A 2.016\text{ g} sample of a hydrated dicarboxylic acid, (\text{COOH})_2 \cdot n\text{H}_2… — JAMB UTME

A $2.016\text{ g}$ sample of a hydrated dicarboxylic acid, $(\text{COOH})_2 \cdot n\text{H}_2\text{O}$ , was dissolved in distilled water and made up to $250.0\text{ cm}^3$ in a volumetric flask. A $25.0\text{ cm}^3$ portion of this solution required exactly $20.0\text{ cm}^3$ of $0.160\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the value of the integer $n$ in the formula of the hydrated acid? $[\text{H} = 1, \text{C} = 12, \text{O} = 16]$

Cevap: 2

Cevap

The value of the integer n is 2.

Titration of 20.0 cm³ of 0.160 mol dm⁻³ NaOH consumes 0.0032 mol of NaOH. Because the dicarboxylic acid is diprotic ((COOH)₂), it reacts in a 1:2 ratio with NaOH, giving 0.0016 mol of acid in the 25.0 cm³ aliquot. Scaling to the full 250.0 cm³ flask yields 0.0160 mol of hydrated acid. The molar mass of the hydrated acid is 2.016 g / 0.0160 mol = 126 g/mol. Since the anhydrous formula mass of (COOH)₂ is 90 g/mol, the water of crystallization contributes 126 - 90 = 36 g/mol. Dividing 36 by 18 (the molar mass of H₂O) gives n = 2.

Adım Adım Çözüm

Calculate the amount in moles of NaOH used in the titration.

Moles of NaOH = 0.0032 mol

Moles = concentration × volume = 0.160 mol dm⁻³ × (20.0 / 1000) dm³ = 0.0032 mol.

Determine the moles of acid present in the 25.0 cm³ titration sample using the stoichiometric mole ratio.

Moles of acid in 25.0 cm³ = 0.0016 mol

Ethanoic/oxalic acid is diprotic ((COOH)₂ + 2NaOH → (COONa)₂ + 2H₂O), requiring 2 moles of NaOH per mole of acid. Moles of acid = 0.0032 / 2 = 0.0016 mol.

Scale up to find the total moles of acid in the original 250.0 cm³ solution.

Total moles of acid = 0.0160 mol

Total moles = 0.0016 mol × (250.0 cm³ / 25.0 cm³) = 0.0160 mol.

Calculate the molar mass of the hydrated dicarboxylic acid.

Molar mass = 126 g mol⁻¹

Molar mass M = sample mass / total moles = 2.016 g / 0.0160 mol = 126 g mol⁻¹.

Calculate the value of integer n by comparing the molar mass to the anhydrous acid mass.

n = 2

Formula mass of anhydrous (COOH)₂ = 2(12) + 2(1) + 4(16) = 90 g mol⁻¹. The water component mass is 18n = 126 - 90 = 36 g mol⁻¹, giving n = 36 / 18 = 2.

Anahtar Kavram

Empirical and Molecular Formula Calculations with Water of Crystallization and Volumetric Stoichiometry

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