Question

The solubility of lead (ii) iodate, $Pb{{(I{{O}_{3}})}_{2}}$, is 0.76g/L at $25{{\text{ }}^{{}^\circ }}C$. How do you calculate the value of ${{K}_{sp}}$ at this same temperature?

Answer 1

The solubility product constant or ${{K}_{sp}}$ of a solid substance is an equilibrium constant which defines the extent of dissolution of a solid compound and its ions in an aqueous solution. It is given by the formula
${{K}_{sp}}={{[{{A}^{+}}]}^{a}}{{[{{B}^{-}}]}^{b}}$
Where ${{K}_{sp}}$ = solubility product constant,
${{A}^{+}}$ = cation in an aqueous solution,
${{B}^{-}}$ = anion in an aqueous solution,
And a, b = relative concentrations.

Complete step-by-step answer: The dissociation of ions of lead (ii) iodate $Pb{{(I{{O}_{3}})}_{2}}$ can be described by the equation
$Pb{{(I{{O}_{3}})}_{2}}(s)\overset{{}}{\leftrightarrows}P{{b}^{2+}}+2IO_{3}^{-}$
So, by applying the solubility product constant formula we get
${{K}_{sp}}=[P{{b}^{2+}}]{{[IO_{3}^{-}]}^{2}}$
From the dissociation equation and the solubility product constant, we can see that when one mole of $Pb{{(I{{O}_{3}})}_{2}}$ dissociates, it produces one mole of $P{{b}^{2+}}$ ions and two moles of $IO_{3}^{-}$ ions.
If we assume the molar solubility for $Pb{{(I{{O}_{3}})}_{2}}$ to be ${{S}_{molar}}$ , then
${{K}_{sp}}=[{{S}_{molar}}]{{[2{{S}_{molar}}]}^{2}}=4S_{molar}^{3}$.
We cannot use the solubility value given directly, as it is the mass solubility which is expressed in g/L. We first have to convert the mass solubility into molar solubility before using the data to calculate the solubility product constant ${{K}_{sp}}$.
Molar solubility can be defined as the number of moles of a solute which can be dissolved in one liter of a solution before it becomes a saturated solution.
Now, it is given to us that the mass solubility of $Pb{{(I{{O}_{3}})}_{2}}$ at $25{{\text{ }}^{{}^\circ }}C$ is 0.76 g/L.
The molar mass of $Pb{{(I{{O}_{3}})}_{2}}$ is 577 g/mol.
So, molar solubility will be

$${{S}_{molar}}=\dfrac{{{S}_{mass}}}{Molar\text{ }Mass} \\\ {{S}_{molar}}=\dfrac{0.76\text{ }g/L}{557\text{ }g/mol} \\\ {{S}_{molar}}=1.37\times {{10}^{-3}}\text{ }mol/L \\\$$

Using this value of molar mass solubility, we can calculate the solubility product.

$${{K}_{sp}}=10.28\times {{10}^{-9}} \\\$$

Hence the value of ${{K}_{sp}}$ of lead (ii) iodate, $Pb{{(I{{O}_{3}})}_{2}}$ at $25{{\text{ }}^{{}^\circ }}C$, when its solubility is 0.76g/L, is $1.028\times {{10}^{-8}}$.

Note: It is important to note that higher the value of ${{K}_{sp}}$ of a product, the more the compound has the tendency to be soluble. Since the $K{{ }_{sp}}$ for $Pb{{(I{{O}_{3}})}_{2}}$ is extremely low, it can be inferred that it is a nearly insoluble compound.