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It's very confusing that while $\ce{H2SO4}$ has an $n$-factor of 2, whereas $\ce{H3PO3}$ has $n$-factor 2 instead of 3. Many tried explaining it to me by molecular structures, but I don't know how to draw them. Could you please try explaining without the help of molecular structures?

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    $\begingroup$ As long as you refuse to know the thing, there is no way you can know it. $\endgroup$ Oct 21 '19 at 8:09
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    $\begingroup$ The hint: the structure of $\ce{H3PO4}$ is $\ce{OP(OH)3}$ , but the structure of $\ce{H3PO3}$ is NOT $\ce{P(OH)3}$ $\endgroup$
    – Poutnik
    Oct 21 '19 at 8:20
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    $\begingroup$ Also, note that there is no zero in $\ce{H2SO4}.$ And I'm going to be the third commentator who is going to stress out that you won't be able to go far in chemistry without molecular structures. $\endgroup$
    – andselisk
    Oct 21 '19 at 8:53
  • $\begingroup$ What is the $n$ factor? $\endgroup$
    – Jan
    Oct 21 '19 at 13:07
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$n$-Factor is basically a method to find out a relationship between the compound and what it is equivalent to in terms of acidic nature or basic nature. Note that the two hydrogen atoms in $\ce{H2SO4}$ are both attached to oxygen. But in $\ce{H3PO3}$ two of the three hydrogens are attached to oxygen and the other hydrogen is directly attached to phosphorous.

Note that only those hydrogens attached to highly electronegative elements can be donated easily in case of inorganic acids. One could explain better by using the concept of stabler structures with the $\ce{H}$ removed from $\ce{O}$ than from $\ce{P}$ by resonance. But for now know it as a thumb rule.

Therefore, since $\ce{H3PO3}$ has only two $\ce{H}$ attached to $\ce{O},$ its $n$-factor is 2 and not 3.

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  • $\begingroup$ This is more or less correct (just bring the formatting in order), but the reasoning is based on the structural aspects which OP tries to evade. Sort of catch 22 in action. $\endgroup$
    – andselisk
    Oct 21 '19 at 8:58
  • $\begingroup$ @andselisk Sorry but I can't use mathjax a lot. Maybe you can help. $\endgroup$ Oct 21 '19 at 9:11
  • $\begingroup$ I tried my best. In the meantime, feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk
    Oct 21 '19 at 9:16
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    $\begingroup$ Thanks andselisk! Ill try to implement the codes little by little from next posts here. And anyway I don't know how the OP wants to get the answer without even the basic Lewis structure? $\endgroup$ Oct 21 '19 at 9:19
  • $\begingroup$ No prob, take your time, we are all learning here. I honestly have no clue either how to give a meaningful answer without getting the structure involved. $\endgroup$
    – andselisk
    Oct 21 '19 at 9:20
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I am afraid the answer if your teacher or textbook has told that the n-factor of phosphoric acid is 2, they are teaching it the wrong way. The concept of equivalent weight is totally dependent on how the reaction equation is written, therefore there is no fixed equivalent weight. Think of the permanganate ion- how many equivalent weights does it have!

I will give an example of phosphoric acid, and you can figure out the same for your compound. So, if you are neutralizing only one proton, in an acid base titration, the n factor is equal to 1, i.e., the equivalent weight is the same as the molecular weight

$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$ n factor is equal to 1 in this case.

However, if you are adding enough base that both are two protons are abstracted.

$\ce{H3PO4 + 2NaOH -> Na2HPO4 + 2H2O}$ n factor is equal to 2 in this case

You added more base:

$\ce{H3PO4 + 3NaOH -> Na3PO4 + 3H2O}$ n factor is equal to 3 in this case.

The main lesson is that equivalent weight is calculated the way you write the equation.

