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

Feb 5, 2015

he solubility product constant would be equal to

${K}_{s p} = {\left[P {O}_{4}^{3 -}\right]}^{4} \cdot {\left[S {n}^{4 +}\right]}^{3}$

#### 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

$S {n}_{3} {\left(P {O}_{4}\right)}_{4} \left(s\right) r i g h t \le f t h a r p \infty n s 3 S {n}^{4 +} \left(a q\right) + 4 P {O}_{4}^{3 -} \left(a q\right)$

The solubility product constant would be equal to

${K}_{s p} = {\left[P {O}_{4}^{3 -}\right]}^{4} \cdot {\left[S {n}^{4 +}\right]}^{3}$

I wasn't able to find any reference for a numerical value, but insoluble compounds usually have ${K}_{s p}$ values that range from ${10}^{- 10}$ to ${10}^{- 50}$, so a conservative possible numerical value for stannic phosphate's solubility product constant would be

${K}_{s p} = 1.0 \cdot {10}^{- 30}$