Problem 70 - Chemistry: An Atoms-Focused Approach

Question

Eutrophication, the rapid growth of algae and the death of fish, may be caused by the presence of an excess of phosphates in water. Treatment plants that process sewage may add \(\mathrm{Ca}(\mathrm{OH})_{2}\) (slaked lime) to water to remove phosphates before returning the water to the environment. Although the phosphates may be present in several forms, we can use HPO \(_{4}^{-}\) as a representative phosphate in the net ionic equation: \(5 \mathrm{Ca}^{2+}(a q)+3 \mathrm{HPO}_{4}^{-}(a q)+4 \mathrm{OH}^{-}(a q) \rightarrow\) $$ \mathrm{Ca}_{5} \mathrm{OH}\left(\mathrm{PO}_{4}\right)_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ If phosphates (as HPO \(_{4}^{-}\) ) are present at a level of \(15.7 \mathrm{mg} / \mathrm{L}\) in wastewater, how much \(\mathrm{Ca}(\mathrm{OH})_{2}\) would need to be added to \(1.00 \times 10^{5} \mathrm{L}\) of water to precipitate \(95 \%\) of the phosphate ion present?

Step-by-Step Solution

Answer

Step 1: Calculate the amount of phosphate ions in 1.00 x 10^5 L water moles of HPO₄⁻ = (15.7 mg/L * 1.00 x 10^5 L) / (97.993 g/mol) * (1 g / 1000 mg) Step 2: Determine the amount of phosphate ions to precipitate x = 0.95 * (moles of HPO₄⁻) Step 3: Calculate the required moles of calcium hydroxide moles of Ca(OH)₂ = (5/3) * x Step 4: Convert moles of calcium hydroxide to mass mass of Ca(OH)₂ = (moles of Ca(OH)₂) * (74.092 g/mol) Calculate the values to find the mass of calcium hydroxide required.

Step 1: Calculate the amount of phosphate ions in 1.00 x 10^5 L water

First, we need to find the number of moles of HPO\(_{4}^{-}\) present in \(1.00 \times 10^{5} \mathrm{L}\). Since concentration is given in mg/L, we will first convert this to moles. The molar mass of HPO\(_{4}^{-}\) is \(1\cdot(1+15.999)+1\cdot(30.974)+4\cdot(15.999)=97.993\,\text{g/mol}\). $$ \text{moles of HPO}_4^- = \frac{15.7\,\text{mg/L} \cdot 1.00 \times 10^5\,\text{L}}{97.993\,\text{g/mol}} \cdot \frac{1\,\text{g}}{1000\,\text{mg}} $$

Step 2: Determine the amount of phosphate ions to precipitate

Next, we find the quantity of phosphate ions required to be precipitated by considering \(95 \%\) of the total phosphate ions. Let's denote this amount as \(x\). $$ x = 0.95 \times (\text{moles of HPO}_4^-) $$

Step 3: Calculate the required moles of calcium hydroxide

Now, we will use the stoichiometry of the net ionic equation to find the required moles of calcium hydroxide (\(\mathrm{Ca}(\mathrm{OH})_{2}\)) needed to precipitate \(x\) moles of HPO\(_{4}^{-}\). From the given equation, we see: $$ 5\,\text{moles of Ca}^{2+} \equiv 3\,\text{moles of HPO}_4^- $$ Now, we need moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\), so we use: $$ \text{moles of Ca(OH)}_2 \equiv \text{moles of Ca}^{2+} $$ Thus, we find moles of \(\mathrm{Ca}(\mathrm{OH})_{2}\) needed to precipitate \(x\) moles of HPO\(_{4}^{-}\): $$ \text{moles of Ca(OH)}_2 = \frac{5}{3} \times x $$

Step 4: Convert moles of calcium hydroxide to mass

Lastly, we need to convert the moles of calcium hydroxide (\(\mathrm{Ca}(\mathrm{OH})_{2}\)) to mass. The molar mass of \(\mathrm{Ca}(\mathrm{OH})_{2}\) is \(1\cdot(40.078)+2\cdot(15.999)+2\cdot(1.008)=74.092\,\text{g/mol}\). $$ \text{mass of Ca(OH)}_2 = (\text{moles of Ca(OH)}_2) \times (74.092\,\text{g/mol}) $$ Now, calculate the values and obtain the mass of calcium hydroxide required.

