A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat

Question

siniylev [52]

3 years ago

14

A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat

e the molar mass of the acid. question 16 options:

Chemistry

1 answer:

Jet001 [13] · Answer 1 · 2021-12-05T17:58:25+03:00

Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.