Question

An analytical chemist weighs out 0.223g of an unknown triprotic acid into a 250mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.1500M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 45.5mL of NaOH solution. Calculate the molar mass of the unknown acid. Be sure your answer has the correct number of significant digits.

Answer 1

Sol:-

At equivalence point

No. of gram equivalent of triprotic acid = No. of gram equivalent of NaOH

i.e

No. of moles of triprotic acid x valency factor of triprotic acid = M₁ x valency factor of NaOH x V₁

given mass / Gram molar mass of triprotic acid x valency factor of triprotic acid = M₁ x valency factor of NaOH x V₁

Gram molar mass of triprotic acid = valency factor of triprotic acid x given mass / M₁ x valency factor of NaOH x V₁

Gram molar mass of triprotic acid = 3 x 0.223 g / 0.1500 M x 0.0455 L

Gram molar mass of triprotic acid = 98.03 g/mol