How many milliliters of 0.20 M HCl is required to neutralize 40.0 mL of 0.80 M NaOH?

Answer 1

The reaction would be -

HCl + NaOH =NaCl +H₂O (balanced reaction)

General formula is V_aS_a/V_bS_b=n_a/n_b. Here, V_a=volume of acid, S_a=Concentration of Acid, n_a=number of moles of acid. similarly V_b=volume of base, S_b=Concentration of base

In this problem you need Va

You have been given S_a=0.2 M, n_a=n_b=1, V_b=40 mL, S_b=0.80 M

Using the above formula you would get V_a=160 mL