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Question: If \(Ksp\)​ for \(HgS{O_4}\) is \(6.4 \times {10^{ - 5}}\) , then solubility of this substance in mo...

If KspKsp​ for HgSO4HgS{O_4} is 6.4×10−56.4 \times {10^{ - 5}} , then solubility of this substance in mole per m3{m^3} is:
Hg=200,S=32,O=16Hg = 200,S = 32,O = 16
A. 8×10−38 \times {10^{ - 3}}
B. 8×10−68 \times {10^{ - 6}}
C. 296296
D. 23682368

Explanation

Solution

For a compound like HgSO4HgS{O_4} , the ratio of Hg+2:SO4−2H{g^{ + 2}}:S{O_4}^{ - 2} is 1:11:1 . That is for every mercury ion there is one sulphate ion. This means that the solubility product for the compound is the square of the solubility that is s2{s^2} where ss is the solubility.

Formula used: KspKsp= s2{s^2}
Where KspKsp is the solubility constant and ss is the solubility.

Complete step by step answer:
In this question, we have two ions, there is Hg+2H{g^{ + 2}}and SO4−2S{O_4}^{ - 2} . Therefore, the reaction of dissociation will look like this:
HgSO4⇄Hg+2+SO4−2HgS{O_4} \rightleftarrows H{g^{ + 2}} + S{O_4}^{ - 2}
now we can consider the concentration of Hg+2H{g^{ + 2}}to be ss and the concentration of SO4−2S{O_4}^{ - 2} to be equal to Hg+2H{g^{ + 2}} .
The equilibrium constant can be represented in the following way:
Ksp=[Hg+2][SO4−2]Ksp = \left[ {H{g^{ + 2}}} \right]\left[ {S{O_4}^{ - 2}} \right]
Concentration of mercury and sulphate ions is represented in the manner shown above.
Ksp=[s][s]Ksp = \left[ s \right]\left[ s \right]
Ksp=[s]2Ksp = {\left[ s \right]^2}
For this reason, the equilibrium constant which is called a solubility product is the product of both the ions to form s2{s^2} . The solubility of a solute can be defined as the amount of a solute that can be dissolved in a given amount of solvent at a particular temperature at a state of equilibrium.
According to the question, we have KspKsp to be 6.4×10−56.4 \times {10^{ - 5}} . This can be used to calculate the ss or solubility using the above formula.
6.4×10−5=s26.4 \times {10^{ - 5}} = {s^2}
Ksp=s2Ksp = {s^2}
64×10−6=s264 \times {10^{ - 6}} = {s^2}
64×10−6=s\sqrt {64 \times {{10}^{ - 6}}} = s
8×10−3=s8 \times {10^{ - 3}} = s

So, the correct answer is Option A.

Note: It is important to remember the relation between the solubility product and the solubility ss . It can vary depending on the number of ions a particular compound can form.
The solubility product KspKsp is a constant and does not depend on temperature change. It also does not have a unit.
Solubility is dependent on change in temperature. It has a unit, molem3\dfrac{{mole}}{{{m^3}}} .
The more the KspKsp is the higher the solubility of the solute in a solution.