Answer:
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
The molar mass of 0.6 g sample of the diprotic acid, H₂X which reacted with 40 mL of 0.2 M NaOH to the equivalence point is 150 g/mol
We'll begin by calculating the number of mole of NaOH in the solution.
Volume = 40 mL = 40 / 1000 = 0.04 L
Molarity of NaOH = 0.2 M
Mole = Molarity x Volume
Mole of NaOH = 0.2 × 0.04
H₂X + 2NaOH → Na₂X + 2H₂O
From the balanced equation above,
2 moles of NaOH reacted with 1 mole of H₂X.
Therefore,
0.008 mole of NaOH will react with = 0.008 / 2 = 0.004 mole of H₂X.
Mole of H₂X = 0.004 mole
Mass of H₂X = 0.6 g
Molar mass = mass / mole
Molar mass of H₂X = 0.6 / 0.004
Therefore, the molar mass of the diprotic acid, H₂X is 150 g/mol
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