Question

What volume (in mL) of 0.3900 M HCl is required to neutralize 50.00 mL of 0.8000 M NaOH?

Answer 1

Answer:

102.6 mL

Explanation:

The reaction that takes place is:

• NaOH + HCl → NaCl + H₂O

First we calculate how many NaOH moles are there in 50.00 mL of a 0.8000 M solution:

• 0.8000 M * 50.00 mL = 40 mmol NaOH

Then we convert NaOH moles into HCl moles:

• 40 mmol NaOH * $\frac{1mmolHCl}{1mmolNaOH}$ = 40 mmol HCl

Finally we calculate the volume of a 0.3900 M solution would contain 40 milimoles:

• 40 mmol / 0.3900 M = 102.6 mL