Question

A 12.8 mL sample of HNO3 was diluted to a volume of 100.52 mL. Then, 26 mL of that diluted solution were needed to neutralize 46.0 mL of 0.47 M KOH. What was the concentration of the original nitric acid?

Answer 1

First calculate the concentration of HNO3 as follows:

HNO3 + KOH= KNO3 + H2O

Mole of KOH = 0.046 L*0.47 M=0.022 mol KOH

Now calculate the moles of HNO3:

0.022 mol KOH * 1.0mol of HNO3/ 1.0mol of KOH

=0.022 mol HNO3

See given that 26 mL of that diluted solution were needed to neutralize 46.0 mL of 0.47 M KOH.

Volume of HNO3 = 26 ml = 0.026 L

Molarity = 0.022 mol / 0.026 L= 0.83 M

Now calculate the concentration of the original nitric acid:

M1V1= M2V2

Here M 1= ? V1=12.8 ml, M2 = 0.83 M V2= 100.52 ml

M1= 0.83*100.52/12.8

M1= 6.52 M

Hecne the concentration of the original nitric acid is 6.52 M