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

1 Answer
Stefan V. · 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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