Chemistry (stoichiometry)

Question

A 2.29 g sample of an unknown acid (monoprotic) is dissolved in 1.0 liter of water. A titration required

25.0 mL of 0.500 M NaOH to completely react with all the acid present. What is the molar mass of the

acid?

Answer 1

To find the molar mass, we first need to find the moles of acid present.

Let HA represent the monoprotic acid.

HA + NaOH ==> NaA + H2O

moles NaOH used = 25.0 ml x 1 L/1000 ml x 0.500 mols/L = 0.0125 mols NaOH

moles HA = 0.0125 mols NaOH x 1 mol HA / mol NaOH = 0.0125 mols HA present

molar mass HA = 2.29 g / 0.0125 mols = 183 g/mol (3 sig. figs.)