Question

What is the concentration of a nitric acid solution if 10.0 mL of the solution is neutralized by 3.6 mL of 0.20 M NaOH? a. 0.072 M b. 53.6 M c. 0.56 M d. 5.6 M

Answer 1

The concentration of the nitric acid solution can be calculated using the properties of neutralization reactions. By multiplying the volume and the molarity of NaOH, we get the moles of NaOH, which equals the moles of HNO₃ in a neutralization. Hence, the concentration of the nitric acid solution is 0.072 M.

Step-by-step explanation:

This question is based on the concept of neutralization reactions in chemistry, which relies on the principle that the number of moles of acid equals the number of moles of base in a neutralization. The reaction here involves nitric acid (HNO₃) and sodium hydroxide (NaOH), producing water and sodium nitrate (NaNO₃).

Given the volume of NaOH is 3.6 mL and its molarity is 0.20 M, the number of moles can be calculated by multiplying the volume (in litres) and molarity.

Moles of NaOH = Volume (L) x Molarity = 3.6/1000 x 0.20 = 0.00072 moles

Since the solution is neutrallized, moles of HNO₃ = moles of NaOH = 0.00072 moles

Then, by dividing the number of moles of HNO₃ by the volume of the nitric acid solution, we find the molarity of HNO₃:

Molarity of HNO₃ = Moles / Volume (L) = 0.00072 / 10/1000 = 0.072 M

Therefore, the concentration of the nitric acid solution is 0.072 M (option a).

Learn more about Neutralization Reactions