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    $\begingroup$ The questions is NOT about phosphorIC acid but about phophorOUS acid, en.wikipedia.org/wiki/Phosphorous_acid which is indeed dibasic $\endgroup$
    – Ian Bush
    Oct 21 '19 at 16:02
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    $\begingroup$ @IanBush which doesn’t exist as is it actually phosphonic acid ;) $\endgroup$
    – Jan
    Oct 21 '19 at 16:49
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    $\begingroup$ @Indeed Ian Bush. Any person who can count from 1 to 10 would also be able to differentiate between H3PO4 and H3PO3, I gave this 9th grader (OP) an example to follow rather than solve this homework for H3PO3. $\endgroup$
    – M. Farooq
    Oct 21 '19 at 17:25
  • $\begingroup$ @IanBush chemistry.stackexchange.com/questions/85253/… $\endgroup$ Oct 21 '19 at 17:58
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If the $n$-factor is what I assume it is (the basicity, or ‘how many moles of sodium hydroxide can one mole of acid neutralise?’) then …

It is really hard to explain this without molecular structures. Sum formula are of little help to chemists once you reach more than four or five atoms or if there are two types of non-hydrogen atoms. Two compounds may have the same sum formula but different structures explaining a very different reactivity and properties. Take for example $\ce{C4H10O}$ which, among others, can be diethyl ether, the very low viscosity liquid with a boiling point of $\pu{32 ^\circ C}$ that was once used as an anaesthetic and a replacement for alcohol, or can be tert-butanol, a solid at room temperature except in Japanese labs in summer, with a quite different but nonetheless recognisable odour that has no common uses except sometimes apparently as a paint remover according to Wikipedia. So over time you will have to accustom yourself to molecular structures.

As long as you don’t resort to molecular structures, the only real answer is that even at $\mathrm{pH\ 14}$ the anion will remain $\ce{HPO3-}$ and the final hydrogen will never be lost. The answer to why lies in molecular structures.

You can approach the experimental result in different ways, too. For example, certain methods allow esterification, meaning the attachment of an organic group (e.g. ethyl, $\ce{C2H5}$) to the oxygens replacing the hydrogens there. If you perform esterification reactions with $\ce{H2SO4}$ under appropriate conditions, your final product is diethyl sulphate, $\ce{(C2H5O)2SO2}$ – two hydrogens were replaced, $n$ factor 2. If you perform it with phosphoric acid $\ce{H3PO4}$, your product is triethyl phosphate $\ce{(C2H5O)3PO}$. If you perform it with ‘phosphorous acid’, the product only contains two ethyl groups; the final hydrogen is not removable by this method.

Again, the real answer to why requires structures which you expressed you didn’t want. So all I am going to say is that ‘phosphorous acid’ should actually be termed phosphonic acid because that predicts the correct structure and reactivity.

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  • $\begingroup$ pubchem.ncbi.nlm.nih.gov/compound/…, I think IUPAC accepts phosphorous acid as its name. $\endgroup$
    – M. Farooq
    Oct 21 '19 at 17:28
  • $\begingroup$ This n-factor is taught is taught in Indian high schools (only, I guess). It is an outdated concept of equivalent weights, which was also taught in US/British universities until the 60s. n-factor not only refers to acidity or basicity, it is also used for redox reactions, and it totally depends on how the equation is written. Sadly, it has very little to do with molecular structures. For redox species, one has to account for the number of electrons gained or lost. For acids and bases it is indeed basicity or acidity, for salts it is based on the charge...complicated story of equivalents. $\endgroup$
    – M. Farooq
    Oct 21 '19 at 17:31
  • $\begingroup$ Yes, I agree. But knowing the structure of $KMnO_4$ does not help at all in calculating its equivalent weights and their n factors. The point is that I am not against knowing structures but the way equivalent weight is being taught to these students. Now you can imagine why teaching equivalents is not a good idea. It is good for chemical history classes and it has no room in modern science. $\endgroup$
    – M. Farooq
    Oct 21 '19 at 20:35
  • $\begingroup$ @M.Farooq I mean … my point here is … as far as I understand, OP wants to understand why phosphonic (‘phosphorous’) acid is dibasic. That is very easily explained with the corresponding molecular structures plus the knowledge of electronegativity values. It cannot really be explained without structures because there is nothing in the sum formula to explain the different hydrogens. But the redox chemistry of permanganate is an entirely different beast altogether, given that it can reduce to $\ce{Mn^2+, MnO2}$ or $\ce{MnO4^2-}$ depending on conditions. $\endgroup$
    – Jan
    Oct 22 '19 at 8:55

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