Problem 52 - Chemistry: The Central Science in SI Units, Global Edition

Question

In the lime soda process once used in large scale municipal water softening, calcium hydroxide prepared from lime and sodium carbonate are added to precipitate \(\mathrm{Ca}^{2+}\) as \(\mathrm{CaCO}_{3}(s)\) and \(\mathrm{Mg}^{2+}\) as \(\mathrm{Mg}(\mathrm{OH})_{2}(s);\) $$ \begin{aligned} \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) & \longrightarrow \mathrm{CaCO}_{3}(s) \\ \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) & \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s) \end{aligned} $$ How many moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) should be added to soften (remove the \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ) 1000 L of water in which $$ \begin{array}{l} {\left[\mathrm{Ca}^{2+}\right]=3.5 \times 10^{-4} \mathrm{M}} \\ {\left[\mathrm{Mg}^{2+}\right]=7.5 \times 10^{-4} \mathrm{M}} \end{array} $$

Step-by-Step Solution

Answer

To soften 1000 L of water, you must add \(3.75 \times 10^{-4}\ \mathrm{moles}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(3.5 \times 10^{-4}\ \mathrm{moles}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\).

Step 1: Calculate moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions in the water

First, we need to find the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions in the 1000 L of water. We do this using the given concentrations and the water volume: Moles of \(\mathrm{Ca}^{2+}\) = Concentration of \(\mathrm{Ca}^{2+}\times\) Volume of water Moles of \(\mathrm{Mg}^{2+}\) = Concentration of \(\mathrm{Mg}^{2+}\times\) Volume of water The volume of water is given in liters, but we need to convert it to meters cubed (m³) for our calculation. To convert liters to cubic meters, we multiply by 0.001: Volume of water = 1000 L \(\times\) 0.001 = 1 m³ Now, we can calculate the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions: Moles of \(\mathrm{Ca}^{2+}\) = \((3.5 \times 10^{-4}\ \mathrm{M}) \times (1\ \mathrm{m}^{3}) = 3.5 \times 10^{-4}\ \mathrm{moles}\) Moles of \(\mathrm{Mg}^{2+}\) = \((7.5 \times 10^{-4}\ \mathrm{M}) \times (1\ \mathrm{m}^{3}) = 7.5 \times 10^{-4}\ \mathrm{moles}\)

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

Now that we have the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions in the water, we can use the stoichiometry of the provided reactions to determine the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) required. From the first reaction, one mole of \(\mathrm{Ca}^{2+}\) reacts with one mole of \(\mathrm{CO}_{3}^{2-}\) to form one mole of \(\mathrm{CaCO}_{3}(s)\). Since \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) provides one mole of \(\mathrm{CO}_{3}^{2-}\) per mole of itself, the moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) required will be equal to the moles of \(\mathrm{Ca}^{2+}\) ions: Moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = Moles of \(\mathrm{Ca}^{2+}\) = \(3.5 \times 10^{-4}\ \mathrm{moles}\) From the second reaction, one mole of \(\mathrm{Mg}^{2+}\) reacts with two moles of \(\mathrm{OH}^{-}\) to form one mole of \(\mathrm{Mg}(\mathrm{OH})_{2}(s)\). Since \(\mathrm{Ca}(\mathrm{OH})_{2}\) provides two moles of \(\mathrm{OH}^{-}\) per mole of itself, the moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) required will be half of the moles of \(\mathrm{Mg}^{2+}\) ions: Moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) = \(\frac{1}{2}\times\) Moles of \(\mathrm{Mg}^{2+}\) = \((\frac{1}{2}) (7.5 \times 10^{-4}\ \mathrm{moles}) = 3.75 \times 10^{-4}\ \mathrm{moles}\) So, to soften the 1000 L of water, \(3.75 \times 10^{-4}\ \mathrm{moles}\) of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(3.5 \times 10^{-4}\ \mathrm{moles}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) should be added.

Key Concepts

Understanding the Lime Soda Process

The lime soda process is a chemical method popularly used for water softening. It's mainly used to remove hardness caused by calcium (Ca²⁺) and magnesium (Mg²⁺) ions in water. These ions can make water unsuitable for various domestic and industrial uses.
Here's how the lime soda process works:

• Calcium hydroxide, derived from lime, and soda ash (sodium carbonate) are added to hard water.


• The sodium carbonate reacts with calcium ions in the water to form calcium carbonate, which is insoluble and precipitates out of the solution.


• Similarly, the hydroxide ions from calcium hydroxide react with magnesium ions to form magnesium hydroxide, which also precipitates out.

This process effectively reduces the hardness of the water, making it cleaner and safer for consumption and preventing scale buildup in pipes and boilers.

Role of Stoichiometry in Water Softening

Stoichiometry is a crucial part of the lime soda process. It deals with calculating the amounts of reactants and products in chemical reactions. By understanding stoichiometry, you can determine the exact quantities of lime (calcium hydroxide) and soda ash (sodium carbonate) needed to treat a specific volume of hard water.
In the exercise, stoichiometry helps answer questions like:

• How many moles of sodium carbonate are needed to react with a given amount of calcium ions?


• How many moles of calcium hydroxide are required to remove the magnesium ions present in water?

Stoichiometry ensures that we add exactly what's needed to softener the water without overuse of chemicals, which could lead to unwanted residues or side effects.

Exploring the Chemical Reactions Involved

During the lime soda process, two key chemical reactions take place, each aimed at removing specific ions.- **Reaction for Calcium:**
The equation \[ \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq) \longrightarrow \mathrm{CaCO}_{3}(s) \] illustrates that calcium ions react with carbonate ions to form calcium carbonate, which is a solid precipitate.
- **Reaction for Magnesium:**
The equation \[ \mathrm{Mg}^{2+}(aq) + 2 \mathrm{OH}^{-}(aq) \longrightarrow \mathrm{Mg}( ext{OH})_{2}(s) \] shows that magnesium ions react with hydroxide ions to form magnesium hydroxide, another solid.
In both cases, the transformations involve turning dissolved ions into insoluble, removable solids, effectively reducing water hardness.

Understanding Precipitation Reactions

Precipitation reactions are fundamental in water softening processes like the lime soda process. These reactions occur when dissolved ions in solution combine to form a solid compound that precipitates, or falls out of the solution.
In our context:

• Calcium ions and carbonate ions combine to form the solid calcium carbonate.
• Magnesium ions and hydroxide ions form the solid magnesium hydroxide.

Precipitation reactions are important because they provide a way to separate and remove unwanted dissolved ions from water, making it suitable for various applications. Without precipitation, these ions would remain dissolved, continuing to contribute to water hardness. The effectiveness of this method depends largely on proper stoichiometry to ensure all ions are adequately targeted and precipitated as solids.