In a titration of 35 mL of HCl is titrated with 32.2 mL of a 0.185M NaOH solution, what is the initial concentration of HCl?

Question

In a titration of 35 mL of HCl is titrated with 32.2 mL of a 0.185M NaOH solution, what is the initial concentration of HCl?

Answer 1

HCl + NaOH ==-> NaCl + H₂O  Balanced equation

From the balanced equation, it can be seen that 1 mole HCl required 1 mole NaOH.

 

moles NaOH used = 0.0322 L x 0.185 mol/L = 0.005957 moles NaOH

moles HCl present = 0.005957 moles HCl (1:1 in balanced equation)

 

Concentration of HCl = 0.005957 moles/0.035 L = 0.1702 moles/L = 0.1702 M = 0.170 M (to 3 significant figures)