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

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.