# What is the molar solubility of AlPO_4?

Sep 28, 2016

Low. Approx. ${10}^{-} 8 \cdot g \cdot {L}^{-} 1$.

#### Explanation:

$A l P {O}_{4} \left(s\right) r i g h t \le f t h a r p \infty n s A {l}^{3 +} + P {O}_{4}^{3 -}$

Now, this equilibrium is governed, i.e. quantified by the ${K}_{\text{sp}}$ expression, so that:

${K}_{\text{sp}}$ $=$ $9.84 \times {10}^{-} 21$ $=$ $\left[A {l}^{3 +}\right] \left[P {O}_{4}^{3 -}\right]$ $=$ ${S}^{2}$

Now, since $\left[A {l}^{3 +}\right]$ $=$ $\left[P {O}_{4}^{3 -}\right]$ $=$ $S$, where $S$ is the solubility of aluminum phosphate, all we have to do is to solve for $S$ in the ${K}_{\text{sp}}$ expression.

$S$ $=$ $\sqrt{9.84 \times {10}^{-} 21}$ $=$ $9.92 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$.

And we can get the solubility in grams, by forming the product:

$9.92 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1 \times 121.95 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g*L^-1