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

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml