Answer :
a. At pH ≈ 6.20, 95% of the phosphorus will be in the form of PO_4^3−.
b. [tex]5.57×10^−20 mg/L[/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]
For more such questions on phosphorus
https://brainly.com/question/30459371
#SPJ11