Q19.88P - Chemistry: Molecular Nature Of Matter And Change

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 ${\mathbf{\text{0.023 M NaIO}}}_{\mathbf{\text{3}}}$ ?

Step-by-Step Solution

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.

Ionic product and solubility product

 Q_sp is 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 $\mathrm{Na}{\text{IO}}_3$ 

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

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.