Question

If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution,what is the concentration of the HCl ?

Answer 1

volume of NaOH = 54 mL / 1000 = 0.054 L

Molarity NaOH = 0.1 M

Number of moles NaOH :

n = M * v

n = 0.1 * 0.054

n = 0.0054 moles of NaOH

Finally we calculate the number of moles of HCl in the solution from the stoichiometry of the reaction :

HCl + NaOH = NaCl + H₂O

1 mole HCl -------- 1 mole NaOH
? moles HCl ------- 0.0054 moles NaOH

moles HCl = 0.0054 * 1 /1

= 0.0054 moles of HCl

Volume of HCl = 125 mL / 1000 = 0.125 L

M ( HCl ) = n / V

M = 0.0054 / 0.125

= 0.0432 M

hope this helps!