Question

How do you calculate Molar solubility in grams/$100$mL of calcium iodate in water at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$? Ksp = $7.1 \times {10^{ - 7}}$

Answer 1

Hint: ${{\text{K}}_{{\text{sp}}}}$ represents the solubility product constant at equilibrium. When a solid substance dissolves into water it dissociates into ion. The product of solubilities of the ions with the number of each ion in power is known as the solubility product. First we will determine the molarity and then the equilibrium constant.

Formula used:${{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\text{x}}{{\text{S}}^{\text{x}}}\,{{ \times }}\,{\text{y}}{{\text{S}}^{\text{y}}}$

Complete step-by-step solution:The solubility product constant represents the product of concentrations of constituting ions of an ionic compound at equilibrium each raised number of ions as power.
An ionic compound dissociates into the water as follows:
${{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\, \rightleftarrows \,\,{\text{x}}{{\text{A}}^{{\text{ + Y}}}}\,{\text{ + }}\,{\text{y}}{{\text{B}}^{ - x}}$
Where,
${\text{AB}}$ is an ionic compound.
The general representation for the solubility product of an ionic compound is as follows:
\[{{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\left( {{\text{x}}{{\text{A}}^{{\text{ + y}}}}\,} \right)^{\text{x}}}\, \times \,\,{\left( {{\text{y}}{{\text{A}}^{ - x}}\,} \right)^{\text{y}}}\]
${{\text{K}}_{{\text{sp}}}}\,{\text{ = }}\,{\text{x}}{{\text{S}}^{\text{x}}}\,{{ \times }}\,{\text{y}}{{\text{S}}^{\text{y}}}$
Where,
${{\text{K}}_{{\text{sp}}}}$is the solubility product constant.
${\text{S}}$is the solubility of each ion.
$x$ and $y$ represents the number of ions.
The compound calcium iodate dissociates into the water as follows:
${\text{Ca(I}}{{\text{O}}_3}{)_2} \rightleftarrows \,\,\,{\text{C}}{{\text{a}}^{{\text{2 + }}}}{\text{ + }}\,\,2\,{\text{IO}}_3^ - $
The solubility product for calcium iodate is represented as follows:
${{\text{K}}_{{\text{sp}}}}\, = \,\,\left[ {{\text{C}}{{\text{a}}^{{\text{2 + }}}}} \right]{\left[ {{\text{2IO}}_4^ - } \right]^2}$
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,\,\left[ {\,{\text{S}}} \right]{\left[ {{\text{2S}}} \right]^2}$
$\Rightarrow {{\text{K}}_{{\text{sp}}}}\, = \,\,{\text{4}}{{\text{S}}^3}$
$\Rightarrow {\text{S}}\,\,{\text{ = }}\,\,\,\sqrt[3]{{\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{4}}}\,$
On substituting $7.1 \times {10^{ - 7}}$ for Ksp in above equation,
${\text{S}}\,\,{\text{ = }}\,\,\,\sqrt[3]{{\dfrac{{7.1 \times {{10}^{ - 7}}}}{4}}}\,$
$\Rightarrow {\text{S}}\,\,{\text{ = }}\,\,5.62 \times {10^{ - 3}}\,$
Solubility tells the concentration of solute compounds in solution in molarity.
Molarity tells the mole of solute dissolve in per litre of the solution.
Here, solubility is $5.62 \times {10^{ - 3}}\,$, it means that $5.62 \times {10^{ - 3}}\,$ moles of calcium iodate is dissolved in one litre of water.
We know, ${\text{1}}\,{\text{L}}\,{\text{ = }}\,{\text{1000}}\,{\text{mL}}$
So, we can also say $5.62 \times {10^{ - 3}}$moles of calcium iodate is dissolved in ${\text{1000}}$ mL of water.
We have to determine the molar solubility in g/$100$mL so, we will convert the mole of calcium iodate into gram as follows
The formula of mole is as follows:
${\text{mole = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
Molar mass of calcium iodate is $389.88$ g/mol.
On substituting $5.62 \times {10^{ - 3}}\,$mol for moles of calcium iodate and $389.88$ g/mol for molar mass.
$5.62 \times {10^{ - 3}}\,{\text{mol}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{389}}{\text{.88}}\,{\text{g/mol}}}}$
$\Rightarrow {\text{mass}}\,{\text{ = }}\,\,5.62 \times {10^{ - 3}}\,{\text{mol}}\, \times \,{\text{389}}{\text{.88}}\,{\text{g/mol}}$
$\Rightarrow {\text{mass}}\,{\text{ = }}\,\,2.19\,{\text{g}}$
So, the mass of calcium iodate is $2.19$gram.
So, the molar solubility of calcium iodate is $2.19$gram/$1000$ mL. So, the molar solubility in $100$mL is,
${\text{1000}}\,{\text{mL}}$water = $2.19$gram calcium iodate
$100$mL water = $0.219$gram calcium iodate
So, the molar solubility of calcium iodate is $0.219$gram/ $100$mL.

Therefore, the Molar solubility in grams/$100$mL of calcium iodate in water at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$is $0.219$gram/ $100$mL.

Note: By the addition of atomic mass the molar mass is determined. Solubility tells the concentration of solid compounds dissolved in water. Concentration in the determination of solubility products is taken in terms of molarity. The formula of solubility product constant depends upon the number of ions produced by the ionic compounds in the water. Molar solubility represents the number of ions dissolved per liter of solution. Here, solubility represents the number of ions dissolved in a given amount of solvent.