Question

A direct titration of .68311 grams sample of diprotic acid (H2X) requires 37.45 ml of 0.1372M NaOH solution. What is the experimental molecular weight of this organic acid: H2X+ 2NaOH -------->Na2X + 2H2O

Answer 1

Given Molarity of NaOH is , M = 0.1372 M

          Volume of NaOH , V = 37.45 mL

                                           = 37.45 x 10^-3 L

Number of moles of NaOH , n = Molarity x volume in L

                                                = 0.1372 M x 37.45 x 10^-3 L

                                                = 5.14 x10 ^-3 mol

Molar mass of NaOH is = At.mass of Na + At.mass of O + At.mass of H

                                      = 23 + 16 + 1

                                      = 40 g/mol

So mass of NaOH , m = Number of moles x molar mass of NaOH

                                    = 5.14 x10 ^-3 mol x 40 (g/mol)

                                    = 0.205 g

H₂X+ 2NaOH    Na₂X + 2H₂O

According to the balanced equation,

1 mole of H₂X reacts with 2 moles of NaOH

                                   OR

M g of H₂X reacts with 2x40 = 80 g of NaOH

0.68311 g of H₂X reacts with 0.205 g of NaOH

M = ( 0.68311 x 80 ) / 0.205

   = 265.9 g

Therefore experimental molecular weight of this organic acid is 265.9 g/mol

So the