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I don't understand the concept of N factor at all. Why is it important?

I know that it is something dependent upon the oxidation number, for example N factor of Na would be 1 since it has oxidation number of +1.

For acid and bases we look at the replaceable $\ce{H+}$ or $\ce{OH-}$ present in them.

Apart from this, I didn't really understand how I can use N factor for calculation or how to find the N factor for whole chemical reactions. Please correct me where I'm wrong.

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There are some different applications.

For one, it tells you how many times an acid can lose a proton. For instance you can have a diprotic ("with 2 H+") molecule, with n=2. For instance, H2SO4, deprotonates to HSO4-, which can then go to SO4(-2).

Why is that important? Take an acidic reaction with H2SO4. If the extent of reaction is strongly forward, you would treat H2SO4 as if it had twice its concentration to take into account that it has two protons it can lose. If you only accounted for the first proton you would be underestimating this effect.

It's also known as equivalent concentration, which might make it sound less foreign to you.

In redox reactions, it shows how many electrons an agent can accept/donate, and in precipitation reactions it measures the # of ions that'll precipitate out of a solution. For all these situations the n-factor or equivalent concentration corrects for multiplicity, since now we know not all molecules are made equal. Some acids can deprotonate twice, some reagents will lose 4 electrons in redox while others may lose just 2. To consider these differences we use this concept.

Sources: Wikipedia

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n factor is atomic weight or molecular weight by equivalent weight. n factor for acids is its basicity, ie, number of replaceable hydrogen ion per molecule. For exampleg $\ce{HCl}$ has n factor 1 n $\ce{H2SO4}$ has 2 $\ce{H3PO3}$ has n factor 2 because when $\ce{H3PO3}$ dissociates to form H+ and HPO3^-2. Same ways for bases its n factor is its acidity or no of replaceable OH ions per molecule. Eg: for $\ce{NaOH}$ it is 1 n for $\ce{Ca(OH)2}$ its 2.

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N factor is very important for numerical problem in chemistry. N-factor of an acid is its basicity. It is a product of molarity\normality. N-factor helps in determining the gram equivalent of compound. Total H+ ion in a compound shown in reaction is its n-factor. Eg:- $\ce{H2SO4}$ has two hydrogen if one hydrogen is utilised in a chemical reaction then n-factor is 1 if both hydrogen is utilised then 2 is n-factor. Similarly in bases too. In $\ce{NaOH}$ acidity of this compound is. This means n-factor is 1. let a reaction be :- NaOH reacts with $\ce{H2SO4}$ to form $\ce{Na2SO4}$ with water. Here both hydrogen of acid is used so n-factor is 2 for acid and always n-factor of $\ce{NaOH}$ is 1.

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protected by Loong May 13 '17 at 14:31

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