What is the molar solubility of #AlPO_4#?

1 Answer
anor277
Sep 28, 2016

Low. Approx. #10^-8*g*L^-1#.

Explanation:

#AlPO_4(s) rightleftharpoons Al^(3+) + PO_4^(3-)#

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

#K_"sp"# #=# #9.84xx10^-21# #=# #[Al^(3+)][PO_4^(3-)]# #=# #S^2#

Now, since #[Al^(3+)]# #=# #[PO_4^(3-)]# #=# #S#, where #S# is the solubility of aluminum phosphate, all we have to do is to solve for #S# in the #K_"sp"# expression.

#S# #=# #sqrt(9.84xx10^-21)# #=# #9.92xx10^-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^-1# #=# #??g*L^-1#

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