What is the expression of the solubility product constant of tin(IV) phosphate?

1 Answer
Stefan V. ยท Dwayne M.
Feb 5, 2015

he solubility product constant would be equal to

K_(sp) = [PO_4^(3-)]^(4) * [Sn^(4+)]^(3)

Explanation:

TIn (IV) phosphate, or stannic phosphate, is an ionic compound that has "Sn"_3("PO"_4)_4 as its formula unit.

Now, according to the solubility rules (more here: http://www.ausetute.com.au/solrules.html), all phosphates are insoluble with the exception of compounds formed with group 1 cations and ammonium.

Since this implies that "Sn"_3("PO"_4)_4 is insoluble in aqueous solution, you can expect its solubility product constant to be very, very small. The equation for stannic phosphate's dissociation in aqueous solution is

Sn_3(PO_4)_4(s) rightleftharpoons 3Sn^(4+)(aq) + 4PO_4^(3-)(aq)

The solubility product constant would be equal to

K_(sp) = [PO_4^(3-)]^(4) * [Sn^(4+)]^(3)

I wasn't able to find any reference for a numerical value, but insoluble compounds usually have K_(sp) values that range from 10^(-10) to 10^(-50), so a conservative possible numerical value for stannic phosphate's solubility product constant would be

K_(sp) = 1.0 * 10^(-30)

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