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I know that pyrophosphite by itself is $\ce {({P_2}{O_5})^{4-}}$. My question is about the hydrogen part- how to know how many hydrogens to add? The answer key for this question says the compound is $\ce {KH_3P_2O_5}$ implying that I should add 3 hydrogens. But why not add 1 hydrogen and have $\ce {K_3}$ instead?

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As phosphorous acid is biprotic:

$$\ce{HO-PH(=O)-OH},$$

the respective pyrophosphorous acid is biprotic too:

$$\ce{HO-PH(=O)-O-PH(=O)-OH}$$

and pyrophosphite is:

$$\ce{^{-}O-PH(=O)-O-PH(=O)-O-}$$

Therefore, potassium hydrogen pyrophosphite would be:

$$\ce{KHP2H2O5}$$

respectively

$$\ce{KH3P2O5}$$

By other words, from 4 hydrogens of pyrophosphorous acid, only 2, bound to O, are acidic. The other 2, bound directly to P, are not.

Similarly, hypophosphorous acid is monoprotic, as only 1 of 3 hydrogen atoms is acidic:

$$\ce{HO-PH2=O}$$


Pyrophosphite is not analogous to pyrosulphite, nor phosphite to sulfite. There is no $\ce{PO3^3-}$ nor $\ce{P2O5^4-}$.

Pyrophosphorous acid is in this context similar to acetic acid. Not all their hydrogen atoms are acidic, so even if fully neutralized, some remain. Sodium acetate is $\ce{CH3COONa}$, not $\ce{CNa3COONa}$.

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  • $\begingroup$ BTW, such an acid salt is hard to achieve due to multiple equilibriums between all phosphites and hydrogenphosphite species. The closest we can have is potassium dihydrogenphosphite, $\ce{KH2PO3}$ which forms the addendum, $\ce{KH2PO3.H2(HPO3)}$. $\endgroup$ Commented Dec 25, 2022 at 13:29
  • $\begingroup$ @NilayGhosh I have quite intentionally avoided physical availability of the acid, considering formal aspect. I suspect the author of the task may have made it up to trick the resolvers, catching them on unusual attributes of acids of phosphorus. $\endgroup$
    – Poutnik
    Commented Dec 25, 2022 at 13:33

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