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

## 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.

Jun 26, 2016

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

#### Explanation:

This is really a two-part question:

1. What is the molarity of the barium hydroxide solution?
2. What is the molarity of the nitric acid?

1. Molarity of ${\text{Ba(OH)}}_{2}$

The formula for molarity is

color(blue)(|bar(ul(color(white)(a/a) "Molarity" = "moles"/"litres"color(white)(a/a)|)))" "

${\text{Moles of Ba(OH)"_2 = 5.68 color(red)(cancel(color(black)("g Ba(OH)"_2))) × "1 mol Ba(OH)"_2/(171.34 color(red)(cancel(color(black)("g Ba(OH)"_2)))) = "0.033 15 mol Ba(OH)}}_{2}$

$\text{Volume" = "250 mL" = "0.250 L}$

$\text{Molarity" = "moles"/"litres" = "0.033 15 mol"/"0.250 L" = "0.1326 mol/L}$

2. Molarity of ${\text{HNO}}_{3}$

The equation for the neutralization reaction is

$\text{Ba(OH)"_2 + "2HNO"_3 → "Ba"("NO"_3)_2 +"2H"_2"O}$

${\text{Moles of Ba(OH)"_2 = "0.013 01" color(red)(cancel(color(black)("L Ba(OH)"_2))) × "0.1326 mol Ba(OH)"_2/(1 color(red)(cancel(color(black)("L Ba(OH)"_2)))) = "0.001 725 mol Ba(OH)}}_{2}$

${\text{Moles of HNO"_3 = "0.0017 25" color(red)(cancel(color(black)("mol Ba(OH)"_2))) × "2 mol HNO"_3/(1 color(red)(cancel(color(black)("mol Ba(OH)"_2)))) = "0.003 450 mol HNO}}_{3}$

$\text{Molarity of HNO"_3 = "moles"/"litres" = "0.003 450 mol"/"0.0297 L" = "0.116 mol/L}$