Question

5. How many milliliters of 0.258 M NaOH are required to completely neutralize 2.00 g of acetic acid HC2H30,? CHCOOH + poolt + CH3 COOHQ + H2O for complete moles of aceho add = Gwen mas Molar mas hal=h = 2-1 moles 30 . molls of acetic aud 40

Answer 1

Mass of acetic acid = 2.00 g.

Molar mass of acetic acid = 60 g/mol

Moles of acetic acid = mass / molar mass = 2.00 / 60.0 = 0.0333 mol

From the balanced equation,

Moles of NaOH = moles of acetic acid = 0.0333 mol

Molarity of NaOH = 0.258 M

Volume of NaOH = Molles / molairty

V = 0.0333 / 0.258

V = 0.129 L

V = 129 mL

Therefore,

Volume of NaOH solution required = 129 mL