Question

A water contains 1 mg/L of PO43, how many mg/L of AlB+ would be needed to lower the P043 concentration to 10-s mg/L? What is the final concentration of Al at equilibrium? Known: AlPO4 (s)AlPO3 pKs 20 *Assume this is the only relevant reaction

Answer 1

10^-5 mg PO₄^3- = (10^-5 X 10^-3/94.97) moles = 1.053X 10^-10 moles  PO43-

1mg PO₄^3- = (10^-3 /94.97) moles = 1.053X 10^-5 moles PO₄^3-

1mole Al³⁺ reacts with 1 mole PO₄^3-

Let x moles Al³⁺ reacts with x mole  PO₄^3-

(1.053X 10^-5 -x) = 1.053X 10^-10

x= 1.053X 10^-5 moles Al³⁺ = 28.4099x 10^-5 g Al³⁺ = 0.284 mg of Al³⁺

So, the Al³⁺ required is 0.284 mg/L.

Let the equlibrium concentration is 'S' moles of Al³⁺,[Al³⁺] = [ PO₄^3-] = S

Solubility product K_sp= 10^-20 = [Al³⁺] [ PO₄^3-] = S²

S= 10^-10 moles of Al³⁺ remain at equilibrium.