Question

Including activity coefficient, find the concentration of ${\mathbf{Ba}}^{\mathbf{2}\mathbf{+}}$in $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{M}{\left({\mathrm{CH}}_3\right)}_{\mathbf{4}}{\mathbf{NIO}}_{\mathbf{3}}$solution saturated with $\mathbf{Ba}{\left({\mathrm{IO}}_3\right)}_{\mathbf{2}}$.

Short answer

The concentration of ${\mathrm{Ba}}^{2+}$in a $0.001\mathrm{M}{\left({\mathrm{CH}}_3\right)}_4{\mathrm{NIO}}_3$solution saturated with $\mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2$is$6.6\times {10}^{-7}\mathrm{M}$

Step-by-step solution

Step 1:Finding the activity coefficient of aqueous solution Ba2+ and IO3-.

In the given problem, we need to find the activity coefficient along with concentration of ${\mathrm{Ba}}^{2+}$in a $0.001\mathrm{M}{\left({\mathrm{CH}}_3\right)}_4{\mathrm{NIO}}_3$solution saturated with $\mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2$.

The solubility of $\mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2$is not quite expressed. So we will assume that $\mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2$contributes negligible ${{\mathrm{IO}}_3}^{-}\mathrm{to}0.100\mathrm{M}{\left({\mathrm{CH}}_3\right)}_4{\mathrm{NIO}}_3$.

$\mathrm{c}\left({{\mathrm{IO}}_3}^{-}\right)=0.1\mathrm{M}$

From the Table $8-1$, we can know that the values are,

$$\begin{gathered} \gamma \left({\mathrm{Ba}}^{2+}\right)=0.380 \\ \gamma \left({{\mathrm{IO}}_3}^{-}\right)=0.775 \end{gathered}$$

Finding the concentration of  Ba2+in a 0.001M(CH3)4NIO3 solution saturated with  Ba(IO3)2 using equilibrium constant equation.

Now, we will use equilibrium constant equation,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=1.5\times {10}^{-9} \\ {\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{c}\left({\mathrm{Ba}}^{2+}\right)\gamma \left({\mathrm{Ba}}^{\mathrm{2}\mathrm{+}}\right)\right)\times \left({\mathrm{c}}^2\left({{\mathrm{IO}}_3}^{-}\right)\right) \\ 1.5\times {10}^{-9}=\left(\mathrm{c}\left({\mathrm{Ba}}^{2+}\right)\times 0.380\right)\times \left(0.1^2\times 0.{775}^2\right) \\ \mathrm{c}\left({\mathrm{Ba}}^{2+}\right)=6.6\times {10}^{-7}\mathrm{M} \end{gathered}$$

Thus the concentration of ${\mathrm{Ba}}^{2+}$in a $0.001\mathrm{M}{\left({\mathrm{CH}}_3\right)}_4{\mathrm{NIO}}_3$solution saturated with $\mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2$is $6.6\times {10}^{-7}\mathrm{M}$