Question

A 3.0 mL sample of `HNO_3` solution is exactly neutralized by 6.0 mL of .50 M `KOH`. What is the molarity of the `HNO_3` solution?

Answer 1

The key in any neutralization problem is to match the number of moles of acid and base.

Explanation:

In this case, the number of moles of base is
`6.0 mL times (1 L)/(1000 mL) times 0.50 (mol)/L = 3.0 times 10^-3 mol`

This must be equal to the number of moles of `HNO_3`, so the concentration of that solution must be:
`(3.0 times 10^-3 mol)/(3.0 mL) times (1000 mL)/(1L) = 1.0 (mol)/L`