Question

a solution 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.?

Answer 1

The answer to your question is Molarity = 0.0708

Step-by-step 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)/(volume)`

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 =
`(0.001062)/(0.015)`

Molarity = 0.0708