How many mL of 1.00M HCl are needed to neutralize 400mL of 0.200M NaOH?

80 mL

Explanation:

M = mol solute / L solution

mol solute = M * L solution

mol solute = (0.200 M * 0.4 L)

mol solute = 0.08 mol NaOH = mol OH

We need the same amount of H mol to neutralize

L solution = mol solute/M

L solution = 0.08 mol HCl / 1.00M

L solution = 0.08 L = 80 mL