A 1.50\text{ g} sample of an impure hydrated dibasic acid, \text{H}_2\text{X}\cdot 2\text{… — JAMB UTME

A $1.50\text{ g}$ sample of an impure hydrated dibasic acid, $\text{H}_2\text{X}\cdot 2\text{H}_2\text{O}$ (molar mass of anhydrous $\text{H}_2\text{X} = 90.0\text{ g mol}^{-1}$ ), was dissolved in distilled water and made up to $250.0\text{ cm}^3$ of solution in a volumetric flask. A $25.0\text{ cm}^3$ portion of this acid solution required $20.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the percentage purity of the hydrated acid sample?

A
$60.0\%$
B
$42.0\%$
$84.0\%$ Answer
D
$8.4\%$

Answer

The percentage purity of the hydrated acid sample is

84.0\%

The correct answer is

84.0\%

. Each mole of dibasic acid reacts with 2 moles of

\text{NaOH}

. The

20.0\text{ cm}^3

0.100\text{ mol dm}^{-3}

\text{NaOH}

contains

0.0020\text{ mol}

of base, neutralizing

0.0010\text{ mol}

of acid in

25.0\text{ cm}^3

. Scaling to the total

250.0\text{ cm}^3

gives

0.010\text{ mol}

of pure acid in the flask. Multiplying by the hydrated molar mass of

126.0\text{ g mol}^{-1}

(

90.0 + 36.0

) yields

1.26\text{ g}

of pure acid, which corresponds to

(1.26 / 1.50) \times 100\% = 84.0\%

purity.

Step-by-Step Solution

Calculate molar mass of the hydrated acid and write balanced neutralization equation

Molar mass of

\text{H}_2\text{X}\cdot 2\text{H}_2\text{O} = 90.0 + 2(18.0) = 126.0\text{ g mol}^{-1}

. Reaction equation:

\text{H}_2\text{X} + 2\text{NaOH} \rightarrow \text{Na}_2\text{X} + 2\text{H}_2\text{O}

, so mole ratio

n_a : n_b = 1 : 2

The acid is dibasic, meaning each mole of acid reacts with two moles of sodium hydroxide, and the molar mass must include the water of crystallization.

Calculate the moles of base reacted and corresponding moles of acid in the

25.0\text{ cm}^3

aliquot

\text{Moles of NaOH} = \frac{20.0\text{ cm}^3}{1000} \times 0.100\text{ mol dm}^{-3} = 0.0020\text{ mol}

. Moles of acid in

25.0\text{ cm}^3 = \frac{0.0020}{2} = 0.0010\text{ mol}

Applying the stoichiometric ratio

n_a/n_b = 1/2

converts the moles of base used to moles of dibasic acid neutralized.

Scale the moles of pure acid to the total

250.0\text{ cm}^3

solution volume and determine pure mass

\text{Moles of pure acid in } 250.0\text{ cm}^3 = 0.0010\text{ mol} \times \left(\frac{250.0}{25.0}\right) = 0.010\text{ mol}

\text{Mass of pure hydrated acid} = 0.010\text{ mol} \times 126.0\text{ g mol}^{-1} = 1.26\text{ g}

The entire sample was dissolved to make 250 cm³, so multiplying by the dilution factor (10) gives the total moles of pure acid in the sample.

Calculate percentage purity of the sample

\text{Percentage purity} = \left(\frac{1.26\text{ g}}{1.50\text{ g}}\right) \times 100\% = 84.0\%

Percentage purity is the ratio of mass of pure substance to total mass of impure sample, expressed as a percentage.

Key Concept

Volumetric Analysis and Percentage Purity Calculation of a Hydrated Dibasic Acid

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