Question

The precipitation of \(\mathrm{A} \mathrm{L}(\mathrm{OH})_{3}\left(K_{\text {sp }}=1.3 \times 10^{-33}\right.\) ) is sometimes used to purify water. (a) Estimate the \(\mathrm{pH}\) at wh?ch precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) will begin if \(5.0 \mathrm{lb}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is added to \(2000 \mathrm{gal}\) of water. (b) Approximately how many pounds of \(\mathrm{CaO}\) must be added to the water to achieve this \(\mathrm{pH}\) ?

Short answer

The approximate pH at which precipitation of Al(OH)₃ will begin is \(3.29\). To achieve this pH, approximately \(0.4808\) pounds of CaO must be added to the water.

Step-by-step solution

Convert the given amount of Al2(SO4)3 to moles

First, we have to convert the given amount of Al2(SO4)3 (5.0 lbs) to moles. The molar mass of Al2(SO4)3 is 2 * 27 (Al) + 3 * (4 * 16 + 1 * 32) (SO4) = 54 + 3 * (64 + 32) = 54 + 3 * 96 = 342 g/mol. So, moles of Al2(SO4)3 = 5.0 lbs * (1 kg / 2.205 lbs) * (1000 g / 1 kg) * (1 mol / 342 g) = 32.54 mol

Determine the concentration of Al3+ ions

Given the volume of water is 2000 gallons, we need to convert it to liters: (2000 gallons) * (3.78541 liters / 1 gallon) = 7570.82 L. Since one mole of Al2(SO4)3 produces 2 moles of Al3+ ions, we have 32.54 mol * 2 = 65.08 mol of Al3+ ions. Now, we can find the concentration of Al3+ ions in water: [Al3+] = (65.08 mol) / (7570.82 L) = 0.0086 M

Calculate the concentration of OH- ions

Using the Ksp expression for Al(OH)3: Ksp = [Al3+][OH-]^3 We can solve for [OH-]: [OH-] = \(\sqrt[3]{\frac{K_{sp}}{[Al^{3+}]}}\) Plugging in the values: [OH-] = \(\sqrt[3]{\frac{1.3 \times 10^{-33}}{0.0086}} = 1.955 \times 10^{-11}\)

Determine the pH of the solution

Since we have the concentration of OH- ions, we can determine the pOH and then the pH of the solution. pOH = -log[OH-] = -log(1.955 * 10^{-11}) = 10.71 Since pH + pOH = 14, we can determine the pH as follows: pH = 14 - 10.71 = 3.29

Calculate the OH- concentration needed

Using the pH obtained in step 4, we can calculate the concentration of H+ ions needed to achieve this pH: [H+] = 10^{-pH} = 10^{-3.29} = 5.15 * 10^{-4} M

Calculate the moles of OH- ions given off by CaO

Since one mole of CaO reacts with one mole of CO2 to produce one mole of OH- ions, the moles of OH- ions needed to achieve this pH is the same as the moles of CO2. Moles of OH- = (5.15 * 10^{-4} M) * 7570.82 L = 3.8962 mol

Determine the amount of CaO in pounds

Now we need to convert the moles of CaO to pounds. The molar mass of CaO is 40 (Ca) + 16 (O) = 56 g/mol. So, the weight of CaO in grams is: 3.8962 mol * 56 g/mol = 218.1872 g Finally, we convert grams to pounds: 218.1872 g * (1 kg / 1000 g) * (2.205 lbs / 1 kg) = 0.4808 lbs Thus, approximately 0.4808 pounds of CaO must be added to the water to achieve the desired pH of 3.29.

Key concepts

Solubility Product (Ksp)

Understanding the solubility product, or Ksp, is crucial when exploring precipitation reactions. The Ksp value is specific to each substance and refers to the level at which a solid dissolves in water to form a saturated solution. For example, the Ksp for aluminum hydroxide, Al(OH)₃, is a very small number, 1.3 × 10^{-33}, suggesting that it is not highly soluble in water.

• The smaller the Ksp, the less soluble the compound.
• Ksp helps predict whether a precipitate will form under given conditions.

To calculate when a precipitate forms, compare the ion product of the solution with Ksp. If the ion product exceeds Ksp, the products form a solid precipitate. In this exercise, the calculation of [OH^-] for Al(OH)₃ underscored how saturation depends on both the presence of ions and how they interact with the Ksp threshold.

pH Calculation

pH is a scale used to determine the acidity or basicity of a solution. The relationship between hydrogen ions (H⁺) and hydroxide ions (OH⁻) in water affects the pH value.

In this example, the concentration of OH⁻ is determined first through calculations involving Al(OH)₃ and its solubility product.

Here's a simple explanation for calculating pH:

• Determine the concentration of OH⁻ as shown by Ksp equations.
• Calculate pOH using the formula pOH = -log[OH⁻].
• Use the pH and pOH relationship (pH + pOH = 14) to find the pH.

For this problem,

starting from a OH⁻ concentration of 1.955 × 10^{-11}, the pOH was 10.71, leading to a pH value of 3.29.

Molarity and Concentration

Molarity, denoted as M, indicates the concentration of a solute in a solution, expressed as moles per liter (mol/L). Understanding molarity is essential to performing calculations related to concentration and volume in chemistry problems.

• The exercise starts by converting aluminum sulfate mass to moles and then calculating the molarity of Al³⁺ in water.
• Molarity is calculated by dividing the number of moles of solute by the volume of solution in liters.

In this scenario, students first converted mass in pounds to moles and then used the large volume of water (2000 gallons) to understand how much solute contributes to the molarity, showing that the calculated Al³⁺ concentration is important to use in further reactions.

Water Purification

Water purification often utilizes chemical reactions, such as precipitation, to remove impurities. In this exercise, adding Al(OH)₃ is a purification technique meant to capture contaminants in large bodies of water.

Key points about the purification reaction:

• Al(OH)₃ forms when aluminum ions react with hydroxide ions, removing contaminants as it precipitates.
• The process requires precise conditions using Ksp values and pH considerations to trigger precipitation.
• The resultant solid can be filtered out, reducing impurities and yielding cleaner water.

By adjusting the pH of the water, the formation of Al(OH)₃ is controlled and optimized for practical application in water treatment, showcasing how chemistry enables us to maintain water quality.