Question

What is the correct representation for the solubility product of $Sn{S_2}$?

Answer 1

We have to know that the equilibrium constant for a chemical reaction in which a solid ionic compound dissolves to yield its ions in solution is known as “solubility product”. It is denoted by ‘${K_{sp}}$’.

Complete step by step answer:
In simple words, we can define solubility products as equilibrium constant. The equilibrium that is between a solid and its respective ions present in the solution is called a solubility product constant.
It indicates the level at which a solute dissolves in solution. If the solubility of a substance is more, the value of ${K_{sp}}$ would be greater.
Let us consider the general dissolution reaction in aqueous solution.
$a{A_{\left( s \right)}} \rightleftharpoons c{C_{\left( {aq} \right)}} + d{D_{\left( {aq} \right)}}$
To know the value of ${K_{sp}}$ we have to take the required concentration of products and multiply them. If there are any coefficients present before any of the products, we have to necessarily raise the product to that coefficient power (and we can also multiply the concentration by that coefficient).
For the general reaction, we can write the ${K_{sp}}$ as,
${K_{sp}} = {\left[ C \right]^c}{\left[ D \right]^d}$
We should not write the reactant in the expression because solids are not included when calculating equilibrium constant expressions, because reactant concentrations do not change the expression; any change in their concentrations is insignificant, and hence omitted.
Now, coming to the question, we have to give the correct representation for the solubility product of $Sn{S_2}$.
$Sn{S_2}$ dissociates into $S{n^{4 + }}$ and ${S^{2 - }}$. We can write the dissociation reaction for $Sn{S_2}$ as,
$Sn{S_{2\left( s \right)}} \rightleftharpoons S{n^{4 + }} + 2{S^{2 - }}$
We have to omit the concentration of $Sn{S_2}$ in the equilibrium expression, as solids are not included. We can write the solubility product expression as,
${K_{sp}} = {\left[ {S{n^{4 + }}} \right]^1}{\left[ {{S^{2 - }}} \right]^2}$
Therefore, we can write the correct representation for the solubility product of $Sn{S_2}$ is ${K_{sp}} = \left[ {S{n^{4 + }}} \right]{\left[ {{S^{2 - }}} \right]^2}$.

Note:
We must know that the solubility product is also known as an “ion product”. The value of solubility products increases with an increase in temperature due increased solubility. Some of the factors that affect the value of solubility products are the common-ion effect, the diverse-ion effect and presence of ion-pairs.