Question

STANDARDIZATION OF SOLUTIONS - Worksheet NaOH Volume (ml) Part I: Standardization of NaOH Trial KHP Mass (9) Potassium biphthalate 0.517 2 0.500 VE 0.51 0.71 27.00 27.20 Calculate the molarity of the NaOH solution:

Answer 1

Formula used- Mole of Substance = Mass of Substance / Molar mass of Substance At end point of titration, Number of moles of KHP = Number of moles of NaOH => 0.00253 mol = Concentration (Molarity) of NaOH x Volume of NaOH in Liter

KHP is used against titration with NaOH (a) For, Trial 1 - Mass of KHP = 0.517 g Molar mass of KHP = 204.22 g/mol So, Mole of KHP = Mass of KHP / Molar mass of KHP = 0.517 g/(204.22 g/mol) = 0.00253 mol Initial volume of NaOH = V = 0.51 mL Final volume of NaOH = V = 27 mL Used volume of NaOH = V = V. - V= 27 mL - 0.51 mL = 26.49 mL = 0.02649 L

At end point of titration, Number of moles of KHP = Number of moles of NaOH => 0.00253 mol = Concentration (Molarity) of NaOH x Volume of NaOH in Liter => 0.00253 mol = Concentration (Molarity) of NaOH x 0.02649 L => Concentration (Molarity) of NaOH = 0.00253 mol / 0.02649 L => Concentration (Molarity) of NaOH = 0.0955 mol/L = 0.0955 M (b) For, Trial 2 - Mass of KHP = 0.500 g Molar mass of KHP = 204.22 g/mol So, Mole of KHP = Mass of KHP / Molar mass of KHP = 0.500 g/(204.22 g/mol) = 0.00244 mol Initial volume of NaOH = V = 0.71 mL Final volume of NaOH = V. = 27.20 mL Used volume of NaOH = V = V - V = 27.20 mL - 0.71 mL = 26.49 mL = 0.02649 L At end point of titration,

Number of moles of KHP = Number of moles of NaOH => 0.00253 mol = Concentration (Molarity) of NaOH x Volume of NaOH in Liter => 0.00253 mol = Concentration (Molarity) of NaOH x 0.02649 L => Concentration (Molarity) of NaOH = 0.00244 mol / 0.02649 L => Concentration (Molarity) of NaOH = 0.0921 mol/L = 0.0921 M