Question

A 10.0 mL sample of HNO3 was diluted to a

volume of 100.00 mL. Then 25 mL of that
diluted solution was needed to neutralize 50.0
mL of 0.60 M KOH. What was the
concentration of the original nitric acid?
1.2 M
12 M
none of these
O 0.12 M
0.0012M

Answer 1

12 M

Step-by-step explanation:

The reaction between HNO₃ and KOH is:

• HNO₃ + KOH → KNO₃ + H₂O

First we calculate how many KOH moles reacted with the diluted HNO₃ sample, using the given volume and concentration:

• 50.0 mL * 0.60 M = 30 mmol KOH

As 1 KOH mol reacts with 1 HNO₃ mol, in 25 mL of the diluted HNO₃ solution there are 30 HNO₃ mmoles.

With that information in mind we can calculate the HNO₃ concentration in the diluted solution:

• 30 mmol HNO₃ / 25 mL = 1.2 M

Finally we can use the C₁V₁=C₂V₂ formula to calculate the concentration of the original solution:

• C₁ * 10.0 mL = 1.2 M * 100.00 mL

• C₁ = 12 M