Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

Step by step solution

01

Writing the balanced chemical equations

The four reactions described in the exercise are: \n1. \(\mathrm{CaO} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-}\), \n2. \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2} \mathrm{O}\), \n3. \(\mathrm{HPO}_{4}^{2-} + \mathrm{OH}^{-} \rightarrow \mathrm{PO}_{4}^{3-} + \mathrm{H}_{2} \mathrm{O}\), \n4. 5\(\mathrm{Ca}^{2+} + 3\mathrm{PO}_{4}^{3-} + \mathrm{OH}^{-} \rightarrow \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\)

02

Convert mass of phosphorus to moles

The total amount of phosphorus in the water is \(10.0 \mathrm{mg}/\mathrm{L} \times 1.00 \times 10^{4} \mathrm{L} = 1.00 \times 10^{5} \mathrm{mg} = 100 \mathrm{g}\). In order to go further, we need to convert this mass into moles using Phosphorus molar mass (31.0 g/mol). Hence, \(100 \mathrm{g} \times \left( \frac{1 \mathrm{mol}}{31.0 \mathrm{g}} \right) = 3.23 \mathrm{mol}\) of phosphorus.

03

Determine moles of \(CaO\) required

From reaction 4, it is seen that each mole of phosphorus (in the form of \(PO_{4}^{3-}\)) requires 5 moles of \(Ca^{2+}\) for precipitation. Since \(Ca^{2+}\) ions come from \(CaO\) (Reaction 1: 1 mole of \(CaO\) gives 1 mole of \(Ca^{2+}\) ion), we require 5 moles of \(CaO\) for each mole of phosphorus. Hence, \(3.23 \mathrm{mol} \times 5 = 16.2 \mathrm{mol}\) of \(CaO\) are required.

04

Convert moles of \(CaO\) to mass

Now we convert the moles of \(CaO\) into kilograms using the molar mass of \(CaO\) which is 56.08 g/mol. So, \(16.2 \mathrm{mol} \times 56.08 \mathrm{g/mol} = 908 \mathrm{g} = 0.908 \mathrm{kg}\)

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