Question

An analytical chemist weighs out 0.092 g of an unknown triprotic acid into 250 ml volumetric flask and dilutes to the mark with distilled water. he then titrates this solution with 0.1900 m naoh solution. when the titration reaches the equivalence point, the chemist finds he added 14.8 ml of naoh solution. calculate the molar mass of the unknown acid. round your answer to 2 significant digits

Answer 1

The molar mass of the unknown triprotic acid is 98.14 g/mol.

Step-by-step explanation:

To calculate the molar mass of the unknown triprotic acid, we can use the information from the titration.

First, let's calculate the number of moles of NaOH added. We know the volume of NaOH solution added is 14.8 mL = 0.0148 L and the concentration is 0.1900 M. So, the number of moles of NaOH is:

moles NaOH = 0.0148 L x 0.1900 M

= 0.002812 moles NaOH

Next, we need to use the balanced chemical equation to determine the mole ratio between NaOH and the triprotic acid.

In the balanced equation:

H3A(aq) + 3NaOH(aq) → Na3A(aq) + 3H2O(l)

The mole ratio between NaOH and H3A is 3:1. So, the number of moles of H3A is:

moles H3A = 0.002812 moles NaOH x (1 mole H3A / 3 moles NaOH) = 0.0009373 moles H3A

Finally, we can calculate the molar mass of the unknown acid using its mass and the number of moles:

Molar mass = mass / moles

= 0.092 g / 0.0009373 moles

= 98.14 g/mol