If a 0.152 g sample of a weak acid requires 25.30 ml of 0.1036 M NaOH to reach an equivalence point, what is the molar mass of the acid? (g/mol)

Answer 1

Assuming the weak acid is a monoprotic acid (HA), we can write

HA + NaOH ==> H2O + NaA

moles NaOH = 25.30 ml x 1 L/1000 ml x 0.1036 mol/L = 0.002621 moles NaOH = 0.002621 moles HA

molar mass HA = 0.152 g / 0.002621 moles = 58.0 g/mole