Answer to Question #155419 in General Chemistry for C

Question

What mass of NaOH(s) must be added to 300 mL of HCl 0.25 M in order to completely neutralize this acid? 

2 HCl + NaOH →  NaCl +  H₂O

Answer 1

Solution:

The balanced chemical equation:

HCl + NaOH → NaCl + H₂O

According to the equation above: n(HCl) = n(NaOH)

Molarity of HCl = Moles of HCl / Solution volume

Hence,

Moles of HCl = n(HCl) = Molarity of HCl × Solution volume

n(HCl) = 0.25 M × 0.3 L = 0.075 mol

n(NaOH) = n(HCl) = 0.075 mol

Moles of NaOH = Mass of NaOH / Molar mass of NaOH

The molar mass of NaOH is 40.0 g mol^-1.

Hence,

Mass of NaOH = Moles of NaOH × Molar mass of NaOH

Mass of NaOH = 0.075 mol × 40.0 g mol^-1 = 3.00 g

Mass of NaOH = 3.00 g

Answer: 3.00 grams of of NaOH must be added.