Question

Does any solid ${\mathbf{\text{Ba(IO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$ form when $\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{mg}$of ${\mathbf{BaCL}}_{\mathbf{2}}$is dissolved in $\mathbf{500}\mathbf{}\mathbf{ml}$ of${\mathbf{\text{0.023 M NaIO}}}_{\mathbf{\text{3}}}$ ?

Short answer

Since K_sp< Q_sp the reaction will shift to the right and will produce $\mathbf{Ba}\mathbf{(}{\mathbf{IO}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$ precipitate.

Step-by-step solution

Ionic product and solubility product

Q_spis the ionic product expression; value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp value is called K_sp value (solubility-product constant).

${\mathbf{MX}}_{\mathbf{2}}\mathbf{\to }{\mathbf{M}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{X}}^{\mathbf{-}}$

Solid and liquid state is not included in the K_sp equations.

${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{= [M}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{][X}}}^{\mathbf{\text{-}}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

S(Molar solubility) is equal to the concentration of one mol of the ion formed by dissolution of some compound.

Concentration of Ions

First write the dissolution equation for –

$Ba(I{O}_3{)}_2\text{(s)}\rightleftharpoons {\text{Ba}}^{\text{2 +}}{\text{(aq) + 2IO}}_{\text{3}}^{\text{-}}\text{(aq)}$

Construct an ICE table for the expression.

	$\mathbf{Ba}\mathbf{(}{\mathbf{IO}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$	${\mathbf{\text{Ba}}}^{\mathbf{\text{2 +}}}$	${\mathbf{\text{2IO}}}_{\mathbf{\text{3}}}^{\mathbf{\text{-}}}$
Initial	-	0	0
Change	-	+s	+2s
Equilibrium	-	s	2s

The Q_sp is represented as –

${\text{Q}}_{\text{sp}}{\text{= [Ba}}^{\text{2 +}}{\text{][IO}}_{\text{3}}^{\text{-}}{\text{]}}^2$

Calculation

Now solve for the concentration of the ions present from their respective salt.

${\text{Ba}}^{\text{2 +}}$ from ${\text{BaCl}}_2$

$$\begin{gathered} {\mathrm{BaCl}}_2\to {\mathrm{Ba}}^{2+}+2{\mathrm{Cl}}^{-} \\ \left[{\mathrm{Ba}}^{2+}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ \left[{\mathrm{Ba}}^{2+}\right]=\frac{7.5\mathrm{mg}{\mathrm{BaCl}}_2\times \frac{1\mathrm{g}}{1000\mathrm{mg}}\times \frac{1\mathrm{mol}{\mathrm{BaCl}}_2}{208.23\mathrm{g}{\mathrm{BaCl}}_2}\times \frac{1\mathrm{mol}{\mathrm{Ba}}^{2+}}{1\mathrm{mol}{\mathrm{BaCl}}_2}}{200\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}} \\ \left[{\mathrm{Ba}}^{2+}\right]=7.20\times {10}^{-5}\mathrm{M} \end{gathered}$$

${\text{IO}}_{\text{3}}^{\text{-}}$from $\mathrm{Na}{\text{IO}}_3$

role="math" localid="1663766553811" $$\begin{gathered} {\mathrm{NaIO}}_3\to {\mathrm{Na}}^{+}+{\mathrm{IO}}_3^{-} \\ \left[{\mathrm{NaIO}}_3\right]=\left[{\mathrm{IO}}_3^{-}\right]=0.023\mathrm{M} \end{gathered}$$

Now solve for the value of –

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right][{\mathrm{IO}}_3^{-}{]}^2 \\ {\mathrm{Q}}_{\mathrm{sp}}=(7.20\times {10}^{-5})\times {\left(0.023\right)}^2 \\ {\mathrm{Q}}_{\mathrm{sp}}=3.97\times {10}^{-8} \end{gathered}$$

The value for the ${\text{K}}_{\text{sp}}$ of $\mathrm{Ba}({\mathrm{IO}}_3{)}_2$ is ${\mathrm{K}}_{\mathrm{sp}}=1.5\times {10}^{-9}$.

Therefore, ${\text{Q}}_{\text{sp}}$ value is greater and solid $\mathrm{Ba}({\mathrm{IO}}_3{)}_2$ will form.