Answer:
80.0 mL of a 0.2 M NaOH solution.
Explanation:
(MV)before dilution = (MV)after dilution.
We need to calculate the concentration after the dilution for each solution:
M after dilution = (MV)before dilution/V after dilution = (0.1 M)(100.0 mL)/(200.0 mL) = 0.05 M.
M after dilution = (MV)before dilution/V after dilution = (0.4 M)(20.0 mL)/(200.0 mL) = 0.04 M.
M after dilution = (MV)before dilution/V after dilution = (0.2 M)(80.0 mL)/(200.0 mL) = 0.08 M.
M after dilution = (MV)before dilution/V after dilution = (0.5 M)(10.0 mL)/(200.0 mL) = 0.025 M.
So, 80.0 mL of a 0.2 M NaOH solution will have the largest concentration after dilution to 200.0 mL.