Key Concepts

Phosphates in Water

Phosphates are compounds containing the phosphate ion, HPO t 4 − , which is a salt or ester of phosphoric acid. These substances are vital for all life forms as they are a key component of DNA, proteins, and cell membranes. However, when they are present in excess in water bodies, they lead to a phenomenon called eutrophication.
Eutrophication is the rapid growth of algae due to excessive nutrients, primarily phosphates and nitrates. This algae bloom blocks sunlight, disrupting aquatic ecosystems and leading to the depletion of oxygen when the algae dies and decomposes.
This process is detrimental to water quality, causing the death of fish and other aquatic organisms. Therefore, controlling phosphate levels in water is crucial to maintaining a balanced aquatic environment. This is particularly important in lakes, rivers, and coastal areas where wastewater can lead to high nutrient concentrations.

Calcium Hydroxide

Calcium hydroxide, commonly known as slaked lime, has the chemical formula Ca(OH) 2 . It is a colorless crystal or white powder obtained by reacting calcium oxide with water. It is used in a variety of industrial applications including construction and wastewater treatment.
In water treatment, calcium hydroxide is utilized to remove phosphates from wastewater. When added to water, it reacts with phosphate ions to form insoluble compounds that precipitate out of the solution. This process reduces the concentration of phosphates, making the water safer for release into the environment.
The role of calcium hydroxide in this process not only supports environmental protection but also ensures compliance with environmental regulations. Furthermore, it is an effective and economical method for controlling water pollution and preventing eutrophication.

Wastewater Treatment

Wastewater treatment is a process of removing contaminants from water before it is returned to the environment. This process is essential for protecting public health and the environment by minimizing water pollution.
Wastewater treatment plants commonly use chemicals such as calcium hydroxide to remove or neutralize harmful substances like phosphates. The treatment typically involves several stages: primary treatment, which removes large particles; secondary treatment, which biologically removes dissolved and suspended organic matter; and tertiary treatment, which involves chemical treatment for nutrient removal.
Control of phosphates is a critical aspect of wastewater management. The precipitation of phosphates using calcium hydroxide is an effective solution that makes use of chemical reactions to convert dissolved phosphates into solid compounds, thereby removing them from the water. This ensures the water is safe for subsequent use or release into natural water bodies.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Understanding stoichiometry is crucial for determining the correct proportions of substances involved in reactions.
In the context of wastewater treatment, stoichiometry helps in calculating the precise amounts of calcium hydroxide needed to react with phosphates in water. From the net ionic equation provided, we know that 5 moles of calcium ions react with 3 moles of phosphate ions to form the precipitate. This relationship is vital for optimizing the chemical reactions and ensuring the efficient use of resources.
Stoichiometry allows for the adjustment of the quantities used in the reaction to achieve the desired level of phosphate removal, thereby making chemical treatment processes more efficient and cost-effective.

Net Ionic Equation

A net ionic equation simplifies a reaction by showing only the species that actually participate in the chemical change. It's a tool chemists use to focus on the core chemistry of a reaction, stripping away the spectators.
For instance, in the phosphate removal process in wastewater treatment, the net ionic equation focuses on the key reactants: calcium ions, phosphate ions, and hydroxide ions. It shows how they produce calcium phosphate and water, highlighting the actual chemical transformation.
By using net ionic equations, you can better understand the reaction mechanisms and the chemical logic behind wastewater treatment processes. This simplified view helps in comprehending how and why certain reactants convert into products, aiding in the effective management and optimization of chemical treatments.