Question

A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1922 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 13.68 mL of the standard solution. What is the concentration of the original sulfuric acid solution?

Answer 1

Explanation:

The given data is as follows.

          Concentration of standard NaOH solution = 0.1922 M

Let the original acid solution concentration be x.

                        $M_{1}V_{1} = M_{2}V_{2}$

                        $x \times 25 = 250 \times M_{2}$

                         $M_{2} = \frac{x \times 25}{250}$

                                     = 0.1 x M

$V_{2}$ = 10.00 mL (given)

The reaction equation is as follows.

             $2NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O$

   Concentration × Volume of $H_{2}SO_{4}$ = Concentration × Volume of NaOH

           $\frac{M_{2}V_{2}}{n_{2}} = \frac{M_{3}V_{3}}{n_{3}}$

     $\frac{0.1 x \times 10 ml}{1} = \frac{0.1922 M \times 13.68 ml}{2}$

                  x = 1.314 M

Therefore, we can conclude that the concentration of the original sulfuric acid solution is 1.314 M.