Question

How do you calculate the molar solubility of `AlPO_4`?

Answer 1

Molar Solubility of 1:1 ionization ratio = `sqrt(K_(sp)`

Explanation:

`AlPO_4 rightleftharpoons Al^(3+) + PO_4^(3-)`
`K_(sp) = [Al^(3+)][PO_4^(3-)] = (x)(x) = x^2`
=> `x` = Solubility = `sqrt(K_(sp))`
`K_(sp)(AlPO_4) = 6.3x10^(-19)`
=> Solubility = `sqrt(6.3x10^(-19))M = 7.94x10^(-10)M`