Question

A solution of an unknown monoprotic acid was prepared by dissolving 72.9 mg of the acid into 50. mL of water. 10.0 mL of 0.10 M NaOH is needed to titrate a 25 mL aliquot of the unknown monoprotic acid. From this info, calculate the molar mass of the unknown acid (g/mol).

Answer 1

Answer:

36.45 g/mol

Explanation:

The titration of a monoprotic acid with a base as NaOH is:

HX + NaOH → NaX + H₂O.

That means 1 mol of acid reacts per 1 mol of base.

10.0mL of 0.10M NaOH are:

0.0100L ₓ (0.10mol / L) = 0.0010 moles NaOH ≡ 0.0010 moles HX

If 72.9mg of the monoprotic acid are dissolved in 50.0 mL of water, 25 mL of the solution will contains:

72.9mg × (25mL / 50mL) = 36.45mg = 0.03645g of HX.

Molar mass is ratio between mass and moles of substance, that is:

0.03645g of HX / 0.0010 moles HX = 36.45 g/mol