Question

How many milliliters of 0.258M NaOH are required to completely neutralize 2.00 g of acetic acid HC2H3O2?

Answer 1

Answer:

0.129 L = 129.0 mL.

Explanation:

• NaOH neutralizes acetic acid (CH₃COOH) according to the balanced reaction:

NaOH + CH₃COOH → CH₃COONa + H₂O.  

• According to the balanced equation: 1.0 mole of NaOH will neutralize 1.0 mole of CH₃COOH.  

no. of moles of CH₃COOH = mass/molar mass = (2.0 g)/(60 g/mol) = 0.033 mole.  

• For NaOH:

no. of moles = (0.258 mol/L)(V)

• At neutralization: no. of moles of NaOH = no. of moles of CH₃COOH  

∴ (0.258 mol/L)(V) = 0.033 mole

∴ The volume of NaOH =  (0.033 mole)/(0.258 mol/L) = 0.129 L = 129.0 mL.