What is the molarity (concentration) of 2.53 L of NaOH which exactly neutralizes 5.0L of 1.25 M HCl?

The concentration of NaOH is 2.47 M

Explanation:

The equation of the reaction is

HCl (aq) + NaOH(aq) > NaCl(aq) + H2O(l)

Given;

Concentration of acid CA= 1.25 M

Volume of acid VA= 5.0L

Concentration of base CB=

Volume of base VB= 2.53 L

Number of moles of acid NA= 1

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CB= CAVANB/VBNA

CB= 1.25 × 5.0 ×1 / 2.53×1

CB= 2.47 M

Therefore the concentration of NaOH is 2.47 M

idk i’m struggling too