Problem 51 - Chemistry The Central Science

Question

How many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) should be added to soften \(1200 \mathrm{~L}\) of water in which \(\left[\mathrm{Ca}^{2+}\right]=5.0 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{HCO}_{3}^{-}\right]=7.0 \times 10^{-4} \mathrm{M}\) ?

Step-by-Step Solution

Answer

To soften \(1200 \mathrm{~L}\) of water with the given concentrations of ions, \(0.42 \mathrm{~mol}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.6 \mathrm{~mol}\) of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) should be added.

Step 1: Find moles of ions in the water

The volume of the water is 1200 L. We can find the moles of \(\left[\mathrm{Ca}^{2+}\right]\) and \(\left[\mathrm{HCO}_{3}^{-}\right]\) ions in the water by using the formula: \(moles = concentration \times volume\) a. Moles of \(\mathrm{Ca}^{2+}\) ions: Moles of $\mathrm{Ca}^{2+}= 5.0 \times 10^{-4} M \times 1200 L = 0.6 \mathrm{~mol}\) b. Moles of \(\mathrm{HCO}_{3}^{-}\) ions: Moles of $\mathrm{HCO}_{3}^{-}= 7.0 \times 10^{-4} M \times 1200 L = 0.84 \mathrm{~mol}\)

Step 2: Determine moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\)

a. Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\): From the reaction between \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{HCO}_{3}^{-}\), we know that 1 mol of \(\mathrm{Ca}(\mathrm{OH})_{2}\) reacts with 2 moles of \(\mathrm{HCO}_{3}^{-}\). Moles of $\mathrm{Ca}(\mathrm{OH})_{2} = \frac{0.84 \mathrm{~mol} \mathrm{HCO}_{3}^-}{2} = 0.42 \mathrm{~mol}\) b. Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\): From the reaction between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Na}_{2}\mathrm{CO}_{3}\), we know that 1 mol of \(\mathrm{Ca}^{2+}\) reacts with 1 mol of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). Moles of $\mathrm{Na}_{2}\mathrm{CO}_{3} = 0.6 \mathrm{~mol} \mathrm{Ca}^{2+} = 0.6 \mathrm{~mol}\) So, to soften \(1200 \mathrm{~L}\) of water, we need 0.42 moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and 0.6 moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\).

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that involves the calculation of the quantities of reactants and products in chemical reactions. This concept is a fundamental aspect of chemistry because it provides a quantitative relationship between the elements and compounds involved in a reaction.

For example, when determining the amount of chemicals needed to soften water, one must understand the stoichiometric ratios of the reactants to the products. In the given exercise, the stoichiometry of the reaction between \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{HCO}_{3}^{-}\) ions indicates that 1 mole of \(\mathrm{Ca}(\mathrm{OH})_{2}\) reacts with 2 moles of \(\mathrm{HCO}_{3}^{-}\). Similarly, the reaction between \(\mathrm{Ca}^{2+}\) and \(\mathrm{Na}_2\mathrm{CO}_3\) is 1:1. Knowing these ratios allows us to calculate the exact amount of each chemical required to fully react with the ions present in the water.

Molar Concentration

Molar concentration, also known as molarity, measures the number of moles of a solute that is present in one liter of solution. It is a critical concept in chemistry for understanding how solutions of different substances interact and react with each other. The formula to calculate molar concentration is expressed as \(\text{molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\).

In the context of water softening, the molar concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{HCO}_{3}^{-}\) ions are given, allowing us to determine the number of moles of each ion in a given volume of water, a crucial step before we can calculate the amount of softening agent needed. For instance, with a water volume of 1200 liters and the given molar concentrations, we can easily calculate the moles of each ion present in the water, forming the foundation for the subsequent stoichiometry calculations.

Chemical Reactions

Chemical reactions are processes where reactants transform into products through breaking and forming of chemical bonds. Each reaction is governed by a unique equation that tells us which reactants combine and in what proportion to form particular products.

In the water softening process, the chemical reactions involved play a crucial role. For instance, \(\mathrm{Ca}^{2+}\) ions in water can cause hardness and are typically removed by reaction with \(\mathrm{Na}_2\mathrm{CO}_3\), which forms calcium carbonate, a compound that precipitates and can be removed from the water. Additionally, bicarbonate ions (\(\mathrm{HCO}_{3}^{-}\)) react with calcium hydroxide (\(\mathrm{Ca}(\mathrm{OH})_{2}\)) to form calcium carbonate and water. This knowledge of chemical reactions allows us to select the appropriate compounds for softening and determine their quantities, ensuring that the treatment process is efficient and complete.