Question

Which will be more soluble (moles of metal dissolved per litre of solution),$\mathbf{Ba}{\left({\mathrm{IO}}_3\right)}_{\mathbf{2}}\left(K_{\mathrm{sp}}=1.5x{10}^{-9}\right)\mathbf{}\mathbf{or}\mathbf{}\mathbf{Ca}{\left({\mathrm{IO}}_3\right)}_{\mathbf{2}}\left(K_{\mathrm{sp}}=7.1x{10}^7\right)$ ? Give an example of a chemical reaction that might occur that would reverse the predicted solubilities.

Short answer

$\mathrm{Ca}{\left({\mathrm{IO}}_3\right)}_2$is found to be more soluble.

Step-by-step solution

solubility product

The more soluble component in the given compounds must be anticipated, and examples must be provided.

Solubility product $\left(K_{\mathrm{sp}}\right)$:

It is defined as the mathematical product of the dissolved ion's concentration raised to the power of its stoichiometric coefficient.

$$\begin{gathered} {\mathrm{M}}_{\mathrm{y}}{\mathrm{X}}_{\mathrm{z}\left(\mathrm{s}\right)}\cancel{}{{\mathrm{yM}}^{\mathrm{z}+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{zX}}^{\mathrm{y}-}}_{\left(\mathrm{aq}\right)} \\ {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{M}}^{\mathrm{z}}\right]}^{\mathrm{y}}{\left[{\mathrm{X}}^{\mathrm{y}-}\right]}^{\mathrm{z}} \end{gathered}$$

check solubility product value

The solubility product $\left({\mathrm{K}}_{\mathrm{sp}}\right)$for,

$$\begin{gathered} \mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2=1.5\mathrm{x}{10}^{-9} \\ \mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2=7.1\mathrm{x}{10}^{-7} \end{gathered}$$

The solubility product value is more for $\mathrm{Ca}{\left({\mathrm{IO}}_3\right)}_2$than for $\mathrm{Ba}{\left({\mathrm{IO}}_3\right)}_2$. As a result, it should be more soluble and have the same stoichiometry. If the stoichiometry is not the same, compare the $\left({\mathrm{K}}_{\mathrm{sp}}\right)$values directly. If, for example, barium salt can form ion pairs, but calcium salt cannot, your forecast may be incorrect.