Phosphate ([tex]PO_4^{3−}[/tex]) is a pollutant from inadequately treated municipal sewage and excess fertilizer usage. Calcium is frequently used to precipitate phosphate in the following reactions:

\[ \text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3−} \]
\[ K_s = 1 \times 10^{−27} \]

\[ \text{HPO}_4^{2−} \rightleftharpoons \text{H}^+ + \text{PO}_4^{3−} \]
\[ K_{a3} = 3.98 \times 10^{−13} \]

Phosphorus needs to be in the form of [tex]PO_4^{3−}[/tex] to precipitate with calcium.

(a) Determine the pH where 95% of the phosphorus in the second equilibrium equation is in the form of [tex]PO_4^{3−}[/tex].

(b) If the phosphate concentration ([tex]PO_4^{3−}[/tex]) is measured to be 8.5 mg/L at the pH from the previous step, what concentration of calcium ([tex]Ca^{2+}[/tex]) would be required in mg/L to remove 95% of the phosphate?

(a) To determine the pH where 95% of the phosphorus in the second equilibrium equation is in the form of PO_4^3−, we need to calculate the concentration of HPO_4^2− and PO_4^3− at equilibrium for various pH values.

The dissociation reaction is: [tex]HPO_4^2− ⇌ H^+ + PO_4^3−[/tex]

We are given the equilibrium constant Ka_3 for this reaction, which is 3.98×10^−13. Since Ka_3 = [H^+][PO_4^3−] / [HPO_4^2−], we can use this equation to calculate the concentration of [PO_4^3−] at different pH values.

At equilibrium, if we assume [HPO_4^2−] = [PO_4^3−] (since both are produced in a 1:1 ratio), we can solve for [PO_4^3−]:

[tex]Ka_3 = [H^+][PO_4^3−] / [PO_4^3−][/tex]

3.98×10^−13 = [H^+]

[tex][H^+] = 6.31×10^−7 M[/tex]

To calculate the pH, we use the equation pH = -log[H^+]:

[tex]pH = -log(6.31×10^−7) ≈ 6.20[/tex]

So, at pH ≈ 6.20, 95% of the phosphorus will be in the form of PO_4^3−.

(b) To remove 95% of the phosphate, you want most of it to be in the form of [tex]PO_4^3−.[/tex] Given that the phosphate concentration ([PO_4^3−]) is 8.5 mg/L at pH 6.20, you can calculate the required calcium concentration ([Ca^2+]) using the solubility product constant K_s:

[tex]K_s = [Ca^2+][PO_4^3−]^2[/tex]

[tex]1×10^−27 = Ca^2+^2[/tex]

Solving for [Ca^2+]:

[tex][Ca^2+] = (1×10^−27) / (8.5×10^−3)^2 ≈ 1.39×10^−23 M[/tex]

Convert this to mg/L:

[tex][Ca^2+] = (1.39×10^−23) * 40.08 g/mol * 1000 mg/g ≈ 5.57×10^−20 mg/L[/tex]

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