a substrate or simply undergoing oxidation itself. This is why H₃PO₂ (two P–H bonds) is a considerably stronger reducing agent than H₃PO₃ (one P–H bond), which is in turn a stronger reductant than H₃PO₄ (zero P–H bonds, and correspondingly essentially non-reducing). The reducing power tracks the P–H bond count almost linearly across the series, a clean structure–property correlation that reinforces why the structural (not formula-based) picture of these acids matters.