If it takes 48 mL of 0.1 M NaOH to neutralize 125 mL of an HCI solution, what is the concentration of the HCI?

Answer 1

HCl + NaOH ==> NaCl + H2O

Concentration of HCl = moles HCl / Liter of solution.

We know we have 125 mls = 0.125 L, so all we need to find is moles of HCl.

moles NaOH = 48 ml x 1 L / 1000 ml x 0.1 mol/L = 0.0048 mols NaOH

moles HCl = 0.0048 mol NaOH x 1 mol HCl / mol NaOH = 0.0048 mols HCl

Molarity of HCl = 0.0048 moles / 0.125 L = 0.0384 M = 0.04 M (1 sig. fig.)