Question

What volume of 0.215 M HCl is required to neutralize 50.0 mL of
0.800 M NaOH?

Answer 1

186 mL HCl

Step-by-step explanation:

M1V1 = M2V2

M1 = 0.215 M HCl

V1 = ?

M2 = 0.800 M NaOH

V2 = 50.0 mL

Solve for V1 --> V1 = M2V2/M1

V1 = (0.800 M)(50.0 mL) / (0.215 M) = 186 mL HCl