A 0.600 g sample of diprotic acid H2X is dissolved in water, and titrated to the equivalence point with 40.0 mL of 0.200 M NaOH. Assume both protons of the acid were titrated. What is the molar mass of H2X

150 g/mol

Explanation:

Let's consider the complete neutralization of a diprotic acid H₂X with NaOH.

H₂X + 2 NaOH → Na₂X + 2 H₂O

40.0 mL of 0.200 M NaOH. were required to reach the endpoint. The reacting moles of NaOH are:

0.0400 L × 0.200 mol/L = 8.00 × 10⁻³ mol

The molar ratio of H₂X to NaOH is 1:2. The reacting moles of H₂X are 1/2 × 8.00 × 10⁻³ mol = 4.00 × 10⁻³ mol.

4.00 × 10⁻³ moles of H₂X have a mass of 0.600 g. The molar mass of H₂X is:

0.600 g/4.00 × 10⁻³ mol = 150 g/mol

a:

by the combination of an unlimited number of elements