Question

Solubility of the $HgC{{l}_{2}}$in a solvent is S moles/litre. Its solubility product will be:
A. $16{{S}^{2}}$
B. $8{{S}^{2}}$
C. $16{{S}^{4}}$
D. $4{{S}^{3}}$

Answer 1

Hint: When a solid salt is dissolved to give the constituent ions in solution, then in such a reaction, the solubility product is the equilibrium constant. It is usually represented by the symbol ${{K}_{sp}}$.

Complete answer:
- Basically, solubility products are used to find the concentration of one ion when the concentration of the other ion is known in a solution.
- If we take mercury chloride along with its saturated solution then equilibrium will be established between the undissolved ions and solids.
-We can write the equation for the ionization of $HgC{{l}_{2}}$ as:
     \[\begin{align}
  & H{{g}_{2}}C{{l}_{2}}\rightleftharpoons H{{g}_{2}}^{+}+2C{{l}^{-}} \\
 & \text{ S 2S} \\
\end{align}\]
-We will write the expression for the solubility product of mercury chloride as:
\[{{K}_{sp}}=\left[ H{{g}_{2}}^{2+} \right]{{\left[ C{{l}^{-}} \right]}^{2}}\]
\[\begin{align}
  & {{K}_{sp}}=\left[ S \right]{{\left[ 2S \right]}^{2}} \\
 & =4{{S}^{3}} \\
\end{align}\]
-This tells us that when mercury chloride is in equilibrium with its saturated solution, then the product of the concentrations of ions of both mercury and chloride is equal to the solubility product.

Hence, we can conclude that the correct option is (D), that is the solubility product of $HgC{{l}_{2}}$is $4{{S}^{3}}$.

Additional information:
- Solubility products have applications like in salting out of soap, and in purification of common salt etc.

Note:
- We should not get confused in the terms solubility and solubility product, as both terms are different.
- Solubility is the total amount of solute dissolved in the solvent at equilibrium, whereas solubility product is the equilibrium constant that tells us about the equilibrium associated between the solid solute and its ions that are dissociated in solution.