Determine the volume in mL of 0.57 M HNO3(aq) needed to reach the half-equivalence

Question

Determine the volume in mL of 0.57 M HNO₃(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 37.9 mL of 0.5 M CH₃NH₂(aq)(aq). The K_b of methylamine is 3.6 x 10^-4. Enter your answer with two decimal places and no units.

Answer is 16.62

Answer 1

Molarity x volume = 0.5 M x 37.9 mL = 18.95 mmol

At half equivalence point, half of base is neutralized.

Amount of methylamine neutralized: 18.95/2 = 9.475 mmol

Volume = moles/ molarity = 9.475mmol/0.57M = 16.62mL