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

Answer 1

Answer:

CH₃COOH + NaOH 

1 mole 1 mole

Molar mass of acetic acid is 60.052 g/mol if we use C - 12.01 g/mol; H - 1.008g/mol, O - 16g/mol

Mole of acetic acid that reacted is mass / molar mass

Mol acetic acid = 2.00 g÷ 60.1 g = 0.0333

Therefore mole of NaOH is 0.0333

Volume of NaOH solution (ml) = mole ÷ molarity = 0.0333 × 1000 ÷ 0.258  = 129

(1000 is to convert L to ml)