How do you calculate the molarity of the #HNO_3# solution?

Question

A pure sample of barium hydroxide of mass 5.68 g was dissolved and diluted to the mark in a 250 mL volumetric flask. It was found that 13.01 mL of this solution was needed to reach the stoichiometric point in a titration of 29.7 mL of a nitric acid solution.

Ellie Andrews · Accepted Answer

The molarity of the nitric acid was 0.116 mol/L.

This is really a two-part question: 1. Molarity of #"Ba(OH)"_2# The formula for molarity is #color(blue)(|bar(ul(color(white)(a/a) "Molarity" = "moles"/"litres"color(white)(a/a)|)))" "# #"Moles of Ba(OH)"_2 = 5.68 color(red)(cancel(color(black)("g Ba(OH)"_2))) × "1 mol Ba(OH)"_2/(171.34 color(red)(cancel(color(black)("g Ba(OH)"_2)))) = "0.033 15 mol Ba(OH)"_2# #"Volume" = "250 mL" = "0.250 L"# #"Molarity" = "moles"/"litres" = "0.033 15 mol"/"0.250 L" = "0.1326 mol/L"# 2. Molarity of #"HNO"_3# The equation for the neutralization reaction is #"Ba(OH)"_2 + "2HNO"_3 → "Ba"("NO"_3)_2 +"2H"_2"O"# #"Moles of Ba(OH)"_2 = "0.013 01" color(red)(cancel(color(black)("L Ba(OH)"_2))) × "0.1326 mol Ba(OH)"_2/(1 color(red)(cancel(color(black)("L Ba(OH)"_2)))) = "0.001 725 mol Ba(OH)"_2# #"Moles of HNO"_3 = "0.0017 25" color(red)(cancel(color(black)("mol Ba(OH)"_2))) × "2 mol HNO"_3/(1 color(red)(cancel(color(black)("mol Ba(OH)"_2)))) = "0.003 450 mol HNO"_3# #"Molarity of HNO"_3 = "moles"/"litres" = "0.003 450 mol"/"0.0297 L" = "0.116 mol/L"#

Serenity Allen · Answer

undefined To calculate the molarity (M) of a solution of HNO3 (nitric acid), you divide the number of moles of HNO3 present by the volume of the solution in liters (L). The formula is:

Molarity (M) = moles of solute (HNO3) / volume of solution (in liters)