Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. It only takes a minute to sign up.
Sign up to join this community
I'm having a hard time understanding what equivalent weight means physically. I've looked at many articles on Google but the definition always seems nebulous to me.
For example, if the equivalent of $\ce{H2SO4}$ is $\pu{49 g/mol}$, what exactly does that mean?
Wow, equivalent weights keep popping up again and again. It is not a nebulous term, the point is you are checking recent webpages. You have to look up chemistry books from 1930 to 60. If corona were not an issue I would have suggested a library trip to read Vogel's Quantitative Chemical Analysis (oldest edition).
Equivalent weight has nothing to do with mol. It predates moles story because the unit originated centuries ago. Mixing mole with equivalents is perhaps the main source of confusion.
Equivalent is the weight of X in grams, that will furnish or combine with 1 gram H or 16 gram O atoms.
Let us take sulfuric acid, ask yourself, what weight (=mass) of sulfuric acid will provide 1 gram H. Note I am not involving any mole concept here. The answer is that if we take 49 g of sulfuric acid we will be able to "generate" 1 g hydrogen atoms.
The only opportunity where the equivalents are still used today is in the measurement of pollution in used waters. As the nature of the dissolved impurities is not known, their amount cannot be precisely known or weighed. But they can be characterized by the amount of electrons consumed by a strong oxidant like $\ce{K2Cr2O7^}$ or $\ce{KMnO4}$ if this oxidizing stuff is used for oxidizing all the pollution (whatever its nature, solid, colloidal or liquid). This amount of used electrons, whatever their organic origin, does characterize the equivalent amount of pollution.
For example, if $\pu{12 mL}$ $\pu{ 0.1 M}$ $ \ce{K2Cr2O7}$ is needed to oxidize all the organic impurities present in $\pu{1 L}$ used water, it means that $\pu{1.2E-3 mol}$ $\ce{K2Cr2O7}$ has been used in this titration. And as the ion $\ce{Cr2O7^{2-}}$ needs $6$ electrons to be reduced to $\ce{2Cr^{3+}}$, the amount of electrons used in this titration is $6$ times the amount of $\ce{Cr2O7^{2-}}$ ions, and this is $\pu{7.2E-3mol} $. So it makes sense to say that this amount of pollution is $\pu{7.2E-3}$ equivalent per liter.
