A student wanted to prepare 500 mL 0.20 mol/L NaOH solution for an acid-base titration lab. If 0.50 mol/L NaOH is the only source, how many mL of 0.50 mol/L NaOH should he use to do the 500-mL 0.20 mol/L NaOH solutuion?

From the principle of dilution,M1V1 = M2V2,where M1 is the concentration in molarity (moles/Liters) of the concentrated solution, V2 is the volume of the concentrated solution, M2 is the concentration in molarity of the dilute solution (after more solvent has been added), and V2 is the volume of the dilute solution.

M1=0.50 mol/L

V1=?

M2=0.20 mol/L

V2=500mL

V1 =M2V2/M1

V1= (0.2 x 500)/0.5 = 200

Therefore 200mL of 0.50 mol/L NaOH should be used.

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