Question

A 0.272 g sample of a diprotic acid H2X is titrated to the equivalence point. What is the molar mass of H2X if 40.0 mL of 0.400 M NaOH is required. Assume both protons of the acid were titrated. Express your answer in g/mol.

Answer 1

Answer:

The molar mass is 8.50 g/mol.

Explanation:

Let's consider the neutralization between a diprotic acid and NaOH.

H₂X + 2 NaOH ⇄ Na₂X +  2 H₂O

The molar ratio of H₂X to NaOH is 1:2. Then, the moles of H₂X are:

$nH_{2}X=2n_{NaOH}=2 \times 40.0 \times 10^{-3} L \times \frac{0.400mol}{L} =0.0320mol$

The molar mass of H₂X is:

$\frac{0.272g}{0.0320mol} =8.50g/mol$

Answer 2

The molar mass of a 0.272 g sample of a diprotic acid, H₂X required to react with 40 mL of a 0.4 M NaOH solution to the equivalence point is 34 g/mol

• We'll begin by calculating the number of mole of NaOH in 40 mL of 0.4 M NaOH. This can be obtained as follow:

Volume = 40 mL = 40 / 1000 = 0.04 L

Molarity = 0.4 M

Mole of NaOH =?

Mole = Molarity x Volume

Mole of NaOH = 0.4 × 0.04

Mole of NaOH = 0.016 mole

• Next, we shall determine the number of mole of H₂X required to react with 0.016 mole of NaOH to the equivalence point. This can be obtained as follow:

H₂X + 2NaOH —> Na₂X +  2H₂O

From the balanced equation above,

2 moles of NaOH reacted with 1 mole of H₂X

Therefore,

0.016 mole of NaOH will react with = $\frac{0.016}{2}$ = 0.008 mole of H₂X

• Finally, we shall determine the molar mass of H₂X. This can be obtained as follow:

Mole of H₂X = 0.008 mole

Mass of H₂X = 0.272 g

Molar mass of H₂X

Molar mass = mass / mole

Molar mass of H₂X = 0.272 / 0.008

Molar mass of H₂X = 34 g/mol

Therefore, the molar mass of H₂X is 34 g/mol

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