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.

the answers would be

the first one and the second one

catya that's not even a b or c