Question

If $ {{\text{K}}_{\text{sp}}} $ of $ \text{HgS}{{\text{O}}_{\text{4}}} $ is $ 6\cdot 4\times {{10}^{-5}} $ M, then solubility of $ HgS{{O}_{4}} $ in water will be-
(A) $ 6.4\times {{10}^{-5}} $ M
(B) $ 8\times {{10}^{-3}} $ M
(C) $ 8\times {{10}^{-4}} $ M
(D) $ 6.4\times {{10}^{-3}} $

Answer 1

Hint: $ HgS{{O}_{4}} $ is known as mercury sulfate. $ {{K}_{sp}} $ Of $ HgS{{O}_{4}} $ defines the constant for its solubility product. It gives the product of the solubility of the ions in mole per liter. To find the solubility of $ HgS{{O}_{4}} $ in water first we will write its ionizing equation and then we will take the product of the concentration/solubility of its ions and will equate them equal to the $ {{K}_{sp}} $ value.

Formula Used: $ {{\text{K}}_{\text{sp}}}\left( \text{HgS}{{\text{O}}_{4}} \right)=\left[ \text{H}{{\text{g}}^{2+}} \right]\left[ \text{SO}_{4}^{2-} \right] $

Complete Step By Step Solution
Given, $ {{\text{K}}_{\text{sp}}} $ of $ \text{HgS}{{\text{O}}_{\text{4}}} $ is $ 6.4\times {{10}^{-5}} $ M
The dissociation equation for $ HgS{{O}_{4}} $ is $ \text{HgS}{{\text{O}}_{\text{4}}}\rightleftharpoons \text{H}{{\text{g}}^{2+}}+\text{SO}_{4}^{2-} $
The solubility product constant will be $ {{\text{K}}_{\text{sp}}}\left( \text{HgS}{{\text{O}}_{4}} \right)=\left[ \text{H}{{\text{g}}^{2+}} \right]\left[ \text{SO}_{4}^{2-} \right] $
Let the solubility be’ since the coefficient present is 1 only in the above equation so one is raised to the power of both mercury ions and the sulfate ions. So, the solubility of both the ions will be‘s’ only. Therefore we get,
 $ {{\text{K}}_{\text{sp}}}=\text{s}\times \text{s} $
 $ 6.4\times {{10}^{-5}}={{\text{s}}^{2}} $
 $ \sqrt{6.4\times }{{10}^{-5}}=\text{s} $
 $ \Rightarrow \text{s}=\sqrt{0.64\times }{{10}^{-6}} $
  $ \text{s}=8\times {{10}^{-3}} $
Hence, the solubility is $ 8\times {{10}^{-3}}\text{M} $ .
So, the correct option is (B).

Additional Information
Solubility products represent the level at which a solute dissolves in the solution. Higher value of solubility product gives higher solubility of the solute. To find the $ {{\text{K}}_{\text{sp}}} $ value we require the concentration or the solubility of the ion products. If there are coefficients in front of the products they need to be raised to the power of the concentrations/solubility of the products. Factors affecting solubility are temperature, pressure and molecular size. Solubility increases with temperature due to an increase in kinetic energy which allows the solvent molecules to effectively break the solute molecules that are held together by intermolecular forces of attraction. Solute dissolves in solvents mainly of the same polarity.

Note
 $ \text{HgS}{{\text{O}}_{4}} $ is odorless liquid that forms white granules or crystalline powder. Its molar mass is $ 296\cdot 65 $ g per mole. It decomposes into water as yellow mercuric subsulfate and sulfuric acid. One should always remember to take the product of the concentrations or the solubility of the respective ions in order to find the solubility product constant. One should know the dissociation of various compounds into their respective ions and the charge on them.