Problem 52 - Chemistry: The Central Science

Question

The concentration of \(\mathrm{Ca}^{2+}\) in a particular water supply is \(5.7 \times 10^{-3} \mathrm{M}\). The concentration of bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\). in the same water is \(1.7 \times 10^{-3} \mathrm{M}\). What masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must be added to \(5.0 \times 10^{7} \mathrm{~L}\) of this water to reduce the level of \(\mathrm{Ca}^{2+}\) to \(20 \%\) of its original level?

Step-by-Step Solution

Answer

The mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed is \(1.69 \times 10^{7} \mathrm{g}\), and the mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed is \(9.01 \times 10^{6} \mathrm{g}\) to reduce the level of \(\mathrm{Ca}^{2+}\) in the water to 20% of its original level.

Step 1: Determine the moles of ions and volume of water

The initial concentration of \(\mathrm{Ca}^{2+}\) ions is \(5.7 \times 10^{-3} \mathrm{M}\) and of \(\mathrm{HCO}_{3}^-\) ions is \(1.7 \times 10^{-3} \mathrm{M}\). The total volume of the water is \(5.0 \times 10^{7} \mathrm{L}\). We need to determine the number of moles of each ion in the water. We can calculate this using the formula: Moles = Concentration × Volume

Step 2: Calculate moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{HCO}_{3}^{-}\) ions

Calculate the moles of \(\mathrm{Ca}^{2+}\) ions: Moles of \(\mathrm{Ca}^{2+}\) = \(5.7 \times 10^{-3} \mathrm{M} \times 5.0 \times 10^{7} \mathrm{L} = 2.85 \times 10^{5} \mathrm{mol}\) Calculate the moles of \(\mathrm{HCO}_{3}^{-}\) ions: Moles of \(\mathrm{HCO}_{3}^{-}\) = \(1.7 \times 10^{-3} \mathrm{M} \times 5.0 \times 10^{7} \mathrm{L} = 8.5 \times 10^{4} \mathrm{mol}\)

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

The problem states that we need to reduce the level of \(\mathrm{Ca}^{2+}\) to 20% of its original level. Therefore, multiply the moles of \(\mathrm{Ca}^{2+}\) by 0.2 to find the final moles: Final moles of \(\mathrm{Ca}^{2+}\) = \(0.2 \times 2.85 \times 10^{5} \mathrm{mol} = 5.7 \times 10^{4} \mathrm{mol}\)

Step 4: Calculate the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) required

We find the difference between the initial and final moles of \(\mathrm{Ca}^{2+}\) ions to deduce the number of moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed to precipitate this difference: Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) required = \(2.85 \times 10^{5} \mathrm{mol} - 5.7 \times 10^{4} \mathrm{mol} = 2.28 \times 10^{5} \mathrm{mol}\)

Step 5: Calculate the mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) required

Using the molar mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) which is \(74.09 \mathrm{g/mol}\), calculate the mass required: Mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) required = \(2.28 \times 10^{5} \mathrm{mol} \times 74.09 \mathrm{g/mol = 1.69 \times 10^{7} \mathrm{g}\)

Step 6: Calculate the moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) required

From stoichiometry, the moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) used for the reaction is equal to the moles of \(\mathrm{HCO}_{3}^-\) ions present. Thus, Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) required = \(8.5 \times 10^{4} \mathrm{mol}\)

Step 7: Calculate the mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) required

Using the molar mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) which is \(105.99 \mathrm{g/mol}\), calculate the required mass: Mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) required = \(8.5 \times 10^{4} \mathrm{mol} \times 105.99 \mathrm{g/mol = 9.01 \times 10^{6} \mathrm{g}\) So, the mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed is \(1.69 \times 10^{7} \mathrm{g}\), and the mass of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) needed is \(9.01 \times 10^{6} \mathrm{g}\) to reduce the level of \(\mathrm{Ca}^{2+}\) in the water to 20% of its original level.

Key Concepts

Calcium Ions

Calcium ions, represented as \( \mathrm{Ca}^{2+} \), play a significant role in water hardness. Water hardness typically results from dissolved minerals like calcium and magnesium. These ions interact with soap, reducing its effectiveness. In the context of the given exercise, calcium ions

• contribute to scaling, which can impair pipes and boilers.
• exist naturally in water due to dissolved limestone or dolomite.

The problem outlines an initial concentration of calcium ions at \( 5.7 \times 10^{-3} \mathrm{M} \), which needs to be reduced. The reduction process is crucial for improving water quality and operational efficiency in industrial settings.
To lower the concentration of calcium ions, chemical precipitation is employed. This involves adding chemicals like \( \mathrm{Ca(OH)}_2 \), which reacts with the calcium ions to form an insoluble compound, thus removing them from the solution.
Understanding how to manipulate calcium ions using chemical additions ensures an effective means of managing water hardness, ultimately leading to better water quality and system longevity.

Bicarbonate Ions

Bicarbonate ions, denoted as \( \mathrm{HCO}_3^- \), are vital in buffering pH levels in water systems. They originate from carbon dioxide dissolution in water, forming carbonic acid, which then dissociates. In the exercise, the concentration of these ions is \( 1.7 \times 10^{-3} \mathrm{M} \). Bicarbonate ions influence water hardness by interacting with calcium ions.

• This interaction can lead to temporary hardness, which can be addressed by boiling or chemical treatment.
• They play a role in neutralizing acids, protecting aquatic life, and maintaining systems at a stable pH.

The problem requires adding \( \mathrm{Na}_2\mathrm{CO}_3 \) (sodium carbonate), which reacts with bicarbonate ions to maintain balance. Here, the stoichiometry is direct: the moles of sodium carbonate needed is equivalent to the moles of bicarbonate ions present.
By understanding bicarbonate ions, students can better grasp water chemistry's complexities, especially in how pH and water hardness are interrelated factors.

Chemical Precipitation

Chemical precipitation is a pivotal method for addressing water hardness, particularly useful for removing calcium and magnesium ions from water systems. By adding specific chemicals, these dissolved ions convert into an insoluble form. In this exercise

• we add \( \mathrm{Ca(OH)}_2 \) to decrease calcium ion concentration.
• \( \mathrm{Na}_2\mathrm{CO}_3 \) is added to transform bicarbonate ions to a removed phase.

The insoluble form typically precipitates out as a solid, simplifying its removal from the water system. The chemical reactions involved follow stoichiometry principles

• allowing precise calculations of needed chemicals to achieve desired water quality.
• ensure minimal excess chemicals in the process, enhancing cost-efficiency and environmental impact.

By utilizing chemical precipitation effectively, not only is water hardness addressed, but it also leads to an overall enhancement of water systems' performance and longevity. Understanding this concept empowers students to apply practical solutions to real-world water management problems.