Question

The precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\left(K_{s p}=1.3 \times 10^{-33}\right)\) is sometimes used to purify water. (a) Estimate the \(\mathrm{pH}\) at which 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 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 precipitation of Al(OH)₃ will begin at a pH of 3.4. To achieve this pH, approximately 0.186 pounds of CaO must be added to the water.

Step-by-step solution

Calculate the concentration of Al³⁺ ions

To find the concentration of \(\mathrm{Al}^{3+}\) ions, we need to convert the amount of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in the water to moles of \(\mathrm{Al}^{3+}\) ions and then divide this value by the total volume of the water. First, we need to convert the mass of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (5.0 lb) to grams and then to moles. The molecular weight of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is 342.15 g/mol. 5.0 lb × 453.592 g/lb = 2267.96 g Now, we can convert the mass to moles: 2267.96 g × (1 mol / 342.15 g) = 6.63 mol of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) Since there are two \(\mathrm{Al}^{3+}\) ions per formula unit of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), we have: 6.63 mol × 2 = 13.26 mol of \(\mathrm{Al}^{3+}\) ions Next, we need to convert the volume of water from gallons to liters: 2000 gal × 3.78541 L/gal = 7570.82 L Now, we can find the concentration of \(\mathrm{Al}^{3+}\) ions in the water: (13.26 mol of \(\mathrm{Al}^{3+}\)) / (7570.82 L) = 1.75 x 10⁻³ mol/L

Calculate the concentration of OH⁻ ions and pH value

Now we can use the solubility product constant, \(K_{sp}\), to determine the concentration of \(\mathrm{OH}^-\) ions at which precipitation occurs. The formula for the solubility product constant of \(\mathrm{Al}(\mathrm{OH})_{3}\) is given by: \(K_{sp} = [\mathrm{Al}^{3+}] [\mathrm{OH}^-]^{3}\) We can rearrange this equation to solve for the concentration of \(\mathrm{OH}^-\) ions: \([\mathrm{OH}^-] = \sqrt[3]{ K_{sp} / [\mathrm{Al}^{3+}] }\) Now we can substitute the known values: \([\mathrm{OH}^-] = \sqrt[3]{ (1.3 \times 10^{-33}) / (1.75 \times 10^{-3}\,\mathrm{mol/L}) }\) \([\mathrm{OH}^-] = 2.47 \times 10^{-11}\,\mathrm{mol/L}\) Next, we can find the \(pOH\) by taking the negative base-10 logarithm of the hydroxide ion concentration: \(pOH = -\log ([\mathrm{OH}^-]) = -\log (2.47 \times 10^{-11}) = 10.6\) Finally, we can find the \(pH\) by using the relation: \(pH = 14 - pOH = 14 - 10.6 = 3.4\)

Calculate the required amount of CaO

To calculate the amount of \(\mathrm{CaO}\) needed to achieve a pH of 3.4, we first need to determine the concentration of \(\mathrm{H}^+\) ions that corresponds to this pH: \([\mathrm{H}^+] = 10^{-pH} = 10^{-3.4} = 3.98 \times 10^{-4}\,\mathrm{mol/L}\) Since 1 mol of \(\mathrm{CaO}\) reacts with 2 mol of \(\mathrm{H}^+\) ions to produce \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^-\) ions, we can determine the required amount of \(\mathrm{CaO}\) in moles: \(\mathrm{moles\, of\, CaO} = (\mathrm{moles\, of\, H^+}) / 2\) \(\mathrm{moles\, of\, CaO} = (3.98 \times 10^{-4}\,\mathrm{mol/L} \times 7570.82\,\mathrm{L}) / 2 = 1.51 \,\text{mol}\) Now we can convert the moles of \(\mathrm{CaO}\) to grams and then to pounds: Mass of \(\mathrm{CaO} = (1.51 \,\mathrm{mol}) \times (56.08\,\text{g/mol}) = 84.6 \mathrm{g}\) Mass of \(\mathrm{CaO}\) in pounds: \(84.6 \,\mathrm{g} \times (1 \,\mathrm{lb} / 453.592 \,\mathrm{g}) = 0.186 \,\mathrm{lb}\) So, approximately 0.186 pounds of \(\mathrm{CaO}\) must be added to the water to achieve a pH of 3.4.

Key concepts

Solubility Product Constant

The solubility product constant, often denoted as \(K_{sp}\), is a crucial concept in understanding precipitation reactions in chemistry. It represents the extent to which a compound can dissociate into its ions in a solution. The lower the \(K_{sp}\) value, the less soluble the compound is in water. For example, \(Al(OH)_3\), with its \(K_{sp}\) of \(1.3 \times 10^{-33}\), is a compound that barely dissolves in water. This is why even a small increase in \(Al^{3+}\) ion concentration can lead to precipitation.

The process begins by calculating the concentrations of the ions involved. When these concentrations are such that the product of ion concentrations equals the \(K_{sp}\), the solution becomes saturated, and any additional ions cause a precipitate to form. The stoichiometry of the precipitation reaction is essential, as it determines how many moles of each ion are involved and directly influences the \(K_{sp}\) expression. This principle is pivotal in understanding how to manipulate conditions to either promote or prevent precipitation.

pH Calculations

pH calculations are fundamental in determining the acidity or basicity of a solution. The relationship between pH, \(H^+\) ions, and \(OH^-\) ions is crucial in various chemical reactions, including precipitation reactions. The formula \(pH = 14 - pOH\) connects the concepts of pH and pOH and is handy when you're handling reactions involving hydroxide ions, like those with \(Al(OH)_3\).

To calculate the pH where precipitation occurs, you must first determine the concentration of \(OH^-\) ions that result in saturation via the \(K_{sp}\) expression. By finding \( [OH^-] \), you can subsequently find \(pOH\), and hence, the \(pH\) of the solution by subtracting \(pOH\) from 14. Understanding this interplay allows you to calculate how much of an acid or base must be added to achieve a specific pH level, which is crucial in controlling chemical reactions and processes.

Water Purification Methods

Water purification often involves the removal of impurities via chemical reactions, such as the precipitation of insoluble compounds. Using compounds like \(Al(OH)_3\) for water purification is effective because it forms a solid precipitate that can be easily removed. The addition of \(Al_2(SO_4)_3\) to water increases \(Al^{3+}\) ion concentration, triggering the precipitation of \(Al(OH)_3\) and subsequent removal of contaminants.

Additionally, controlling the pH is vital in these purification methods. By adjusting the pH, often achieved by adding substances like \(CaO\), you can influence the precipitation threshold and optimize the removal process. Understanding the chemistry behind pH adjustments and precipitation helps in designing efficient methods for treating water and ensuring it meets quality standards before consumption or use.