Question

Standard solutions of calcium ion used to test for water hardness are prepared by dissolving pure calcium carbonate, \(\mathrm{CaCO}_{3}\), in dilute hydrochloric acid. A 1.745 -g sample of \(\mathrm{CaCO}_{3}\) is placed in a 250.0 -mL volumetric flask and dissolved in HCl. Then the solution is diluted to the calibration mark of the volumetric flask. Calculate the resulting molarity of calcium ion.

Short answer

The molarity of the calcium ion solution is calculated as follows: first, find the moles of \(\mathrm{CaCO}_{3}\) (1.745 g / 100.09 g/mol) which equals 0.01744 mol. Then, convert the volume of the solution to liters (250.0 mL / 1000), which equals 0.25 L. Finally, divide the moles of \(\mathrm{Ca}^{2+}\) (0.01744 mol) by the volume of the solution (0.25 L), resulting in a molarity of 0.06976 mol/L or 6.976 × 10\(^{-2}\) M for the calcium ion solution.

Step-by-step solution

Calculate the molar mass of \(\mathrm{CaCO}_{3}\)

To determine the moles of \(\mathrm{CaCO}_{3}\), we need to know its molar mass. The molar mass of \(\mathrm{CaCO}_{3}\) is calculated as follows: Molar mass of \(\mathrm{CaCO}_{3}\) = Molar mass of Ca + Molar mass of C + (3 × Molar mass of O) where: - Molar mass of Ca = 40.08 g/mol - Molar mass of C = 12.01 g/mol - Molar mass of O = 16.00 g/mol

Calculate the moles of \(\mathrm{CaCO}_{3}\)

Now, we can use the molar mass and the mass of the \(\mathrm{CaCO}_{3}\) sample to find the moles of \(\mathrm{CaCO}_{3}\): Moles of \(\mathrm{CaCO}_{3}\) = Mass of \(\mathrm{CaCO}_{3}\) / Molar mass of \(\mathrm{CaCO}_{3}\)

Find the moles of \(\mathrm{Ca}^{2+}\) in solution

The stoichiometry between \(\mathrm{CaCO}_{3}\) and \(\mathrm{Ca}^{2+}\) is 1:1 because each mole of \(\mathrm{CaCO}_{3}\) produces one mole of \(\mathrm{Ca}^{2+}\). This means that the moles of \(\mathrm{Ca}^{2+}\) ions are equal to the moles of \(\mathrm{CaCO}_{3}\).

Convert the volume of the solution to liters

To find the molarity, we'll need the volume of the solution in liters. So convert the given volume from mL to L: Volume in liters = Volume in mL / 1000

Calculate the molarity of \(\mathrm{Ca}^{2+}\) ions

The molarity of the calcium ion is found by dividing the moles of \(\mathrm{Ca}^{2+}\) by the volume of the solution (in liters): Molarity of \(\mathrm{Ca}^{2+}\) = Moles of \(\mathrm{Ca}^{2+}\)/ Volume of solution (L) Now, plug in the values from steps 2 and 4 and calculate the molarity of the calcium ion solution.

Key concepts

Calcium Carbonate

Calcium carbonate, known chemically as \textbf{CaCO}\(_3\), is a common substance that appears in a variety of forms, from chalk and limestone to marble. As a chemical compound, it plays an essential role in the buffering of ocean pH and is a critical component of biological hard tissues, such as shells and coral.

From an educational perspective, calcium carbonate is an exemplary case study for students learning about molar mass calculations and stoichiometry. Understanding the various properties of calcium carbonate, such as its solubility and reaction with acids, can aid students in developing a solid foundation in chemical analysis. For instance, when \textbf{CaCO}\(_3\) interacts with hydrochloric acid (\textbf{HCl}), it undergoes a chemical reaction where calcium ions (\textbf{Ca}^{2+}) and carbon dioxide (\textbf{CO}\(_2\)) are produced.

To facilitate the comprehension of exercises involving calcium carbonate, such as the determination of water hardness or the preparation of standard solutions, it is imperative to recognize that its role in reactions often links to the availability of \textbf{Ca}^{2+} ions. Thus, these practical applications of calcium carbonate underscore its significance in both natural processes and analytical chemistry.

Stoichiometry

Stoichiometry is a branch of chemistry that quantitatively examines the ratios of reactants and products in chemical reactions. At the heart of stoichiometry lies the concept of the mole, which allows chemists to count particles as a unit, much like a dozen eggs represent 12 individual items.

When solving problems involving stoichiometry, it is essential to:

• Calculate molar masses of the reactants and products.
• Convert mass of a substance to moles using its molar mass.
• Use the coefficients from the balanced chemical equation to relate moles of one substance to moles of another.

Through these critical steps, it becomes possible to predict how much product will form from given reactants or how much of one reactant is needed to react completely with another. This proves particularly useful when dealing with exercises that incorporate mixtures and reactions, such as the dissolution of calcium carbonate to prepare a solution for water hardness testing.

Strengthening stoichiometric skills is invaluable for students, enabling them to tackle complex problems across various fields within science, including environmental studies, medicinal chemistry, and materials science.

Volumetric Analysis

Volumetric analysis is a technique used in chemistry to determine the concentration of a solute in a solution. It involves the careful measurement of the volume of a solution of known concentration (a standard solution) to react completely with a solution of the substance being analyzed.

The process typically follows these steps:

1. Choose a standard solution with a known molarity.
2. Measure a precise volume of the sample solution containing the unknown concentration.
3. Carefully conduct a titration, if necessary, to reach the end point of the reaction, indicating stoichiometric equivalence.
4. Calculate the unknown concentration using the volume-to-volume ratio and stoichiometry of the reaction.

For students dealing with exercises related to volumetric analysis, understanding the method's principles is paramount. Whether to determine water hardness, as in our calcium carbonate exercise, or to analyze the purity of a substance, precision in measurement and calculations is crucial. Practicing volumetric analysis teaches students accurate lab techniques and fosters an appreciation for the meticulous nature of analytical chemistry.