Asolution of sodium hydroxide, 0.0500 m, is used to titrate a 15.00 ml sample of hydrochloric acid to the endpoint. the initial buret reading was 3.87 m; the final reading was 25.11 ml. what is the molarity of the acid solution.?

The answer to your question is Molarity = 0.0708

Explanation:

Data

NaOH  0.05 M     Volume 1 = 3.87 ml    Volume 2 = 25.11 ml

HCl   15 ml

Process

1.- Find the volume used of NaOH

                              25.11 - 3.87 = 21.24 ml = 0.02124 l

2.- Write the balanced equation of the reaction

                   NaOH  +  HCl   ⇒   NaCl + H₂O

3.- Calculate the moles of NaOH in the solution

Molarity =

moles = Molarity x volume

moles = 0.05 x 0.02124

moles = 0.001062

4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.

                   1 mol of NaOH -------------  1 mol of HCl

 0.001062 moles of NaOH ------------    x

                  x = (0.001062 x 1) / 1

                  x = 0.001062 moles of HCl

5.- Find the molarity of HCl

Molarity =

Molarity = 0.0708

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