Question

If ${{\text{K}}_{\text{sp}}}$ for $HgS{{O}_{4}}$ is $6.4\times {{10}^{-5}}$, then solubility of this substance in mole per ${{\text{m}}^{\text{3}}}$ is:
(A) $8\times {{10}^{-3}}$
(B) $6.4\times {{10}^{-5}}$
(C)$8\times {{10}^{-6}}$
(D) None of the above

Answer 1

The solubility product constant can be simply defined as the product of the solubility of the ions in the solute which gets ionised on dissolution. The value of solubility product constant represents the level at which a solute dissolves in solution.
- Here the question can be solved using the equation,${{K}_{sp}} = s\times s$

Complete Solution :
So in the question we are provided with the value of solubility product constant of $HgS{{O}_{4}}$and we have to find the solubility of this substance in mole per${{\text{m}}^{\text{3}}}$.
First let’s have an idea of how the solubility product constant and the solubility of the ions are related to each other.
- The solubility product ${{\text{K}}_{\text{sp}}}$ for a solid substance dissolved in water is its equilibrium constant. It gives us an idea of the level of dissociation of the solute in the solution. The more soluble a substance is, the higher the value of ${{\text{K}}_{\text{sp}}}$ it has.
- Or we can say that the ${{\text{K}}_{\text{sp}}}$ is the product of the solubility of the ions present in the solution.
For any general reversible reaction:
$\text{aA}\to \text{bB+cC}$
- We can write the equation of ${{\text{K}}_{\text{sp}}}$ as the products of the concentration of the ions present in the solution and if any coefficient is there related to the concentration of the ions then we write the coefficient as the power raised to the respective concentration.
- Hence we can write ${{\text{K}}_{\text{sp}}}$ for the above equation as, ${{K}_{sp}}={{\left[ B \right]}^{b}}{{\left[ C \right]}^{c}}$
Now we now that $HgS{{O}_{4}}$ dissolves in water and ionises as $H{{g}^{+2}}$ and $SO_{4}^{-2}$
We can write the equation as,$HgS{{O}_{4}}\to H{{g}^{+2}}+SO_{4}^{-2}$
The solubility product, ${{K}_{sp}}=\left[ H{{g}^{+2}} \right]\left[ SO_{4}^{-2} \right]$
We will take the solubility of each ion as s.
Then, ${{K}_{sp}}=s\times s={{s}^{2}}$
We know, ${{K}_{sp}}=6.4\times {{10}^{-5}}$
Substitute the above value in the equation.
We get, ${{K}_{sp}}={{s}^{2}}=6.4\times {{10}^{-5}}$
$s=\sqrt{6.4\times {{10}^{-5}}}$
$\text{s=8 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{mol/L}$
$\text{s=8 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{-3}}}\times \text{1}{{\text{0}}^{3}}\text{mol/}{{\text{m}}^{3}}$
$\text{s=8mol/}{{\text{m}}^{\text{3}}}$
Hence we got the solubility as,$\text{s=8mol/}{{\text{m}}^{\text{3}}}$
So, the correct answer is “Option D”.

Note: Ensure that the solids do not appear in the equilibrium equation of solubility product. Since the active mass of solids is unity (1). Sometimes, a student can mistakenly take their concentration into consideration.