What is the molar solubility of $A l P O_{4}$ $AlPO_4$ ?

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What is the molar solubility of $A l P O_{4}$ $AlPO_4$ ?

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Answer 1 · 2016-09-28T20:32:33

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

Explanation:

$A l P O_{4} (s) ⇌ A l^{3 +} + P O_{4}^{3 -}$ $AlPO_4(s) rightleftharpoons Al^(3+) + PO_4^(3-)$

Now, this equilibrium is governed, i.e. quantified by the $K_{sp}$ $K_"sp"$ expression, so that:

$K_{sp}$ $K_"sp"$ $=$ $=$ $9.84 \times 10^{- 21}$ $9.84xx10^-21$ $=$ $=$ $[A l^{3 +}] [P O_{4}^{3 -}]$ $[Al^(3+)][PO_4^(3-)]$ $=$ $=$ $S^{2}$ $S^2$

Now, since $[A l^{3 +}]$ $[Al^(3+)]$ $=$ $=$ $[P O_{4}^{3 -}]$ $[PO_4^(3-)]$ $=$ $=$ $S$ $S$ , where $S$ $S$ is the solubility of aluminum phosphate, all we have to do is to solve for $S$ $S$ in the $K_{sp}$ $K_"sp"$ expression.

$S$ $S$ $=$ $=$ $\sqrt{9.84 \times 10^{- 21}}$ $sqrt(9.84xx10^-21)$ $=$ $=$ $9.92 \times 10^{- 11} \cdot m o l \cdot L^{- 1}$ $9.92xx10^-11*mol*L^-1$ .

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

$9.92 \times 10^{- 11} \cdot m o l \cdot L^{- 1} \times 121.95 \cdot g \cdot m o l^{- 1}$ $9.92xx10^-11*mol*L^-1xx121.95*g*mol^-1$ $=$ $=$ $? ? g \cdot L^{- 1}$ $??g*L^-1$

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What is the molar solubility of $A l P O_{4}$ $AlPO_4$ ?

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