Question

Calcium carbonate, \(\mathrm{CaCO}_{3}\), can be obtained in a very pure state. Standard solutions of calcium ion are usually prepared by dissolving calcium carbonate in acid. What mass of \(\mathrm{CaCO}_{3}\) should be taken to prepare \(500 . \mathrm{mL}\) of \(0.0200 \mathrm{M}\) calcium ion solution?

Short answer

To prepare 500 mL of a 0.0200 M calcium ion solution, approximately 1.001 g of calcium carbonate is needed.

Step-by-step solution

Understand Molarity

Molarity (M) is defined as the moles of solute per liter of solution. In this problem, we are given the molarity of the calcium ion solution as 0.0200 M. We need to find the mass of calcium carbonate required to prepare this solution.

Convert Volume to Liters

We are given the volume of the solution in milliliters (500 mL). To use it with molarity, we need to convert it to liters: 500 mL × (1 L / 1000 mL) = 0.5 L

Calculate Moles of Calcium Carbonate

Using the given molarity (0.0200 M) and the converted volume in liters (0.5 L), we can find the moles of calcium ions present in the solution since moles = Molarity × Volume. Moles of calcium ions = (0.0200 M) × (0.5 L) = 0.0100 moles As calcium carbonate ionizes completely in the solution, the 1 mole of the calcium ion in the solution is equal to 1 mole of calcium carbonate. Therefore, there are 0.0100 moles of calcium carbonate needed to prepare the 500 mL of 0.0200 M calcium ion solution.

Calculate Mass of Calcium Carbonate

Now that we have the number of moles of calcium carbonate required, we can find the mass of calcium carbonate by using its molar mass. The molar mass of calcium carbonate (CaCO3) is approximately 40.08 g/mol (for Ca) + 12.01 g/mol (for C) + 3 × 16.00 g/mol (for O) = 100.09 g/mol. Therefore, the mass of calcium carbonate needed = (0.0100 moles) × (100.09 g/mol) = 1.0009 g (approximated to 1.001 g when considering significant figures). The mass of calcium carbonate required to prepare 500 mL of 0.0200 M calcium ion solution is approximately 1.001 g.

Key concepts

Chemistry Stoichiometry

Stoichiometry is the section of chemistry that pertains to the calculation of the quantities of reactants and products in chemical reactions. In the context of our exercise, stoichiometry allows us to determine the amount of calcium carbonate (\textbf{CaCO}\(_3\)) needed to produce a specific concentration of calcium ions (\textbf{Ca}^{2+}) in solution. To do this, we must first understand that stoichiometry is based on the principle of the conservation of mass and the notion of the mole, which is a unit that represents a certain number of particles, ions, or molecules.
In our example, the stoichiometric calculations revolve around the relationship between the molarity of the calcium ion solution and the mass of calcium carbonate required. As each mole of \textbf{CaCO}\(_3\) produces one mole of \textbf{Ca}^{2+} ions upon dissolution, the stoichiometry is straightforward. It's vital to note that stoichiometric calculations often begin with a balanced chemical equation, but in this scenario, we didn't need the full equation since the dissolution of calcium carbonate was assumed to yield solely calcium ions.

Molar Mass

The molar mass is an essential constant in stoichiometric calculations; it's the mass of one mole of a substance usually expressed in grams per mole (g/mol). When you’re solving for the mass of a reactant, like calcium carbonate (\textbf{CaCO}\(_3\)), needed to prepare a solution, you convert moles to grams using the molar mass. The molar mass of \textbf{CaCO}\(_3\) is derived from the atomic masses of its constituent atoms, shown in the periodic table: calcium (Ca), carbon (C), and oxygen (O).
To calculate the molar mass of \textbf{CaCO}\(_3\), each atomic mass is multiplied by the number of respective atoms in the formula and then added together:
\(h4> Molar Mass of \textbf{CaCO}\)_3\(:

• Calcium (Ca): 40.08 g/mol,
• Carbon (C): 12.01 g/mol,
• Oxygen (O): 16.00 g/mol × 3 atoms,

Adding these values gives the molar mass of \textbf{CaCO}\)_3\( as about 100.09 g/mol. This value is critical for converting moles of \textbf{CaCO}\)_3\( to grams.

Solution Concentration

Solution concentration is a measure of how much solute is present in a given quantity of solvent. Molarity (M), the metric mentioned in the exercise, is a common way to express concentration, defined as moles of solute per liter of solution. To find out how much solute to use, you need the final volume of the solution and the desired molarity.
In the example problem, the goal is to prepare a 0.0200 M solution of calcium ions. The conversion of volume from milliliters to liters is critical because molarity depends on liters of solution. After the conversion, you use the formula:
\[\begin{equation}\text{Moles of solute} = \text{Molarity} \times \text{Volume of solution}\br\br\br\br\right).\br\br\brThe result gives us the moles of calcium ions needed, and thus the moles of \textbf{CaCO}\)_3\(, since one mole of \textbf{CaCO}\)_3$ yields one mole of \textbf{Ca}^{2+} ions. Understanding molarity is essential for precise preparation of solutions in both laboratory and industrial settings, making it a foundational concept in chemistry.