A standard solution is prepared by dissolving 1.575\text{ g} of hydrated ethanedioic acid … — JAMB UTME

A standard solution is prepared by dissolving $1.575\text{ g}$ of hydrated ethanedioic acid ( $\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}$ ) in distilled water to make $250.0\text{ cm}^3$ of solution. A $25.0\text{ cm}^3$ sample of this acid solution requires $25.0\text{ cm}^3$ of a $0.100\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the integer value of $x$ in the formula of the hydrated acid? [Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

Answer: 2

Answer

The integer value of

x

is 2.

The value of

x

is calculated as

2

. Based on the reaction stoichiometry,

25.0\text{ cm}^3

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

\text{NaOH}

contains

0.0025\text{ mol}

of base, which neutralizes

0.00125\text{ mol}

of the diprotic acid in the

25.0\text{ cm}^3

sample. The entire

250.0\text{ cm}^3

solution therefore contains

0.0125\text{ mol}

of acid. Dividing the mass (

1.575\text{ g}

) by

0.0125\text{ mol}

gives a molar mass of

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

for

\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}

. Subtracting the molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4

(

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

) gives

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

for water, which corresponds to

x = 36 / 18 = 2

Step-by-Step Solution

Determine the mole ratio from the balanced chemical neutralization equation.

\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + (x+2)\text{H}_2\text{O}

. The stoichiometric ratio of acid to base is

1 : 2

Ethanedioic acid is a diprotic acid requiring two moles of hydroxide ions for complete neutralization per mole of acid.

Calculate the amount in moles of sodium hydroxide solution used in the titration.

\text{Moles of NaOH} = 0.100\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.0025\text{ mol}

Number of moles is equal to molar concentration multiplied by volume in cubic decimeters.

Calculate the total moles of hydrated acid present in the

250.0\text{ cm}^3

volumetric flask.

\text{Moles in } 25.0\text{ cm}^3 \text{ aliquot} = \frac{0.0025}{2} = 0.00125\text{ mol}

. Total moles in

250.0\text{ cm}^3 = 0.00125 \times \frac{250.0}{25.0} = 0.0125\text{ mol}

Using the stoichiometric ratio (

n_a/n_b = 1/2

) and scaling up from the

25.0\text{ cm}^3

aliquot to the full

250.0\text{ cm}^3

solution volume.

Compute the molar mass of the hydrated acid and solve for

x

\text{Molar mass} = \frac{1.575\text{ g}}{0.0125\text{ mol}} = 126\text{ g mol}^{-1}

. Molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4 = 2(1) + 2(12) + 4(16) = 90\text{ g mol}^{-1}

. Mass of

x\text{H}_2\text{O} = 126 - 90 = 36\text{ g mol}^{-1}

. Thus,

x = \frac{36}{18} = 2

Subtracting the molar mass of the anhydrous acid from the total molar mass gives the mass of the water of crystallization, which is divided by the molar mass of water (

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

) to find

x

Key Concept

Volumetric Analysis and Water of Crystallization Determination

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