12.0 of .500 M NaOH neutralized 6.0 ml of HCl solution. What was the molarity of the HCl

The molarity of the HCl is 1 M when 12.0 of .500 M NaOH neutralized 6.0 ml of HCl solution.

Explanation:

Data given:

molarity of the base NaOH, Mbase =0. 5 M

volume of the base NaOH, Vbase = 12 ml

volume of the acid, Vacid = 6 ml

molarity of the acid, Macid = ?

The titration formula for acid and base is given as:

Mbase Vbase = Macid Vacid

Macid =

Macid = 1 M

we can see that 1 M solution of HCl was used to neutralize the basic solution of NaOH. The volume of NaOH is 12 ml and volume of HCl used is 6ml.

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