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.

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.

$M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M$

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.

$Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol$

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