Question

Find the solubility of ${\text{Ca}}{\left( {{\text{I}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ (in mol/L) in a solution containing 0.1 M ${\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$, at ${18^ \circ }{\text{C}}$ is, [${{\text{K}}_{{\text{sp}}}}$ ${\text{Ca}}{\left( {{\text{I}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ = $6.3 \times {10^{ - 7}}$, at ${18^ \circ }{\text{C}}$]

Answer 1

Hint: Solubility of a compound is defined as, the ability of the compound to get dissolved in a solvent and convert into its constituent ions. When a solid is in chemical equilibrium with its solution, the equilibrium is called the solubility equilibrium and ${{\text{K}}_{{\text{sp}}}}$ is the solubility product constant of that equilibrium.

Complete step by step solution: Let’s write the balanced chemical reaction for the given question,
${\text{Ca}}{\left( {{\text{I}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{(s)}} \rightleftharpoons {\text{C}}{{\text{a}}^{{\text{2 + }}}}{\text{(aq) + 2IO}}_{\text{3}}^{\text{ - }}{\text{(aq)}}$
Now, consider ‘x’ to be the solubility of ${\text{Ca}}{\left( {{\text{I}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ , then we can write the solubility at equilibrium as
$
{\text{Ca}}{\left( {{\text{I}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}{\text{(s)}} \rightleftharpoons {\text{C}}{{\text{a}}^{{\text{2 + }}}}{\text{(aq) + 2IO}}_{\text{3}}^{\text{ - }}{\text{(aq)}} \\
{{\text{E}}_q}{\text{ x 0}}{\text{.1 2x}} \\
$ (where ${{\text{E}}_q}$ = equilibrium and 0.1M of ${\text{C}}{{\text{a}}^{{\text{2 + }}}}$ is given in the question.)
From the balanced chemical equation, we can say that 1 mole ${\text{Ca}}{\left( {{\text{I}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ gives, 1 mole ${\text{C}}{{\text{a}}^{{\text{2 + }}}}$, and 2 moles of ${\text{IO}}_{\text{3}}^{\text{ - }}$ ( 1 and 2 moles of
${\text{C}}{{\text{a}}^{{\text{2 + }}}}$ and ${\text{IO}}_{\text{3}}^{\text{ - }}$ respectively, can also be called as the stoichiometric coefficient). Hence, we can write the solubility product constant as;
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {{\text{C}}{{\text{a}}^{{\text{2 + }}}}} \right]^1}{\left[ {{\text{I}}{{\text{O}}_3}} \right]^2}$
$ \Rightarrow $$6.3 \times {10^{ - 7}}$ = 0.1 $ \times $${\left( {\text{x}} \right)^2}$(the stoichiometric coefficients are raised to the power in the ${{\text{K}}_{{\text{sp}}}}$equation)
$ \Rightarrow $$6.3 \times {10^{ - 7}}$ - 0.1 = ${{\text{x}}^2}$
$ \Rightarrow $ ${{\text{x}}^2}$= $6.2 \times {10^{ - 7}}$ ( now we will square root on both sides, to get the value of ‘x’)

$ \Rightarrow $$\sqrt {{{\text{x}}^2}} $ = $\sqrt {6.2 \times {{10}^{ - 7}}} $ (the value of $\sqrt {{{\text{x}}^2}} $ = x )
$ \Rightarrow $x = 0.000787 (since the first non zero number ‘7’ is after 3 zeros, we will multiply and divide the equation with 1000, to get the answer in 3 decimals points)
$ \Rightarrow $ x = $0.000787 \times \dfrac{{1000}}{{1000}}$
$ \Rightarrow $ x = $\dfrac{{0.787}}{{{{10}^3}}}$
$ \Rightarrow $x = $0.787 \times {10^{ - 3}}$m/L
Hence, the solubility of ${\text{Ca}}{\left( {{\text{I}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ in the solution containing 0.1 M ${\text{Ca}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$, at ${18^ \circ }{\text{C}}$ is $0.787 \times {10^{ - 3}}$mol/L

Note: While calculating the solubility constant, remember that the concentration of solids and liquids are considered as unity (1). Also, pay attention to the S.I units, the question specifies to get the solubility in mol/L, the number of moles of solute dissolved in per liter of solution is also called molarity and the S.I unit is M.