Question

In an acid-base titration, it takes 28 mls of .65M NaOH to neutralize 35 mls of HCl. What is the molarity of the HCl?

Answer 1

`"0.52 M"`

Explanation:

For given acid base titration

`"M"_1"V"_1 = "M"_2"V"_2`

`"M"_2 = ("M"_1"V"_1)/"V"_2 = ("0.65 M" × 28 cancel"mL")/(35 cancel"mL") = "0.52 M"`

Click here for more details of the fromula.