Question

If 26.25 ml of 0.1850 m naoh solution reacts with 25.00 ml of h2so4, what is the molarity of the acid solution?

Answer 1

Explanation:

The given data is as follows.

       $V_{NaOH}$ = 26.25 ml,        $M_{NaOH}$ = 0.1850 m

       $V_{H_{2}SO_{4}}$ = 25.00 ml,   $M_{NaOH}$ = ?

It is known that normality is n times molarity where "n" signifies the number of hydrogen or hydroxide ions.

Therefore, normality of NaOH is calculated as follows.

           $N_{NaOH} = n \times M_{NaOH}$

                                     = $1 \times 0.1850$

                                     = 0.1850 N

Normality of $H_{2}SO_{4}$ is calculated as follows.

          $N_{NaOH}V_{NaOH} = N_{H_{2}SO_{4}}V_{H_{2}SO_{4}}$

              $0.1850 N \times 26.25 ml = N_{H_{2}SO_{4}} \times 25.00 ml$

                      $N_{H_{2}SO_{4}}$ = 0.194 N      

Hence, molarity of $H_{2}SO_{4}$ will be as follows.

                     $N_{H_{2}SO_{4}} = n \times M_{H_{2}SO_{4}}$

                    $M_{H_{2}SO_{4}} = \frac{N_{H_{2}SO_{4}}}{n}$

                                         = $\frac{0.194 N}{2}$

                                         = 0.097 M

Thus, we can conclude that molarity of the acid solution is 0.097 M.