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

Step-by-step 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 * M_(NaOH)`

=
`1 * 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 * 26.25 ml = N_{H_(2)SO_(4)} * 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 * M_{H_(2)SO_(4)}`

`M_{H_(2)SO_(4)} = \frac{N_{H_(2)SO_(4)}}{n}`

=
`(0.194 N)/(2)`

= 0.097 M

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