Problem 69 - Chemistry and Chemical Reactivity

Question

If 38.55 mL of \(\mathrm{HCl}\) is required to titrate \(2.150 \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) according to the following equation, what is the concentration \((\mathrm{mol} / \mathrm{L})\) of the HCl solution? \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow\) \(2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

The concentration of the HCl solution is approximately 1.053 mol/L.

Step 1: Calculate moles of Na2CO3

First, we need to calculate the number of moles of \(\mathrm{Na}_2\mathrm{CO}_3\). To do this, we use the formula: \[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\] The molar mass of \(\mathrm{Na}_2\mathrm{CO}_3\) is calculated as follows: \[2 \times \text{(molar mass of Na)} + \text{(molar mass of C)} + 3 \times \text{(molar mass of O)} = 2 \times 22.99 + 12.01 + 3 \times 16.00 = 105.99 \text{ g/mol}\] Now, calculate the moles of \(\mathrm{Na}_2\mathrm{CO}_3\): \[\text{moles of } \mathrm{Na}_2\mathrm{CO}_3 = \frac{2.150}{105.99} \approx 0.0203 \text{ mol}\]

Step 2: Determine moles of HCl required

According to the balanced chemical equation, 1 mole of \(\mathrm{Na}_2\mathrm{CO}_3\) reacts with 2 moles of \(\mathrm{HCl}\). Therefore, the moles of \(\mathrm{HCl}\) required are calculated by doubling the moles of \(\mathrm{Na}_2\mathrm{CO}_3\): \[\text{moles of } \mathrm{HCl} = 2 \times 0.0203 = 0.0406 \text{ mol}\]

Step 3: Calculate the concentration of the HCl solution

We know the volume of \(\mathrm{HCl}\) used is 38.55 mL. To calculate the concentration, first convert this volume to liters: \[38.55 \text{ mL} = 0.03855 \text{ L}\] The concentration of \(\mathrm{HCl}\) is given by: \[\text{concentration} (\text{mol/L}) = \frac{\text{moles of } \mathrm{HCl}}{\text{volume in liters}} = \frac{0.0406}{0.03855} \approx 1.053 \text{ mol/L}\]

Key Concepts

Molarity Calculations

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Molarity is commonly represented by the symbol \(M\) and is calculated using the formula: M = \(\frac{\text{moles of solute}}{\text{volume of solution in liters}}\).Understanding the molarity of a solution is essential in various applications, such as titrations, which require precise concentrations to determine unknown quantities.In our exercise, we calculated the molarity of an HCl solution. We used 0.0406 moles of HCl in a volume of 0.03855 liters. By dividing the number of moles by the volume of the solution in liters, we found the molarity to be approximately 1.053 mol/L.Molarity also helps in understanding the concept of dilution. A solution can be diluted by adding more solvent, which decreases the molarity but keeps the overall amount of solute constant.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants involved on the left side and the products formed on the right side, with an arrow in between implying the direction of the reaction. These equations must be balanced, meaning the number of atoms of each element should be equal on both sides.In the given exercise, the chemical equation is:\(\mathrm{Na}_2\mathrm{CO}_3(\mathrm{aq}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow 2\mathrm{NaCl}(\mathrm{aq}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\ell)\). This balanced equation shows that one mole of \(\mathrm{Na}_2\mathrm{CO}_3\) reacts with two moles of \(\mathrm{HCl}\) to produce two moles of \(\mathrm{NaCl}\), one mole of \(\mathrm{CO}_2\), and one mole of \(\mathrm{H}_2\mathrm{O}\).Balancing chemical equations is crucial for the correct quantification of reactants and products, and they also enable us to conduct stoichiometric calculations, such as the one we performed in this exercise to find out how much HCl was required for the reaction.

Moles

The mole is a fundamental unit in chemistry that measures the amount of a substance. One mole contains exactly \(6.022 \times 10^{23}\) entities (Avogadro's number), whether they are atoms, molecules, or ions. This allows chemists to count particles by weighing them.Calculating moles involves using the formula:\(\text{moles} = \frac{\text{mass of substance}}{\text{molar mass}}\).In the exercise, we calculated the moles of \(\mathrm{Na}_2\mathrm{CO}_3\) by dividing its mass, 2.150 grams, by its molar mass, 105.99 g/mol, resulting in approximately 0.0203 moles.Understanding moles is crucial because it helps us relate the macroscopic world we see to the microscopic world of atoms and molecules. It is essential in converting between mass and moles and for performing calculations involving gas volumes, solutions, and other chemical quantities.