If the titration of 25.0 mL sample of hydrochloric acid (HCl) requires 22.15 mL of 0.155M sodium hydroxide (NaOH), what is the molarity of the acid?

Given:

Volume of HCl (VHCl) = 25,0 ml = 0.025 L

Volume of NaOH (VNaOH) = 22.15 ml = 0.02215 L

Molarity of NaOH (MNaOH) = 0.155 M

To determine:

Molarity of HCl

Explanation:

Titration reaction-

HCl + NaOH → NaCl + H2O

Based on the reaction stoichiometry:

1 mole NaOH requires 1 mole HCl

# moles of NaOH = VNaOH * MNaOH = 0.02215*0.155 = 0.003433 moles

# moles of HCl = # moles of NaOH = 0.003433 moles

Molarity of NaOH = 0.003433/0.025 = 0.137 M