How many moles of HX have been added at the equivalence point?

Assume that 30.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solution of the monoprotic strong acid HX.

1 Answer
Jun 21, 2016


#"0.0030 moles HX"#


You're titrating a weak base #"B"# that can accept one proton with a strong acid #"HX"# that can donate one proton, so right from the start you can say that the number of moles of strong acid must match the number of moles of weak base in order to have a complete neutralization.

That is what you're looking for here -- the number of moles of strong acid needed to completely neutralize the weak base. This is what the equivalence point is.

So, the balanced chemical equation for this reaction can be written like this

#"B"_ ((aq)) + "HX"_ ((aq)) -> "BH"_ ((aq))^(+) + "X"_((aq))^(-)#

So, if every mole of weak base requires #1# mole of strong acid, it follows that two solutions that have equal molarities must be combined in equal volumes in order to ensure that equal numbers of moles take part in the reaction.

You know that both solutions have a molarity of #"0.10 M"#. The weak base solution has a volume of #"30.0 mL"#, which means that a complete neutralization would require mixing this solution with #"30.0 mL"# of strong acid solution.

All you have to do now is use the molarity of the strong acid solution as a conversion factor to find the number of moles of #"HX"# present in your sample.

A #"0.10 M"# solution will contain #"0.10# moles of solute per liter of solution. Since #"1 L" = 10^3"mL"#, you will have

#30.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.10 moles HX"/(1color(red)(cancel(color(black)("L")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.0030 moles HX")color(white)(a/a)|)))#

The answer is rounded to two sig figs.

SIDE NOTE Notice that the neutralization reaction produces #"BH"^(+)#, the conjugate acid of #"B"#.

This tells you that the pH at equivalence point will actually be lower than #7# because of the presence of #"BH"^(+)#.