Question

Sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) is available in very pure form and can be used to standardize acid solutions. What is the molarity of an \(\mathrm{HCl}\) solution if \(28.3 \mathrm{~mL}\) of the solution is required to react with \(0.256 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Short answer

The molarity of the HCl solution is approximately 0.1707 M.

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction between sodium carbonate and hydrochloric acid is: \[ \mathrm{Na}_2\mathrm{CO}_3 (aq) + 2 \mathrm{HCl} (aq) \rightarrow 2 \mathrm{NaCl} (aq) + \mathrm{H}_2\mathrm{O} (l) + \mathrm{CO}_2 (g) \]This equation shows that 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid.

Calculate moles of Na2CO3

First, find the molar mass of sodium carbonate \(\mathrm{Na}_2\mathrm{CO}_3\). Molar mass is calculated as follows:- Sodium (Na): 22.99 g/mol, Carbon (C): 12.01 g/mol, Oxygen (O): 16.00 g/mol - \[ \mathrm{Molar \ mass \ of} \ \mathrm{Na}_2\mathrm{CO}_3 = (2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99 \ g/mol \]Calculate moles of \(\mathrm{Na}_2\mathrm{CO}_3\):\[ \mathrm{moles} = \frac{0.256 \ g}{105.99 \ g/mol} \approx 0.002414 \ moles \]

Determine moles of HCl needed

According to the balanced equation, 2 moles of \(\mathrm{HCl}\) react with 1 mole of \(\mathrm{Na}_2\mathrm{CO}_3\). Thus, moles of \(\mathrm{HCl}\) needed:\[ \mathrm{moles \ of \ HCl} = 2 \times 0.002414 \ moles \approx 0.004828 \ moles \]

Calculate molarity of the HCl solution

Molarity is defined as moles of solute per liter of solution. Calculate molarity \(M\) using moles of \(\mathrm{HCl}\) and the volume of the solution in liters:\[ \text{Volume in liters} = \frac{28.3 \ mL}{1000 \ mL/L} = 0.0283 \ L \]\[ M = \frac{0.004828 \ moles}{0.0283 \ L} \approx 0.1707 \ M \]

Key concepts

Molarity Calculation

Molarity is a measure of concentration, representing the number of moles of solute present in a liter of solution. This is a fundamental concept in chemistry, especially important in titrations. To calculate the molarity of a solution, the formula used is: \[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \] Let's consider our exercise where we have a solution that required 0.004828 moles of HCl to react with sodium carbonate. The volume of this HCl solution was 28.3 mL, which we first convert to liters:

• Volume in liters = \( \frac{28.3 \text{ mL}}{1000} = 0.0283 \text{ L} \)

Using these values in our molarity formula,

• \( M = \frac{0.004828 \text{ moles}}{0.0283 \text{ L}} \approx 0.1707 \text{ M} \)

The molarity of the HCl solution is approximately 0.1707 M. This tells us the concentration of the acid in the solution.

Balanced Chemical Equation

Writing a balanced chemical equation is crucial to understanding how reactants turn into products and to ensure that mass is conserved during the reaction. In acid-base titrations, it helps identify the stoichiometry—the relationship between quantities of reactants—and calculate required moles for reaction.Let's look at our equation: - \( \mathrm{Na}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{NaCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \)This equation indicates that one mole of sodium carbonate reacts with two moles of hydrochloric acid:

• 1 mole \( \mathrm{Na}_2\mathrm{CO}_3 \) requires 2 moles \( \mathrm{HCl} \)
• This stems from the stoichiometry of reaction, maintaining equal balance on both sides in terms of atoms of each element.

Balanced equations are vital for calculating the precise amount of reagents needed, leading to correct determinations of concentration, as in our exercise.

Moles Calculation

Calculating the number of moles is a fundamental step in determining the amount of a substance present in a given mass. It involves using the molar mass of the substance to convert from grams to moles. The formula used is:\[ \text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \]Here's how it applies to our exercise with sodium carbonate \( \mathrm{Na}_2\mathrm{CO}_3 \):

• Given mass = 0.256 g
• Molar mass of \( \mathrm{Na}_2\mathrm{CO}_3 \) = 105.99 g/mol
• \( \text{moles of } \mathrm{Na}_2\mathrm{CO}_3 = \frac{0.256 \text{ g}}{105.99 \text{ g/mol}} \approx 0.002414 \text{ moles} \)

This calculation is essential because it translates the measured mass into a chemical quantity needed for further computations in titrations.

Sodium Carbonate Standardization

Sodium carbonate (\( \mathrm{Na}_2\mathrm{CO}_3 \)) is often used for standardizing acid solutions due to its high purity and stability. This process involves accurately determining the concentration of an acid by reacting it with a substance of known purity and concentration.In this particular exercise, sodium carbonate serves as a primary standard to determine the molarity of an HCl solution. Here's how this technique is advantageous:

• \( \mathrm{Na}_2\mathrm{CO}_3 \) has a reliable and stable molar mass, making it ideal for precise calculations.
• The reaction with HCl is straightforward and produces easily detectable products.
• By knowing the exact amount of \( \mathrm{Na}_2\mathrm{CO}_3 \), you can calculate the precise molarity of the HCl solution, ensuring accurate titration results.

Using sodium carbonate for standardization leads to high accuracy and reliability in determining the concentration of acidic solutions.