My chemistry textbook [1], chapter Periodic Properties of Elements, section p-Block elements — formation of oxides suggests that the reaction is

$$\ce{P2O3 + 2 NaOH -> 2 NaHPO3},\tag{R1}$$

but surfing over the internet I couldn't find that compound anywhere (at least in any verified places) which got me thinking whether it was a typo which should be corrected as $\ce{Na2HPO3}$. In fact, the overall reaction I think should be

$$\ce{P2O3 + 2NaOH -> 2Na2HPO3 + H2O},\tag{R2}$$

but that's just my attempt of making a balanced equation. So, which one is correct and why?


  1. Sanjit Kumar Guha. Chemistry Class XI–XII, New and Revised Ed.; Lecture Publications Ltd.: Bangladesh, 2019; Vol. 1. ISBN 978-984-90642-1-3. (in Bengali)

2 Answers 2


Phosphorus(III) oxide $\ce{P4O6}$ (molecular formula for “real” cluster of $T_\mathrm d$ symmetry is preferred over historically inaccurate $\ce{P2O3})$ is hydrophobic and undergoes hydrolysis over time when dispersed in water in solid or liquid form (m.p. ~ 24 °C). Addition of an inorganic base is an nucleophilic attack on $\ce{P4O6}$ that promotes hydrolysis to form corresponding inorganic phosphites.

There is a summary of phosphorus(III) oxide chemistry by D. Heinz [1]. Under ambient conditions $\ce{P4O6}$ hydrolyses over the course of weeks to phosphorous acid $\ce{H3PO3}$, which is contaminated with phosphorus(V) acids and phosphorus suboxides. However, it has been shown that under mild basic conditions, such as sodium bicarbonate aqueous solution, and lower temperatures (0 to 10 °C) about 70 % of $\ce{P4O6}$ is hydrolysed to diphosphite $\ce{H2P2O5^2-}$:

$$\ce{P4O6(s) + 4 H2O(l) -> 2 H2[H2P2O5](aq)}$$

Heinz further points out that depending on temperature, ionic strength, pH and other factors there are numerous reaction paths leading to different hydrolysis rates and condensation grades. The most common products, however, are still phosphites and diphosphites. Products with condensation grade greater than 2 are only intermediate products, and pretty much nonexistent under temperatures over 50 °C due to instability of chain-linked bridged $\ce{P-O-P}$ bonds. To sum it up, the higher the temperature, the greater the hydrolysis rate and the smaller the diphosphites content among the products.

Phosphorus chemistry is not all that primitive, and there is no way to provide an exact answer without knowing the conditions the author of your textbook implies. The only fact that is certain is that hydrolysis with $\ce{NaOH}$ will pretty much always yield disodium hydrophosphite $\ce{Na2HPO3},$ but your textbook made a typo and left incorrect stoichiometry (as other answers pointed out).

If you are still looking for the referenced reaction, there is one listed in Lidin's Reactivity of Inorganic Substances [2, p. 167] for concentrated solution and, as I would speculate, room temperature (albeit the reaction is going to be exothermic):

$$\ce{P4O6 + 6 NaOH (conc)-> 2 Na2HPO3 + Na2H2P2O5 + H2O}$$


  1. Heinz, D. Zur Chemie von Phosphor(III)-Oxid. Pure and Applied Chemistry 1975, 44 (2), 141–172. DOI: 10.1351/pac197544020141. (Open Access, in German)
  2. R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Your reaction scheme is correct, but the equation isn't correctly balanced and the stoichiometry written in your book is written with a typo. It should be $\ce{Na2HPO3}$. While I haven't found the exact equation in Inorganic Chemistry by Housecroft and Sharpe (my go-to inorganic chemistry reference), they say that $\ce{P4O6}$ reacts with water to form $\ce{H3PO3}$. And I assume that the reaction is carried out with aqueous $\ce{NaOH}$, not the molten version (I don't know if that would work too).

The $\mathrm{p}K_\mathrm{a}$ values of phosphonic acid are $2.00$ and $6.59$, so both will be deprotonated by a hydroxide solution, leading to $\ce{Na2HPO3}$.

It is interesting to note that the third hydrogen is not attached to oxygen, but to phosphorus, so the $\ce{PO3^3-}$ ion does not form.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.