hannahwiggins23 hannahwiggins23
  • 17-04-2021
  • Chemistry
contestada

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

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maacastrobr
maacastrobr maacastrobr
  • 22-04-2021

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 * [tex]\frac{1mmolHCl}{1mmolNaOH}[/tex] = 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

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