Question

Natural waters often contain relatively high levels of calcium ion, \(\mathrm{Ca}^{2+},\) and hydrogen carbonate ion (bicarbonate), \(\mathrm{HCO}_{3}^{-}\), from the leaching of minerals into the water. When such water is used commercially or in the home, heating of the water leads to the formation of solid calcium carbonate, \(\mathrm{CaCO}_{3}\), which forms a deposit ("scale") on the interior of boilers, pipes, and other plumbing fixtures. $$ \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If a sample of well water contains \(2.0 \times 10^{-3} \mathrm{mg}\) of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) per milliliter, what mass of \(\mathrm{CaCO}_{3}\) scale would \(1.0 \mathrm{~mL}\) of this water be capable of depositing?

Short answer

The mass of \(\mathrm{CaCO}_{3}\) scale that would be formed when 1.0 mL of this water is used and heated is \(1.234 \times 10^{-6}\ \mathrm{g}\).

Step-by-step solution

Write down the balanced chemical equation.

The balanced chemical equation is given as: $$ \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(a q) \rightarrow \mathrm{CaCO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

Find the concentration of calcium hydrogen carbonate.

The concentration of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) is given as \(2.0 \times 10^{-3}\ \mathrm{mg/mL}\).

Calculate the number of moles of calcium hydrogen carbonate per milliliter of water.

First, we need to convert the mass concentration to moles per milliliter. To do this, we'll use the molar mass of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) (162.11 g/mol). Number of moles = \( \frac{Mass}{Molar\ Mass} \) Number of moles of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\)/mL = \( \frac{2.0 \times 10^{-3}\ \mathrm{mg}}{162.11\ \mathrm{g/mol}} \) Note: Convert mg to grams: \(2.0 \times 10^{-3}\ \mathrm{mg} = 2.0 \times 10^{-6}\ \mathrm{g} \) Number of moles of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\)/mL = \( \frac{2.0 \times 10^{-6}\ \mathrm{g}}{162.11\ \mathrm{g/mol}} \) = \( 1.234 \times 10^{-8}\ mol/mL\)

Calculate the moles of calcium carbonate produced per milliliter of water.

Using the stoichiometry of the balanced equation, we know that for every 1 mole of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\), 1 mole of \(\mathrm{CaCO}_{3}\) is produced. So, the number of moles of \(\mathrm{CaCO}_{3}\) per mL = Number of moles of \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}\) per mL = \(1.234 \times 10^{-8}\ mol/mL\)

Calculate the mass of calcium carbonate produced per milliliter of water.

Use the molar mass of \(\mathrm{CaCO}_{3}\) (100.09 g/mol) to convert moles of \(\mathrm{CaCO}_{3}\) to mass. Mass of \(\mathrm{CaCO}_{3}\) = Number of moles of \(\mathrm{CaCO}_{3} * Molar\ Mass\) Mass of \(\mathrm{CaCO}_{3}\)/mL = \(1.234 \times 10^{-8}\ \mathrm{mol/mL}\) * \(100.09\ \mathrm{g/mol}\) = \(1.234 \times 10^{-6} \ \mathrm{g/mL}\)

Calculate the mass of calcium carbonate per 1.0 mL of water.

Since we know the mass of \(\mathrm{CaCO}_{3}\) produced per milliliter of water, the mass of \(\mathrm{CaCO}_{3}\) for 1.0 mL of water is the same. Mass of \(\mathrm{CaCO}_{3}\) for 1.0 mL of water = \(1.234 \times 10^{-6}\ \mathrm{g}\) The mass of \(\mathrm{CaCO}_{3}\) scale that would be formed when 1.0 mL of this water is used and heated is \(1.234 \times 10^{-6}\ \mathrm{g}\).

Key concepts

Chemical Stoichiometry

Chemical stoichiometry is one of the foundational concepts in chemistry, which deals with the quantitative relationships between the reactants and products in a chemical reaction. It enables chemists to predict the amounts of substances consumed and produced in a reaction.

To grasp this concept, it’s helpful to first understand what a mole is. In chemistry, a mole is a unit of measurement used to express amounts of a chemical substance. One mole is equal to Avogadro's number (\(6.022 \times 10^{23}\)) of atoms, molecules, or ions.

Stoichiometry is used to calculate the moles of reactants and products using the balanced chemical equation. In our example, the equation shows that one mole of calcium hydrogen carbonate yields one mole of calcium carbonate, carbon dioxide, and water. This 1:1:1:1 ratio means that the number of moles of calcium carbonate produced will be equal to the number of moles of calcium hydrogen carbonate that reacted. Understanding these ratios allows us to perform calculations that translate to the real world, like computing how much scale builds up in pipes and boilers.

Molar Mass Calculation

The molar mass of a compound is the mass in grams of one mole of that substance. It is a crucial concept for converting between mass and moles, which is necessary when applying stoichiometry in a chemical reaction. The molar mass is determined by adding up the atomic masses of all the atoms in a molecule. These atomic masses can be found on the periodic table.

For example, in the exercise, we calculated the molar mass of calcium carbonate ((CaCO_3). Calcium ((Ca)) has a molar mass of about 40.08 g/mol, carbon ((C)) about 12.01 g/mol, and oxygen ((O)), which is present as three atoms in the compound, has a molar mass of about 16.00 g/mol per atom. To find the molar mass of calcium carbonate, we sum these values, yielding approximately 100.09 g/mol.

Knowing the molar mass, we can convert between the mass of a substance (in grams) and the number of moles, as demonstrated in the step-by-step solution of the exercise. This allows us to predict how much calcium carbonate will precipitate out of solution when the water is used and heated.

Aqueous Solution Reactions

Reactions in an aqueous solution are common in both nature and industrial processes. An aqueous solution is where one or more substances are dissolved in water. The dissolved substances, known as solutes, may interact and undergo chemical reactions, leading to the production of new substances.

In our example, the reaction involves calcium hydrogen carbonate ((Ca(HCO_3)_2)) in solution reacting upon heating to form solid calcium carbonate ((CaCO_3)), which precipitates out of the solution, along with the formation of carbon dioxide ((CO_2)) gas and water ((H_2O)).

Predicting the outcomes of such reactions, including the type and amount of precipitate that will form, involves understanding chemical stoichiometry and molar mass calculations, as well as the solubility rules for ionic compounds. These reactions are not only important for scientific predictions but also have practical implications, impacting equipment efficiency and requiring regular maintenance to manage the scale deposits.