Problem 118 - Chemistry The Molecular Nature of Matter

Question

A sample of hard water was found to have 278 ppm \(\mathrm{Ca}^{2+}\) ion. Into \(1.00 \mathrm{~L}\) of this water, \(1.00 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) was dissolved. What is the new concentration of \(\mathrm{Ca}^{2+}\) in parts per million? (Assume that the addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) does not change the volume, and assume that the density of the aqueous solutions involved are all \(1.00 \mathrm{~g} \mathrm{~mL}^{-1}\).)

Step-by-Step Solution

Answer

The new concentration of \(\mathrm{Ca}^{2+}\) in the water after the reaction is 0 ppm.

Step 1: Calculate moles of \(\mathrm{Na}_2\mathrm{CO}_3\)

Use the molar mass of \(\mathrm{Na}_2\mathrm{CO}_3\) to convert grams to moles. Molar mass of \(\mathrm{Na}_2\mathrm{CO}_3\) is approximately 106 g/mol, so \(1.00 \text{ g} \div 106 \text{ g/mol} = 0.00943 \text{ moles}.\)

Step 2: Determine moles of \(\mathrm{Ca}^{2+}\) ions that can react

The chemical reaction between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Na}_2\mathrm{CO}_3\) can be represented as \(\mathrm{Ca}^{2+} + \mathrm{CO}_3^{2-} \rightarrow \mathrm{CaCO}_3(s)\). Because the reaction is in a 1:1 mole ratio, 0.00943 moles of \(\mathrm{Na}_2\mathrm{CO}_3\) will react with 0.00943 moles of \(\mathrm{Ca}^{2+}\) ions.

Step 3: Calculate the initial moles of \(\mathrm{Ca}^{2+}\) ions in the sample

Convert the initial concentration from ppm to moles/L. Since there are 278 ppm \(\mathrm{Ca}^{2+}\) and the molecular weight of \(\mathrm{Ca}\) is 40.08 g/mol, the initial moles in 1 L are \(278 \text{ mg/L} \div 40.08 \text{ mg/mmol} = 6.93 \text{ mmol/L} = 0.00693 \text{ moles}.\)

Step 4: Subtract the moles of \(\mathrm{Ca}^{2+}\) that reacted

Subtract the moles of \(\mathrm{Ca}^{2+}\) that reacted with \(\mathrm{Na}_2\mathrm{CO}_3\) (0.00943 moles) from the initial moles (0.00693 moles) to find the remaining moles of \(\mathrm{Ca}^{2+}\), which yields a negative value. Because this value cannot be negative, it means all \(\mathrm{Ca}^{2+}\) reacted and precipitated as \(\mathrm{CaCO}_3\), so the remaining \(\mathrm{Ca}^{2+}\) concentration is 0.

Step 5: Convert the remaining moles back to ppm

Since there are no remaining moles of \(\mathrm{Ca}^{2+}\), the concentration in ppm is also 0.

Key Concepts

Parts per Million (ppm)

Understanding 'parts per million' or ppm is crucial when dealing with concentrations in chemistry, especially in solutions. Ppm is a unit that describes the amount of one substance in a million parts of another. It is widely used to measure levels of pollutants in air, soil, or water. In our exercise, the concentration of calcium ions, \(\mathrm{Ca}^{2+}\), is initially given in ppm. It's important to note that 1 ppm is equivalent to 1 milligram of solute per liter of solution (mg/L), assuming the density of water to be 1 g/mL. This unit allows scientists and engineers to express very dilute concentrations, such as those commonly found in water hardness measurements.

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the quantitative relationships between the substances as they participate in various chemical reactions. It allows chemists to predict the amounts of products and reactants that will be consumed or produced in a given reaction. In our water hardness exercise, stoichiometry is used to determine how many moles of \(\mathrm{Ca}^{2+}\) ions will react with a given amount of \(\mathrm{Na}_2\mathrm{CO}_3\) based on the balanced chemical equation. The key concept here is the mole ratio between reactants and products, crucial for calculating the new concentration of \(\mathrm{Ca}^{2+}\) ions in solution.

Molar Mass

The molar mass of a substance is the mass of one mole of that substance, measured in grams per mole (g/mol). It is a key property of a substance that we need for converting between grams and moles, and it's directly related to the substance's molecular or formula weight. In our example, the molar mass of \(\mathrm{Na}_2\mathrm{CO}_3\) is around 106 g/mol. This is used in the first step of the solution to convert the mass of \(\mathrm{Na}_2\mathrm{CO}_3\) added to the water into moles, a necessary step for applying stoichiometry in the subsequent calculations.

Chemical Reactions

Chemical reactions involve the conversion of reactants into products through a process that includes breaking and forming chemical bonds. In our scenario with hard water, the key chemical reaction is between \(\mathrm{Ca}^{2+}\) ions and carbonate ions \(\mathrm{CO}_3^{2-}\) to form \(\mathrm{CaCO}_3\), a precipitate. Understanding the nature of this reaction is fundamental to calculating the change in water hardness after the addition of \(\mathrm{Na}_2\mathrm{CO}_3\).

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where an insoluble solid, known as a precipitate, forms when two aqueous solutions are mixed. The formation of \(\mathrm{CaCO}_3\) in our water hardness example is a classic case of a precipitation reaction. By accurately predicting the outcome of such reactions, one can deduce the composition of the resulting mixture—in our case, whether any \(\mathrm{Ca}^{2+}\) ions remain in the water after reaction.

Moles to Grams Conversion

The process of converting moles to grams and vice versa is an essential skill in chemistry. This conversion relies on the molar mass of the substance. For any substance, one mole is equal to its molar mass in grams. The exercise provided showcases how to convert the given mass of \(\mathrm{Na}_2\mathrm{CO}_3\) into moles before proceeding with the stoichiometric calculations. Understanding the mole concept is instrumental for precisely handling and measuring chemical substances.

Ionic Equations

Ionic equations help to provide a clearer picture of the species that are actually participating in the reaction in a solution, as opposed to full molecular equations. They show only the species that change form during the reaction—the ions. In our water hardness problem, the ionic equation simplifies down to \(\mathrm{Ca}^{2+} + \mathrm{CO}_3^{2-} \rightarrow \mathrm{CaCO}_3(s)\), making it evident that \(\mathrm{Ca}^{2+}\) and \(\mathrm{CO}_3^{2-}\) ions react to form the precipitate \(\mathrm{CaCO}_3\). This simplification is particularly helpful when dealing with reactions in aqueous solutions, such as in the water hardness calculation.