Problem 93 - General Chemistry: Principles and Modern Applications

Question

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

Step-by-Step Solution

Answer

The balanced chemical equations are (1) \(\mathrm{CaO} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\), (2) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2} \mathrm{O}\), (3) \(\mathrm{HPO}_{4}^{2-} + \mathrm{OH}^{-} \rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2} \mathrm{O}\), (4) 5\(\mathrm{Ca}^{2+} + 3\mathrm{PO}_{4}^{3-} + \mathrm{OH}^{-} \rightarrow \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\), and 0.908 kg of lime is needed to remove the phosphorus from the contaminated water.

Step 1: Writing the balanced chemical equations

The four reactions described in the exercise are: \n1. \(\mathrm{CaO} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\), \n2. \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2} \mathrm{O}\), \n3. \(\mathrm{HPO}_{4}^{2-} + \mathrm{OH}^{-} \rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2} \mathrm{O}\), \n4. 5\(\mathrm{Ca}^{2+} + 3\mathrm{PO}_{4}^{3-} + \mathrm{OH}^{-} \rightarrow \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\)

Step 2: Convert mass of phosphorus to moles

The total amount of phosphorus in the water is \(10.0 \mathrm{mg}/\mathrm{L} \times 1.00 \times 10^{4} \mathrm{L} = 1.00 \times 10^{5} \mathrm{mg} = 100 \mathrm{g}\). In order to go further, we need to convert this mass into moles using Phosphorus molar mass (31.0 g/mol). Hence, \(100 \mathrm{g} \times \left( \frac{1 \mathrm{mol}}{31.0 \mathrm{g}} \right) = 3.23 \mathrm{mol}\) of phosphorus.

Step 3: Determine moles of \(CaO\) required

From reaction 4, it is seen that each mole of phosphorus (in the form of \(PO_{4}^{3-}\)) requires 5 moles of \(Ca^{2+}\) for precipitation. Since \(Ca^{2+}\) ions come from \(CaO\) (Reaction 1: 1 mole of \(CaO\) gives 1 mole of \(Ca^{2+}\) ion), we require 5 moles of \(CaO\) for each mole of phosphorus. Hence, \(3.23 \mathrm{mol} \times 5 = 16.2 \mathrm{mol}\) of \(CaO\) are required.

Step 4: Convert moles of \(CaO\) to mass

Now we convert the moles of \(CaO\) into kilograms using the molar mass of \(CaO\) which is 56.08 g/mol. So, \(16.2 \mathrm{mol} \times 56.08 \mathrm{g/mol} = 908 \mathrm{g} = 0.908 \mathrm{kg}\)

Key Concepts

Balanced Chemical Equations

Chemical equations describe the connection of reactants turning into products. In phosphorus removal from wastewater, balanced chemical equations are vital. They ensure the correct proportions of substances are used, avoiding waste or excess.

For example, when lime (\(\text{CaO}\)) reacts with water, it forms calcium ions (\(\text{Ca}^{2+}\)) and hydroxide ions (\(\text{OH}^{-}\)). This reaction is:
\[\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca}^{2+} + 2\text{OH}^{-}\]

Next, the \(\text{OH}^{-}\) ions convert \(\text{H}_2\text{PO}_4^{-}\) ions to \(\text{HPO}_4^{2-}\):
\[\text{H}_2\text{PO}_4^{-} + \text{OH}^{-} \rightarrow \text{HPO}_4^{2-} + \text{H}_2\text{O}\]

Then, \(\text{HPO}_4^{2-}\) becomes \(\text{PO}_4^{3-}\) using another \(\text{OH}^{-}\):
\[\text{HPO}_4^{2-} + \text{OH}^{-} \rightarrow \text{PO}_4^{3-} + \text{H}_2\text{O}\]

Finally, calcium and phosphate ions form a solid precipitate:
\[5\text{Ca}^{2+} + 3\text{PO}_4^{3-} + \text{OH}^{-} \rightarrow \text{Ca}_5(\text{PO}_4)_3\text{OH}(s)\]

Balancing these equations is crucial as it highlights the stoichiometric relationships, allowing us to predict the quantities needed or produced during the reactions.

Aqueous Ecosystems

Aqueous ecosystems are water-based environments like lakes, rivers, and oceans. These precious ecosystems rely on balanced nutrient levels to flourish.

Phosphorus, while necessary for aquatic plants, can become detrimental when in excess.

This nutrient often comes from human activities, particularly from sewage and detergents. When phosphorus levels rise, they can spur rapid algae growth, known as "algal blooms". These blooms consume oxygen, suffocating other aquatic organisms.

Maintaining a balance is key since it preserves:

• Healthy fish populations
• Water quality
• Biodiversity

Treating water to remove extra phosphorus mitigates these issues, helping sustain life in aqueous ecosystems.

Chemical Precipitation

Chemical precipitation is a method to remove dissolved ions from solutions by converting them into solid particles.

In the context of phosphorus removal, lime is added to create a chemical reaction. Here’s how it works:

• First, the water's pH is altered when lime releases \(\text{OH}^{-}\) ions.
• These \(\text{OH}^{-}\) ions convert the phosphate ions \(\text{H}_2\text{PO}_4^{-}\) and \(\text{HPO}_4^{2-}\) into \(\text{PO}_4^{3-}\).
• Next, \(\text{Ca}^{2+}\) ions from lime bind with \(\text{PO}_4^{3-}\) ions, forming the insoluble compound \(\text{Ca}_5(\text{PO}_4)_3\text{OH}(s)\).

This precipitate settles at the bottom of treatment tanks, separating phosphorus from the water.

Effectively, chemical precipitation is crucial in sewage treatment plants to control phosphorus levels, protecting the environment.