An unknown weak base (0.5673 g) was dissolved in 25.0 mL of water and titrated with 0.1876 M HCl. If it took 16.51 mL to reach the equivalence point, what is the molar mass of the weak base?

Answer 1

Let the weak base be B^-

B^- + HCl ==> HB + Cl-

moles HCl needed = 25.0 ml x 1 L / 1000 ml x 0.1876 mol/L = 0.00469 mols HCl

moles B^- present = 0.00469 mol HCl x 1 mol B- / mol HCl = 0.00469 mols B^-

molar mass B^- = g / mols = 0.5673 g / 0.00469 mols

Molar mass = 121 g / mole