Question

What is the expression of the solubility product constant of tin $ \left( {IV} \right) $ phosphate?

Answer 1

Hint: Tin $ \left( {IV} \right) $ phosphate is a chemical compound with the molecular formula of $ S{n_3}{\left( {P{O_4}} \right)_4} $ . It can dissociate into ions in presence of water as $ S{n^{4 + }} $ and $ P{O_4}^{3 - } $ . The solubility product of this compound can be written as the product of concentration of these two ions.

Complete answer:
Tin is a chemical element with a chemical symbol Sn which can lose electrons and acts as cation, phosphate is a group that can act as an anion. Both these involve bond formation and form a compound named as tin $ \left( {IV} \right) $ phosphate.
As the aqueous solution of tin $ \left( {IV} \right) $ phosphate can dissociate into ions, it can be considered as an ionic salt.
The chemical dissociation of tin $ \left( {IV} \right) $ phosphate will be as follows:
 $ S{n_3}{\left( {P{O_4}} \right)_4} \rightleftharpoons 3S{n^{4 + }} + 4P{O_4}^{3 - } $
The solubility product can be represented by $ {K_{sp}} $ , it can be defined as the equilibrium constant of a dissociation of a solid substance into an aqueous solution. It can be written as the product of concentrations of the cation and anion represented in the above dissociation chemical equation.
The expression of the solubility product constant of tin $ \left( {IV} \right) $ phosphate can be written as
 $ {K_{sp}} = {\left[ {S{n^{4 + }}} \right]^3}{\left[ {P{O_4}^{3 - }} \right]^4} $
The number of moles of ions present should be written as the power of concentration of ions.

Note:
The number of electrons lost by a tin atom must be equal to the number of electrons gained by the phosphate group. Thus, the number of moles of tin and phosphate groups should be written according to the number of electrons involved. Here, the number of electrons involved are $ 12 $ .