Question #b073d

FULL QUESTION

To find equivalent weight of any compound, the formula is Equivalents of solute/valency then valency for acids is number of replaceable hydrogen ions.

#H_2SO_4 + NaOH -> NaHSO_4 + H_2O#

but why number of replaceable hydrogen in #H_2SO_4# is equal to #1# and not #2#?

#stackrel("--------------------------------------------------------------------------------------------------------------------")#

Sulfuric acid, #"H"_2"SO"_4#, is a diprotic acid that ionizes in two steps.

#"H"_2"SO"_text(4(aq]) -> "H"_text((aq])^(+) + "HSO"_text(4(aq])^(-)#

and

#"HSO"_text(4(aq])^(-) rightleftharpoons "H"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

Only the first ionization is actually considered characteristic of a strong acid, i.e. going to completion. The second ionization is actually characteristic of a weak acid, hence the equilibrium sign.

Now, when you titrate sulfuric acid with sodium hydroxide, #"NaOH"#, a strong base, you will see that the titration curve shows two end-points.

The first end-point occurs when equal number of moles of sulfuric acid and sodium hydroxide are added. This is what you have in your case

#"H"_2"SO"_text(4(aq]) + "NaOH"_text((aq]) -> "NaHSO"_text(4(aq]) + "H"_2"O"_text((l])#

This means that, for this reaction, the number of equivalents produced by one mole of sulfuric acid will indeed be equal to #1#, since you have one mole of protons coming from the acid reacting with one mole of hydroxide ions coming from the base.

If you stop the titration at this point, then you can say that sulfurci acid only produces one equivalent per mole.

As you know, an acid's equivalent weight is defined as the weight of that acid that produces one equivalent of protons.

If you take sulfuric acid to produce only one equivalent of protons, you have

#98.08"g"/color(red)(cancel(color(black)("mol H"_2"SO"_4))) * (1color(red)(cancel(color(black)("mol H"_2"SO"_4))))/("1 equiv. H"^(+)) = "98.08 g/equiv. H"^(+)#

What will happen if you're interested in the complete neutralization of the acid?

If you continue to add sodium hydroxide, you will get

#"NaHSO"_text(4(aq]) + "NaOH"_text((aq]) -> "Na"_2"SO"_text(4(aq]) + "H"_2"O"_text((l])#

Now the acid is completely deprotonated and you're left with a neutral solution.

So what would be the equivalent weight of #"NaHSO"_4#? Once again, because it produces one equivalent of protons per mole, you can say that it's equal to its molar mass.

Now, if you're interested in the complete reaction, then you have

#{("H"_2"SO"_text(4(aq]) + "NaOH"_text((aq]) -> "NaHSO"_text(4(aq]) + "H"_2"O"_text((l])), ("NaHSO"_text(4(aq]) + "NaOH"_text((aq]) -> "Na"_2"SO"_text(4(aq]) + "H"_2"O"_text((l])) :}#

Add these two reactions to get

#"H"_2"SO"_text(4(aq]) + color(red)(2)"NaOH"_text((aq]) + color(red)(cancel(color(black)("NaHSO"_text(4(aq])))) -> color(red)(cancel(color(black)("NaHSO"_text(4(aq])))) + "Na"_2"SO"_text(4(aq]) + 2"H"_2"O"_text((l])#

Finally,

#"H"_2"SO"_text(4(aq]) + color(red)(2)"NaOH"_text((aq]) -> "Na"_2"SO"_text(4(aq]) + 2"H"_2"O"_text((l])#

Now, in this case, sulfuric acid reacts with #color(red)(2)# moles of sodium hydroxide, which means that it will produce two equivalents of protons per mole.

This time, its equivalent weight will be

#98.08"g"/color(red)(cancel(color(black)("mol H"_2"SO"_4))) * (1color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(color(red)(2) " equiv. H"^(+)) = "49.04 g/equiv. H"^(+)#

So, the take-home message here is that the number of active units, i.e. equivalents, depends on the context of the reaction.

For the complete neutralization, sulfuric acid indeed produces two equivalents of protons per mole. For its first ionization, however, it only produces one equivalent of protons per mole.

So always keep in mind what your reaction looks like when trying to figure out the number of equivalents or the equivalent weight of an acid/base.