Question

In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are in the flask?

Answer 1

Answer: There are 0.006 moles of acid in the flask.

Step-by-step explanation:

Given:
`V_(1)` = 21.35 mL,
`M_(1)` = 0.150 M

`V_(2)` = 25.0 mL,
`M_(2)` = ?

Formula used to calculate molarity of
`H_(2)SO_(4)` is as follows.

`M_(1)V_(1) = M_(2)V_(2)`

Substitute the values into above formula as follows.

`M_(1)V_(1) = M_(2)V_(2)\\0.15 M * 21.35 mL = M_(2) * 25.0 mL\\M_(2) = 0.1281 M`

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution =
`V_(1) + V_(2)`

= 21.35 mL + 25.0 mL

= 46.36 mL (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

`Molarity = (no. of moles)/(Volume (in L))\\0.1281 M = (no. of moles)/(0.04635 L)\\no. of moles = 0.006 mol`

Thus, we can conclude that there are 0.006 moles of acid in the flask.