Question

The solubility of $A{{g}_{2}}S$ is:
$[{{K}_{sp}} = 256\times {{10}^{-6}}]$
A. $4\times {{10}^{-2}}$
B. $4\times {{10}^{-3}}$
C. $0.4\times {{10}^{-2}}$
D. $0.4\times {{10}^{-3}}$

Answer 1

As we know that solubility product constant is basically simplified equilibrium constant, which is denoted as ${{K}_{sp}}$. It shows the equilibrium between a solid and its respective ions in a given solution. It is found that the expression of ${{K}_{sp}}$ for a given salt is the product of the concentrations of the ions.

Complete Solution :
- First of all let us take the solubility of $A{{g}_{2}}S$ as x moles/L.
- We can write the reaction of $A{{g}_{2}}S$ as:

& A{{g}_{2}}S\rightleftarrows 2A{{g}^{+}}+{{S}^{2-}} \\\ & {{k}_{sp}}={{[A{{g}^{+}}]}^{2}}+{{[S]}^{2-}} \\\ & ={{\left( 2x \right)}^{2}}\left( x \right) \\\ & \therefore {{k}_{sp}} = 4{{x}^{3}} \\\ \end{aligned}$$ We are being provided with the value of $[{{K}_{sp}} = 256\times {{10}^{-6}}]$ Or we can write it as: 256 X ${{10}^{-6}}$ = $4{{x}^{3}}$ Or $\begin{aligned} & x=\sqrt[3]{\dfrac{256\times {{10}^{-6}}}{4}} \\\ & \therefore x = 4\times {{10}^{-2}} \\\ \end{aligned}$ **So, the correct answer is “Option A”.** **Additional information** - \- Let us discuss about the significance of solubility product in brief: \- It is found that the solvation enthalpy of ions is always negative and we can say that this means the energy is released during the process. \- We can see that when a salt is being dissolved in a solvent, then the strong forces of attraction of solute need to be overcome by interactions that were present in between ions and the solvent. **Note:** \- It is found that the greater the value of solubility product constant, the more soluble is the compound. \- As we know that solubility product constant can be affected by various factors such as temperature, molecular size, pressure. Its main importance is in determining the solubility, understanding the common ion effect and for predicting if a precipitate is formed.