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 \(\mathrm{pH}\) ?

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

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: \(pH = 14 - pOH = 14 - 10.6 = 3.4\)

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.