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. (how is the answer 16.62?)

Answer 1

Half equivalence point menas half of base is neutralize.

Mmol of base =37.9×0.50

= 18.95 mmol

So mmol of HNO3= 18.95/2= 9.475 mmol

Mmol = M × V

9.475 = 0.57 × V

V = 9.475/0.57

= 16.62 mL