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 Kb of methylamine is 3.6 x 10-4. Enter your answer with two decimal places and no units. (how is the answer 16.62?)
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
Determine the volume in mL of 0.57 M HNO3(aq) needed to reach the half-equivalence (stoichiometric) point...
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