I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products. But I'm facing difficulties in understanding double titrations. Moreover, this doesn't work for equations where the n-factor of all the reactants and products don't match.

Let's take an example:

$$\ce{KIO3 + KI + H2SO4 \to I2 + K2SO4 + H2O}$$

Here, the gram equivalence of $\ce{H2O}$ is $2$ which should match with that of $\ce{KIO3}$. But the gram equivalence of $\ce{KIO3}$ is $5$.

Similarly, $\text{g.eq}_{\ce{H2SO4}} \neq \text{g.eq}_{\ce{I2}}$ and so on. When the n-factors don't match, the gram equivalents don't match. Here, we can consider gram equivalent to be $\text{number of moles} \times \text{n-factor}$. Again, this n-factor is what is posing a problem.

Can someone define the Law of Equivalence and state its limitations because of which I seem to find problems as I've mentioned above?

  • $\begingroup$ Did you even try reading about this in a book? Which book did you refer to? Is there a particular paragraph in that book you didn't understand (and what specifically did you not understand)? Did you search the internet? Which websites have you browsed? Did you ask your teachers? Did they say something you didn't understand (or that you're unsure of what they said was correct)? $\endgroup$ – Berry Holmes May 24 '17 at 4:30
  • $\begingroup$ You really need to show your efforts. What did you do before you came to us for help. What do you already know about the topic? If you take the trouble to ask a good question, we would be more than happy to help you. As of now, this question still is likely to be closed. $\endgroup$ – Berry Holmes May 24 '17 at 4:32
  • $\begingroup$ That'll be a better way, you can always ask separate questions, asking too many questions in a single question is generally frowned upon. $\endgroup$ – Berry Holmes May 24 '17 at 5:39
  • 2
    $\begingroup$ I honestly don't understand why people are clamoring to get this question closed as "Too Broad". We have plenty of other posts here on Chem.SE that are pretty broad (but don't get closed). @Mathbg Why don't you try asking around at The Periodic Table chat room? You might find someone willing to help ;) chat.stackexchange.com/rooms/3229/the-periodic-table $\endgroup$ – paracetamol May 24 '17 at 15:34
  • $\begingroup$ @paracetamol Thanks a lot.. Hope that helps or maybe I'll get it clear from my teacher $\endgroup$ – Mathejunior May 24 '17 at 15:36

Basically the law of equivalence wants you to balance the equivalents that are involved in your reaction. I'd like you to view you reactions in two parts, the oxidation and reduction halves. I see:

$$\begin{align} \ce{I- -> I2} \tag{oxidation} \\ \ce{IO3- -> I2}\tag{reduction} \end{align}$$

In other words, we've just discovered a (reverse disproportionation?) comproportionation reaction. The change in the oxidation state of iodine in the oxidation part is $-1 \to 0 $ ($\Delta_{o.s.} = 1$) and for the reduction half it's $+5 \to 0$ ($\Delta_{o.s.} = 5$). So far so good.

Usually, the questioner would provide you with the number of moles of each species and you're done with number of moles times the n-factor. Nevertheless, let's try to give it a shot. I'm balancing this equation:

$$\ce{1KIO3 + 5KI + 3H2SO4 \to 3I2 + 3K2SO4 + 3H2O}$$

This means for $1$ mole of $\ce{KIO3}$ to react, you'll need $5$ moles of $\ce{KI}$. As this is a proportion, converting it to an absolute value, I'd say you've $1n$ moles of $\ce{KIO3}$ and $5n$ moles of $\ce{KI}$.

So if I try to equate the gram-equivalents, I'll have $1n \times 5 = 5n \times 1$. In other words, $\text{Eq.}_{\ce{KIO3}} = \text{Eq.}_{\ce{KI}}$

I've heard that the law of chemical equivalence states that the gram equivalence of each of the reactants equals the gram equivalence of each of the products.

Law of equivalence is even more fun, for a balanced chemical reaction:

$$ \ce{aA + bB -> cC + dD}$$

Equivalents of every species will be equal. Mathematically,

Equivalents of A = Equivalents of B = Equivalents of C = Equivalents of D

will hold.

Why aren't the gram equivalents of $\ce{H2SO4}$ equal to that of water?

You need to strip your equation of the spectating species. For instance $\ce{H2SO4}$ is primarily here because it acts as a medium-provider, it makes the solution acidic. Neither $\ce{SO4^2-}$ nor $\ce{H+}$ underwent any oxidation/reduction so applying the equivalent concept makes no sense in this case.

  • $\begingroup$ Your answer is really helpful. But I have a doubt, aren't the equivalents of each and every reactant and product same irrespective of whether they belong to a balanced chemical equation or not? $\endgroup$ – Guru Vishnu Sep 2 '19 at 14:30

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