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Chapter 18: Problem 90

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 pH at which precipitation of \(\mathrm{Al}(\mathrm{OH})_{3}\) will begin if $5.0 \mathrm{~kg}\( of \)\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is added to 10,000 L of water. (b) Approximately how many pounds of CaO must be added to the water to achieve this \(\mathrm{pH}\) ?

Short Answer

Expert verified
To purify water using the precipitation of Al(OH)₃, and given 5.0 kg of Al₂(SO₄)₃ in 10,000 L of water, the pH at which precipitation will begin is approximately 3.34. To achieve this pH, approximately \(1.34 \times 10^{-8}\) pounds of CaO must be added to the water.

Step by step solution

01

Calculate moles of Al from Al2(SO4)3

First, let's determine the moles of Al present in 5.0 kg of Al2(SO4)3. We can do this by dividing the mass of the compound by its molar mass, and then multiplying by the number of moles of Al per mole of Al2(SO4)3. Molar mass of Al2(SO4)3 = (2 x 26.98) + (3 x (4 x 16.00 + 32.07)) = 342.15 g/mol Moles of Al2(SO4)3 = (5000 g) / (342.15 g/mol) = 14.61 mol Since there are 2 moles of Al for every mole of Al2(SO4)3, the moles of Al = 2 x 14.61 mol = 29.22 mol.
02

Determine the concentration of Al in water

Next, we need to determine the concentration of Al in the water. Divide the moles of Al by the volume of the water in liters to get the concentration. Concentration of Al = (29.22 mol) / (10000 L) = 0.002922 mol/L
03

Write the equation for dissociation of Al(OH)3 and Ksp expression

The chemical equation for the dissociation of Al(OH)3 is: \(\text{Al}(\text{OH})_3 \rightleftharpoons \text{Al}^{3+} + 3\text{OH}^-\) The solubility product constant, \(K_{sp}\), can be written as: \(K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3\)
04

Calculate the concentration of OH- ions for precipitation

Now, we can determine the concentration of OH- ions required for precipitation to begin. Since we have the value of \(K_{sp}\) and the concentration of Al, we can solve for the concentration of OH- ions. \(1.3 \times 10^{-33} = (0.002922)[\text{OH}^-]^3\) Solving for [OH-], we get: [OH-] = \(2.16 \times 10^{-11}\) M
05

Calculate the pH

To calculate the pH, we first need to find the pOH using the [OH-] concentration we just calculated. Then, we can calculate the pH using the relationship between pH and pOH: pOH = \(-\log(2.16 \times 10^{-11}) = 10.66\) pH = 14 - pOH = 14 - 10.66 = 3.34
06

Calculate moles of CaO required to reach this pH

Now, we need to determine the amount of CaO needed to achieve this pH. The reaction for CaO in water is: \(\text{CaO} + \text{H}_2\text{O} \rightarrow \text{Ca}^{2+} + 2\text{OH}^-\) To find the required moles of CaO, we need to find moles of OH- needed first. Moles of OH- = (Concentration of OH-) x (Volume of water) = \((2.16 \times 10^{-11}) \times 10000 = 2.16 \times 10^{-7}\) mol Since one mole of CaO produces two moles of OH- ions: Moles of CaO = \(\frac{2.16 \times 10^{-7}}{2} = 1.08 \times 10^{-7}\) mol
07

Convert moles of CaO to pounds

Finally, we need to convert the moles of CaO required to achieve this pH to pounds. We will use the molar mass of CaO and the conversion factor from grams to pounds. Molar mass of CaO = (40.08 + 16.00) g/mol = 56.08 g/mol Mass of CaO in grams = \((1.08 \times 10^{-7}) \times 56.08 = 6.06 \times 10^{-6}\) g Mass of CaO in pounds = \((6.06 \times 10^{-6}) \times \frac{1}{453.59} =\) \(1.34 \times 10^{-8}\) lbs

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