If 22.4 mL of 0.0100 M HCI are required to completely react with 15.0 mL NaOH, what was the initial molarity of the NaOH solution?

NaOH → 0.015 M

Explanation:

We propose the neutralization formula to solve the problem:

N acid . Volume acid = N base . Volume base

N means normality. A sort of concentration that is defined as:

M / val where the val means the number of H⁺ and OH⁻ from the acid or the base.

In this case, we have NaOH and HCl, so the M = N

Let's replace data:

0.01 M . 22.4 mL = 15 mL . M NaOH

0.01 M . 22.4 mL / 15 mL = M NaOH → 0.015 M