Question

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

Answer 1

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)`