What volume (in mL) of 0.2550 M HCI is required to neutralize 50.00 mL of 0.8000 M NaOH?

Answer 1

The reaction taking place is

HCl + NaOH ==> NaCl + H2O (a typical acid-base neutralization reaction)

V1M1 = V2M2

V1 = volume HCl = ?

M1 = molarity HCl = 0.2550 M

V2 = volume NaOH = 50.00 ml

M2 = molarity NaOH = 0.8000 M

(V1)(0.2550) = (50.00)(0.8000)

V1 = 156.9 mls HCl