What volume (in mL) of 0.2000 M HCl is required to neutralize 50.00 mL of 0.8000 M NaOH?

Answer 1

Use the dilution formula: M₁V₁=M₂V₂. So M₁=0.800M, V₁=0.050L, M₂=0.200M HCl, V₂=? V₂=M₁V₁/M₂=(0.8)(0.050)/0.2=0.2L=200 mL.