How do you calculate the molar solubility of #AlPO_4#?

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How do you calculate the molar solubility of #AlPO_4#?

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Answer 1 · 2017-05-18T03:30:04

Molar Solubility of 1:1 ionization ratio = #sqrt(K_(sp)#

Explanation:

#AlPO_4 rightleftharpoons Al^(3+) + PO_4^(3-)#
#K_(sp) = [Al^(3+)][PO_4^(3-)] = (x)(x) = x^2#
=> #x# = Solubility = #sqrt(K_(sp))#
#K_(sp)(AlPO_4) = 6.3x10^(-19)#
=> Solubility = #sqrt(6.3x10^(-19))M = 7.94x10^(-10)M#

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How do you calculate the molar solubility of #AlPO_4#?

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