# The Curious Case of the equivalent weight

I understand the following about equivalent weight. Firstly it is given by,

$$Equivalent\:Weight= \frac{Molar\:Weight}{Valence\:Factor}$$

And secondly, the Valence Factor is equal to the acidity/basicity in case of a acid/base and the change in oxidation number of the compounds in the most cases.

However I encountered the following equation, $$\ce{Cu + HNO3(aq, dil) -> Cu(NO3)2 + NO(g) + H2O(l)}$$ Clearly, in this case we see two roles of nitric acid. I tried breaking down the equation into half reactions as follows:

1. Oxidation of copper: $$\ce{Cu -> Cu^{+2} + 2e-}.....(1)$$
2. Dissociation of Nitric acid $$\ce{HNO3(aq) -> H+(aq) + NO3-(aq)}.....(2)$$
3. Reduction of Nitric acid to Nitrogen Monoxide $$\ce{HNO3(aq) + 3H+(aq) + 3e- -> NO(g) + 2H2O(l)}.....(3)$$

Performing $$(1)\times3,\,(2)\times6,\,(3)\times2$$ we get the balanced equation which is: $$\ce{3Cu + 8HNO3(aq,dil) -> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)}$$ Now I started getting confused, as to how do we calculate the valence factor and hence the equivalent weight for this reaction. I tried doing as follows: $$Valence\,factor,\,n\:= n_{acid}\:+\:n_{reduction}$$ Where after multiplication of the equation we get $$n_{acid}\:=\:n(H^+)\:=6$$ and $$\:n_{reduction}\:=\:n(e^-)\:=6\:$$

Now the total number of nitric acid molecules used up is $$8$$. So using the above mentioned formula we can state $$Equivalent\:Weight\:=\:\frac{8\times63}{12}$$ But this doesn't seem to be right. Can someone please explain this scenario.

Okay, now here are a few reference points:

1. I know that the system of equivalents, normality,valence factor etc is now obsolete and discouraged by the IUPAC, but even then it is of importance theoretically at many junior levels of chemistry where theoretical knowledge is more focussed on.
2. A similar question that I referred: The equivalent weight of HNO3 in the following reaction, however this question firstly doesn't have a satisfactory answer and secondly does not address the methodology involved in the calculation of equivalent weight in such cases. Moreover the accepted answer says that no nitrate ion is formed on the RHS, which is completely wrong as discussed above.
• If you want to solve this problem, you must use and stay with ions, and not present the equation with neutral molecules. The equation to be used will be :$$\ce{3 Cu + 2 NO3^- + 8 H+ -> 3 Cu^{2+} + 2 NO + 4 H2O}$$ And then the rest of the calculation will go without problem. – Maurice Sep 6 at 19:42
• @Maurice, I don't understand how the rest of the calculation would go without problem. Are you saying that we just have to look at the conversion of NO3- to NO? – Tesla's Coil Sep 7 at 4:01

Just to add some more information for you with some clarification. I gather that you are asked to calculate the equivalent weight of nitric acid here.

$$\ce{Cu + HNO3(aq, dil) -> Cu(NO3)2 + NO(g) + H2O(l)}$$

So first thing to keep in mind that the equivalent weight is determined with respect to a single role of the molecule of interest. An exception is a disproportionation reaction, where the normality calculation is slightly tricky. In your case, the only role nitric acid is playing is that of an oxidizing agent not like an acid. The nitrate anion of the copper nitrate has to be ignored.

The equivalent weight of nitric acid= Molecular weight of nitric acid/ 3

The equivalent weight of copper = Atomic weight/2

I have never heard of "valence factor" or so-called n-factor- all these obsolete terms are coined by the local textbook writers. Delete these terms once you pass the exams from your mind.

• Thank you for the answer. But how do you determine which role to pick? – Tesla's Coil Sep 7 at 4:01
• There is no dual role, nitric acid is only acting as an oxidizing agent here. – M. Farooq Sep 7 at 4:43

Please forget about equivalents and normal solutions ! It is an old theory, that was abandoned in the middle of the $$20$$th century. It has been replaced by moles, molarity, and similar parameters. A solution containing $$n$$ moles of a solute in one liter water has a concentration expressed in mole per liter, that can be printed on the label of the flask. The concentration can also be calculated in equivalents per liter, and the result is called $$n'$$ times normal. But this value of concentration depends on the particular choice of reaction in which the solute will be used. For example, if a solution of $$\ce{KMnO4}$$ $$0.1$$ M is used to titrate some sulfite solution, its concentration in equivalents units is $$0.5$$ N if used in acidic medium, and only $$\ce{0.3}$$ N in a basic solution. So please forget about equivalents and normal solutions. Prefer using moles and molar solutions !

• As a student the OP doesn't have a choice. The teacher/book uses the concept and the student must strive to understand the concept. – MaxW Sep 6 at 19:47
• How will the teacher react if the student comes to him saying that a $0.1$ M $\ce{KMnO4}$ solution is sometimes $0.5$ N or sometimes $0.3$ N ? – Maurice Sep 6 at 20:07
• Maurice -- LOL... Isn't that due to quantum mechanics? // I truly understand your point. However if the student is in a class where the teacher is grading the student, then the student doesn't have a choice. So if I'm in a geography class and the teacher tells me that the earth is flat, then for that class the earth is flat. // Teaching using equivalent weights still seems to have a firm following in India. – MaxW Sep 6 at 20:15
• @ MaxW. There are plenty of ways for proving that the Earth is not flat. If you want it, I can explain you how I proceed so that my own students in my own high school has to admit by their eyes that the Earth is round. And the quantum mechanics has nothing to do with stoichiometry, Nothing. – Maurice Sep 6 at 20:30
• Moreover, Rutherford's model, Plum-Pudding Model etc are examples of theories that were discarded, even bohr's model cannot explain everything and is vague, but we still study that and not go on to the schrödinger's model directly do we? So I suppose that it is taught just so we know how things proceeded in the past, and how they are what they are today. And best way to know the shortcomings of a method is to break your head over it and understand where it fails. But what to discard what to not, is not a student's choice. – Tesla's Coil Sep 7 at 3:59

As you noted, IUPAC does discourage using the equivalent weight concept because the value is fuzzy. Consider the following reactions:

1. Dissociation of Nitric acid $$\ce{HNO3 ->[aq] H+ + NO3-}$$
2. Reduction of Nitrate acid to Nitrogen dioxide $$\ce{NO3^- + 2H+ + e- ->[aq] NO2(g) + H2O}$$
3. Reduction of Nitrate acid to Nitrogen Monoxide $$\ce{NO3^- + 4H+ + 3e- ->[aq] NO(g) + 2H2O}$$

For reaction 1, $$\ce{HNO3}$$ loses one proton, thus for the acid base reaction the acid/base equivalence is 1.

For reaction 2, $$\ce{HNO3}$$ loses one electron, thus for that reduction reaction the redox equivalence is 1.

For reaction 3, $$\ce{HNO3}$$ loses three electrons, thus for that reduction reaction the redox equivalence is 3.

So the problem here is what number, 1 or 3, do you pick for the equivalent factor?

• So back when this concept was in practice, what did people do for such reactions? Did they leave it out or did they actually use some method of judging the equivalence factor? Or did this one remain forever answered and added to the disowning of Normality notion? – Tesla's Coil Sep 7 at 3:51