50.0mL of an #H_2SO_3# solution were titrated with 36.90mL of a 0.100M #LiOH# solution to reach the equivalence point What is the molarity of the #H_2SO_3# solution?

1 Answer
Ernest Z.
Jul 2, 2016

The molarity of the #"H"_2"SO"_3# is 0.0369 mol/L.

Explanation:

The equation for the neutralization reaction is:

#"H"_2"SO"_3 + "2LiOH" → "Li"_2"SO"_3 + "2H"_2"O"#

#"Moles of LiOH" = "0.036 90" color(red)(cancel(color(black)("L LiOH"))) × "0.100 mol LiOH"/(1 color(red)(cancel(color(black)("L LiOH")))) = "0.003 69 mol LiOH"#

#"Moles of H"_2"SO"_3 = "0.003 690" color(red)(cancel(color(black)("mol LiOH"))) × ("1 mol H"_2"SO"_3)/(2 color(red)(cancel(color(black)("mol LiOH")))) = "0.001 845 mol H"_2"SO"_3#

#"Molarity of H"_2"SO"_3 = "moles"/"litres" = "0.001 845 mol"/"0.0500 L" = "0.0369 mol/L"#

Related questions
See all questions in Titration Calculations
Impact of this question
2191 views around the world
You can reuse this answer
Creative Commons License