Question

A 5.00 g quantity of a diprotic acid was dissolved in water and made up exactly to 225 mL. Calculate the molar mass of the acid is 25.0mL of this solution required 11.6 mL of 1.00 M KOH for neutralization. Assume that both protons of the acid were titrated.

Answer 1

Volume of KOH 11.6 mL Molarity of KOH= 1.0014 Moles ofKOH= 11.6mL×1.00M× mol/L1L × - 11.6x103 mol M 1000 mL here, the acid is diprotic acid, 2 moles ofH ions neutralizes 1 mole of ion. the moles of aci d in 25.0 mL solution is equal to the half of moles of KOH moles of acid in 25 mL = 11 6x103 mol_ 8x103 mol - 11.6x103 225 mL 25 mL In 225 mL of soluti on, moles of the acid- ×5.8×10-3 mol 52.2x10-3 mol mass of the acid= 5.00 g mass of the acid m olar mass of acid = moles of acid in 225 mL 5.00 g 52.2×10-s mol = 95.8g/mol Therefore, the required mol ar mass is |95.8 g/mol