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.Calculate the molar mass of the acid.

Question

Quizplus · Accepted Answer

The molar mass of the acid is calculated by first finding the moles of NaOH used, which is $0.106 \, \text{M} \times 0.0352 \, \text{L} = 0.0037312 \, \text{mol}$. Since it's a triprotic acid, the moles of acid will be a third of the moles of NaOH, so $0.0037312 \, \text{mol} / 3 = 0.00124373 \, \text{mol}$. The molar mass is then the mass of the sample divided by the moles of acid, $0.307 \, \text{g} / 0.00124373 \, \text{mol} = 247 \, \text{g/mol}$.

A 0307-G Sample of an Unknown Triprotic Acid Is Titrated to to the Third