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""To neutralize completely $\pu{20 mL}$ of $\pu{0.1 M}$ aqueous solution of phosphorus $\ce{H3PO3}$ acid the volume of $\pu{0.1 M}$ aqueous $\ce{KOH}$ solution required is: ____""

To solve this question a source used the reaction: $$\ce{H3PO3 + 2 KOH -> K2HPO3 + 2 H2O}$$

While other source used the reaction $$\ce{H3PO3 + 3 KOH -> K3PO3 + 3 H2O}$$

I know that $\ce{H3PO3}$ is dibasic. So I thought definitely two $\ce{H+}$ will get replaced by $\ce{K}$ atoms (As shown in the 1st reaction). But later I learnt that if $\ce{KOH}$ is in excess, then $\ce{H3PO3}$ will react with $\ce{KOH}$ to produce $\ce{K3PO3}$ and $\ce{H2O}$. However, if $\ce{KOH}$ is limited, then $\ce{H3PO3}$ will react with $\ce{KOH}$ to produce $\ce{K2HPO3}$ and $\ce{H2O}$.

So, my question that arose was do we take $\ce{KOH}$ as excess when we solve questions like this or do we take a limited value?

Edit:The 2nd reaction in the question does not take place as described in the accepted answer

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  • $\begingroup$ @DrMoishePippik It is a phosphite. The last H is not acidic. $\endgroup$
    – Poutnik
    Commented Dec 29, 2023 at 21:46
  • $\begingroup$ @Poutnik Sorry I clarified what I meant to ask.I think now one will be able to understand what I meant $\endgroup$
    – Rajesh Paul
    Commented Dec 30, 2023 at 14:48
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    $\begingroup$ Does this answer your question? Why is phosphorous acid diprotic and not triprotic? $\endgroup$
    – Mithoron
    Commented Dec 30, 2023 at 15:13
  • $\begingroup$ Hello @Mithoron .Thanks for your reply but sorry my question was not why phosphorous acid is diprotic.My question was do we take NaOH in excess or in limited while we solve questions like that $\endgroup$
    – Rajesh Paul
    Commented Dec 30, 2023 at 15:18
  • $\begingroup$ Chem+Math formatting reference: MathJax Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Commented Dec 30, 2023 at 16:46

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Phosphorous acid $\ce{H3PO3}$ is not an ordinary acid. It has extremely unusual reactions.

First, although it has three $\ce{H}$ atoms, only two are related to oxygen atoms, and only these two $\ce{H}$ atoms can be replaced by metallic atoms like $\ce{K}$ or $\ce{Na}$. The third $\ce{H}$ atom is related directly to the central phosphorus atom, and cannot be replaced by alcaline atoms. The acid structure is probably better described by $\ce{HPO(OH)2}$ or $\ce{H2[HPO3]}$. So its neutralization by $\ce{NaOH}$ occurs through one of the two following reactions : $$\ce{H3PO3 + NaOH -> NaH2PO3 + H2O}$$ $$\ce{H3PO3 + 2 NaOH -> Na2HPO3 + 2 H2O}$$ These two sodium salts should be better described as :

$\ce{Na[H(PO)(OH)O]}$ or $\ce{NaH[HPO3]}$ instead of $\ce{NaH2PO3}$

$\ce{Na2[H(PO)O2]}$ or $\ce{Na2[HPO3]}$ instead of $\ce{Na2HPO3}$

The "expected" third salt $\ce{Na3PO3}$ is never produced by neutralization and does not exist.

Second, the acid and its salt have strong reducing properties. For example they reduce $\ce{Hg^{2+} and Ag+}$ ions into the corresponding metal : $$\ce{H3PO3 + 2 Ag+ + H2O -> 2 Ag + H3PO4 + 2 H+}$$

Phosphorous acid has the specific property of reducing sulfur dioxide in solution into elementary sulfur at room temperature : $$\ce{2H3PO3 + SO2 -> 2H3PO4 + S}$$

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  • $\begingroup$ I learnt that if NaOH is in excess, then H3PO3 will react with NaOH to produce Na3PO3 and H2O.However, if NaOH is limited, then H3PO3 will react with NaOH to produce Na2HPO3 and H2O.Is this correct? @Maurice $\endgroup$
    – Rajesh Paul
    Commented Dec 30, 2023 at 14:36
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    $\begingroup$ No. The third $\ce{H}$ atom of $\ce{H3PO3}$ is not able to react with a hydroxyde, in excess or not, whatever the circumstances. To be able to react with $\ce{NaOH}$ or $\ce{KOH}$, the $\ce{H}$ atom has to be bound to an oxygen atom. This third $\ce{H}$ atom is not bound to an oxygen atom. So it does not react as an acid. In the similar molecule $\ce{PH3}$, the $\ce{H}$ atoms don't react with hydroxides, even in great excess. $\endgroup$
    – Maurice
    Commented Dec 30, 2023 at 15:38
  • $\begingroup$ Please see the production part in this page en.m.wikipedia.org/wiki/Tripotassium_phosphate @Maurice $\endgroup$
    – Rajesh Paul
    Commented Dec 30, 2023 at 15:55
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    $\begingroup$ @Rajesh Paul. Your reference is related to tripotassium phosphate $\ce{K3PO4}$ which exists. But this well-known phosphate has nothing to do with derivates of phosphorous acid $\ce{H3PO3}$. There are $4$ oxygen atoms in tripotassium phosphate, and not $3$ like in derivates of phosphorous acid. $\ce{H3PO3}$ ≠ $\ce{H3PO4}$ ! $\endgroup$
    – Maurice
    Commented Dec 30, 2023 at 16:51
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    $\begingroup$ Technical note: do not use references to mobile versions of Wikipedia pages. If posting it from a phone, modify en.m.wikipedia.org to en.wikipedia.org $\endgroup$
    – Poutnik
    Commented Dec 30, 2023 at 18:33

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