Question

If $Ksp$​ for $HgS{O_4}$ is $6.4 \times {10^{ - 5}}$ , then solubility of this substance in mole per ${m^3}$ is:
$Hg = 200,S = 32,O = 16$
A. $8 \times {10^{ - 3}}$
B. $8 \times {10^{ - 6}}$
C. $296$
D. $2368$

Answer 1

Hint: For a compound like $HgS{O_4}$ , the ratio of $H{g^{ + 2}}:S{O_4}^{ - 2}$ is $1:1$ . 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 ${s^2}$ where $s$ is the solubility.

Formula used: $Ksp$= ${s^2}$
Where $Ksp$ is the solubility constant and $s$ is the solubility.

Complete step by step answer:
In this question, we have two ions, there is $H{g^{ + 2}}$and $S{O_4}^{ - 2}$ . Therefore, the reaction of dissociation will look like this:
$HgS{O_4} \rightleftarrows H{g^{ + 2}} + S{O_4}^{ - 2}$
now we can consider the concentration of $H{g^{ + 2}}$to be $s$ and the concentration of $S{O_4}^{ - 2}$ to be equal to $H{g^{ + 2}}$ .
The equilibrium constant can be represented in the following way:
$Ksp = \left[ {H{g^{ + 2}}} \right]\left[ {S{O_4}^{ - 2}} \right]$
Concentration of mercury and sulphate ions is represented in the manner shown above.
$Ksp = \left[ s \right]\left[ s \right]$
$Ksp = {\left[ s \right]^2}$
For this reason, the equilibrium constant which is called a solubility product is the product of both the ions to form ${s^2}$ . 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 $Ksp$ to be $6.4 \times {10^{ - 5}}$ . This can be used to calculate the $s$ or solubility using the above formula.
$6.4 \times {10^{ - 5}} = {s^2}$
$Ksp = {s^2}$
$64 \times {10^{ - 6}} = {s^2}$
$\sqrt {64 \times {{10}^{ - 6}}} = s$
$8 \times {10^{ - 3}} = s$

So, the correct answer is Option A.

Note: It is important to remember the relation between the solubility product and the solubility $s$ . It can vary depending on the number of ions a particular compound can form.
The solubility product $Ksp$ 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, $\dfrac{{mole}}{{{m^3}}}$ .
The more the $Ksp$ is the higher the solubility of the solute in a solution.