Question

\(0.7 \mathrm{~g}\) of a sample of \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{xH}_{2} \mathrm{O}\) were dissolved in water and the volume was made to \(100 \mathrm{ml} .20 \mathrm{ml}\) of this solution required \(19.8 \mathrm{ml}\) of \(\mathrm{N} / 10 \mathrm{HCl}\) for complete neutralization. The value of ' \(x\) ' is: a. 3 b. 2 c. 4 d. 6

Short answer

x = 2

Step-by-step solution

Calculate the amount of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \) that reacts

First, calculate the amount of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \) involved in the neutralization. Since 100 ml solution contains 0.7 g of the compound, 20 ml contains \( \frac{0.7}{100} \times 20 = 0.14 \) g.

Calculate the amount of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) equivalent used

The equivalent reaction with HCl is \( \mathrm{Na}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \). Each mole of \( \mathrm{Na}_2\mathrm{CO}_3 \) requires 2 equivalents of HCl. With 19.8 ml of \( \frac{N}{10} \mathrm{HCl} \) (or equivalent 0.198 eq), the moles of \( \mathrm{Na}_2\mathrm{CO}_3 \) reacting is \( \frac{0.198}{2} = 0.099 \) eq of \( \mathrm{Na}_2\mathrm{CO}_3 \).

Calculate the molecular weight of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \)

The molecular weight \( M \) of \( \mathrm{Na}_2\mathrm{CO}_3 \) is 106 g/mol, and the molar mass of water is 18 g/mol. Calculate the equivalent factor using: \[ \frac{0.14}{M} = 0.099 \]. Rearrange to solve for \( M \), \( M = \frac{0.14}{0.099} = 1.414 \).

Solve for the number of water molecules, \( x \)

The molecular weight of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \) is \( 1.414 \times 106 + 18x \). Set this equal to \( M = 142 \, \text{g/mol} \) and solve for \( x \): \( 142 = 106 + 18x \). Thus, \( 18x = 36 \). Solving gives \( x = 2 \).

Determine the correct answer

Compare the calculated \( x \) to the options provided. The correct value of \( x \) should allow the compound to have a total weight of 142 g/mol, where \( x = 2 \) satisfies this condition.

Key concepts

Stoichiometry

Stoichiometry is crucial for calculating the relationships between reactants and products in chemical reactions. It involves using balanced chemical equations to understand how much of each substance is involved.

In this exercise, the reaction between sodium carbonate and hydrochloric acid is at the core. The balanced equation is:

• \( \mathrm{Na}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \)

From the equation, you can tell that one mole of \( \mathrm{Na}_2\mathrm{CO}_3 \) reacts with two moles of \( \mathrm{HCl} \).

This relationship is essential to calculate the moles of sodium carbonate that are involved when you know the amount of hydrochloric acid that was used. By this stoichiometric relationship, we understand that one equivalent of sodium carbonate will consume two equivalents of hydrochloric acid, helping to solve for the amount of \( \mathrm{Na}_2\mathrm{CO}_3 \) present.

Mole Concept

The mole concept is a fundamental chemistry concept that relates the mass of a substance to the number of particles it contains. The mole allows chemists to convert between the mass of a substance and the number of particles or units it represents, making it invaluable for chemical calculations.

In the exercise, we focus on finding the amount of substance in moles, specifically using the compound \( \mathrm{Na}_2\mathrm{CO}_3 \cdot \mathrm{xH}_2\mathrm{O} \). The solution already calculated the equivalents of \( \mathrm{HCl} \) used, resulting in equivalents of sodium carbonate, which is converted to moles.

• Equivalent reaction for \( 0.198 \) eq of \( \mathrm{HCl} \): implies \( 0.099 \) eq of \( \mathrm{Na}_2\mathrm{CO}_3 \)

This conversion allows us to eventually find the mole ratio needed to solve for the unknown \( x \) in \( \mathrm{Na}_2\mathrm{CO}_3 \cdot \mathrm{xH}_2\mathrm{O} \), relating mass to molecular structure.

Neutralization Reaction

Neutralization reactions involve an acid reacting with a base to produce a salt and water. Here, the neutralization occurs between sodium carbonate (a base) and hydrochloric acid (an acid).

These reactions are used in the exercise to determine the amount of water in the hydrated sodium carbonate. By measuring how much hydrochloric acid is required to fully neutralize a specific portion of the sodium carbonate solution, one can figure out the composition of the hydrate.

The calculation of moles and equivalents extinguished during the reaction provides insights into the chemical formula of the compound. The balanced equation shows how the reactants convert into products alongside water, thus completing the neutralization. This examination of the reaction further supports our stoichiometry and mole calculations, connecting to the coefficient of water, \( x \), in the hydrate.