Question

Solubility Problem:

How many mg/L PO₄³⁺ would be in solution at equilibrium with AlPO₄?

The solubility product is K_sp = 10^-20.0 and relationship is

AlPO₄ <--> Al³⁺ + PO₄^3-

X_aY_b = aXⁿ⁺ + bY^m- and K_sp = [Xⁿ⁺]^a[Y^m-]^b

1. Determine a & b, a = ?, b = ?

2. Determine the Molarity of the reaction product [PO₄³⁺] :

3. Convert Moles PO₄³⁺ to mg/L:

Answer 1

AlPO₄ <--> Al³⁺ + PO₄^3-

Solubility product , Ksp = [Al³⁺ ][ PO₄^3- ]

Where a = 1 & b = 1

Given Ksp = 10^-20

Let S ( mol/L) be the solubility of AlPO₄

Then [Al³⁺ ]= S & [ PO₄^3- ] = S

10^-20 = S x S

S2 = 10^-20

S = 10^-10 mol/L

Therefore the molarity of [ PO₄^3- ] = S = 10^-10 mol/L

The molar mass of AlPO₄ = 122 g/mol

So S = ( mass/ molar mass) / Volume in L

10^-10 mol/L = ( mass / 122(g/mol)) / 1 L

mass = 1.22x10^-8 g

         = 1.22x10^-5 mg

So the concentration is 1.22x10^-5 mg /L