# How do you calculate the molar solubility of AlPO_4?

Molar Solubility of 1:1 ionization ratio = sqrt(K_(sp)
$A l P {O}_{4} r i g h t \le f t h a r p \infty n s A {l}^{3 +} + P {O}_{4}^{3 -}$
${K}_{s p} = \left[A {l}^{3 +}\right] \left[P {O}_{4}^{3 -}\right] = \left(x\right) \left(x\right) = {x}^{2}$
=> $x$ = Solubility = $\sqrt{{K}_{s p}}$
${K}_{s p} \left(A l P {O}_{4}\right) = 6.3 x {10}^{- 19}$
=> Solubility = $\sqrt{6.3 x {10}^{- 19}} M = 7.94 x {10}^{- 10} M$