Question #93d51

1 Answer
Sep 12, 2016

#"3.306 mmoles H"^(+)#


The trick here is to realize that the problem is actually asking for the number of hydrogen ions, #"H"^(+)#, that can be delivered to a reaction by a solution of oxalic acid.

Oxalic acid, #"H"_2"C"_2"O"_4#, is a diprotic acid, meaning that it can deliver two hydrogen ions in a neutralization reaction. For example, you need #2# moles of sodium hydroxide to neutralize #1# mole of oxalic acid

#"H"_ 2"C"_ 2"O"_ (4(aq)) + 2"OH"_ ((aq))^(-) -> "C"_ 2"O"_ (4(aq))^(2-) + 2"H"_ 2"O"_ ((l))#

Now, oxalic acid dihydrate, #"H"_2"C"_2"O"_4 * color(red)(2)"H"_2"O"#, is formed when water of crystallization is added to the structure of the acid.

In other words, every mole of oxalic acid dihydrate will contain

  • one mole of oxalic acid, #1 xx "H"_2"C"_2"O"_4#
  • two moles of water, #color(red)(2) xx "H"_2"O"#

However, you should not count the hydrogen atoms present in the water of crystallization as being potential hydrogen ions.

Those hydrogen atoms are stuck in their respective water molecules and are not going anywhere. When you dissolve oxalic acid dihydrate in water, the water of crystallization will no longer be part of the structure of the acid.

When that happens, you get anhydrous oxalic acid, which is #"H"_2"C"_2"O"_4#.

Therefore, you can say that for every mole of oxalic acid dihydrate that dissolves in water, you get #1# mole of anhydrous oxalic acid, which in turn can deliver #2# moles of hydrogen ions in a neutralization reaction.

The number of millimoles of hydrogen ions present in your sample will thus be

#1.653 color(red)(cancel(color(black)("mmol H"_2"C"_2"O"_4))) * overbrace("2 mmoles H"^(+)/(1color(red)(cancel(color(black)("mmol H"_2"C"_2"O"_4)))))^(color(blue)("because the acid is diprotic")) = color(green)(bar(ul(|color(white)(a/a)color(black)("3.306 mmoles H"^(+))color(white)(a/a)|)))#

The answer has four sig figs.