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Question

A 5.0 mL sample of a stock solution of concentrated HNO3, a strong acid, is diluted with distilled water to prepare 100.0 mL of its dilute solution. A 10.0 mL sample of this dilute solution is then taken, and a pellet of sodium hydroxide, NaOH, with a mass of 250 mg is added to it. This pellet completely neutralizes the dilute acid solution. Calculate the molarity of the original stock solution of concentrated HNO3.

Answer 1

Work backwards.

The neutralization reaction of HNO₃ with NaOH is

HNO₃ + NaOH ==> NaNO₃ + H₂O ... balanced equation

mols NaOH used = 250 mg NaOH x 1 g / 1000 mg x 1 mol NaOH / 40 g = 0.00625 mols NaOH

mols HNO₃ in the 10.0 ml sample = 0.00625 mols HNO₃ since the mole ratio in balanced eq. is 1 : 1

So, you have 0.00625 mols HNO₃ in the 10.0 ml sample.

Concentration HNO₃ = 0.00625 moles / 10.0 mls x 1000 mls / L = 0.625 mols / L = 0.625 M

This is also the concentration of HNO₃ in the 100.0 mls sample from which the 10.0 ml sample was taken.

To find the concentration in the original 5.0 ml, we need to multiply by 20, since 5 mls diluted to 100 mls is a 20x dilution.

Molarity of original stock solution of HNO₃ = 20 x 0.625 M = 12.5 M HNO₃