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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.