AlPO_4(s) rightleftharpoons Al^(3+) + PO_4^(3-)AlPO4(s)⇌Al3++PO3−4
Now, this equilibrium is governed, i.e. quantified by the K_"sp"Ksp expression, so that:
K_"sp"Ksp == 9.84xx10^-219.84×10−21 == [Al^(3+)][PO_4^(3-)][Al3+][PO3−4] == S^2S2
Now, since [Al^(3+)][Al3+] == [PO_4^(3-)][PO3−4] == SS, where SS is the solubility of aluminum phosphate, all we have to do is to solve for SS in the K_"sp"Ksp expression.
SS == sqrt(9.84xx10^-21)√9.84×10−21 == 9.92xx10^-11*mol*L^-19.92×10−11⋅mol⋅L−1.
And we can get the solubility in grams, by forming the product:
9.92xx10^-11*mol*L^-1xx121.95*g*mol^-19.92×10−11⋅mol⋅L−1×121.95⋅g⋅mol−1 == ??g*L^-1??g⋅L−1
