15 P - Chemistry: Molecular Nature Of Matter And Change

Question

How many total moles of ions are released when each of the following samples dissolves completely in water? 

(a) 0.805 mol of  ${\mathbf{\text{Rb}}}_{\mathbf{\text{2}}}{\mathbf{\text{SO}}}_{\mathbf{\text{4}}}$ (b) ${\mathbf{\text{3.85×10}}}^{\mathbf{\text{-3}}}$  g of  $\mathbf{\text{Ca }}{\left({\text{NO}}_{\text{3}}\right)}_{\mathbf{\text{2}}}$

(c)  ${\mathbf{\text{4.03×10}}}^{\mathbf{\text{19}}}$ Formula units of $\mathbf{\text{Sr }}{\left({\text{HCO}}_{\text{3}}\right)}_{\mathbf{\text{2}}}$

Step-by-Step Solution

Answer

Total moles of ions are released when each of the following samples dissolves completely in the water:

(a)  $2.415\text{\hspace{0.17em}mol}$ of ions

(b) $7.038\times {10}^{-5}\mathrm{mol}$ of ions

(c) $2.008\times {10}^{-4}\mathrm{mol}$ of ions 

Determine the solubility of Ionic compounds in water

The ionic compounds and the water molecules are polar in nature therefore both attract each other by the force of attraction.

Most ionic compounds dissociate and dissolve in water easily.

Calculate total moles of ions released when dissolve in water

The dissociation reaction of ${\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4}}$  is:

       ${\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{2Rb}}^{\text{+}}{{\text{ + SO}}_{\text{4}}}^{\text{2-}}$                                 

Since each mole of  ${\text{Rb}}_{\text{2}}{\text{SO}}_{\text{4}}$ when dissolved in water gives 2 moles of  ${\text{Rb}}^{\text{+}}$ ions and 1 mole of ${{\text{SO}}_{\text{4}}}^{\text{2-}}$  ions, therefore a total of 3 moles of ions are produced.

Therefore, number of moles of ions formed when $0.805\text{\hspace{0.17em}mol}$  is dissolved in water is:

 $\left({\text{0.805 mol Rb}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{ x }\frac{\text{3 mol ion}}{{\text{1 mol Rb}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{ = 2.415 mol of ions }$

Calculate total moles of ions released when dissolve in water

The dissociation reaction of $\text{Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$  is:

    $\text{Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\to {\text{Ca}}^{\text{2+}}{{\text{ + 2NO}}_{\text{3}}}^{\text{-}}$                                  

Since each mole of $\text{Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$  when dissolved in water gives 1 mole of ${\text{Ca}}^{\text{2+}}$  ions and 2 moles of  ${{\text{NO}}_{\text{3}}}^{\text{-}}$ ions, therefore a total of 3 moles of ions are produced.

Number of moles of ions formed when  $3.85\times {10}^3\text{\hspace{0.17em}g}$ of $\text{Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is dissolved in water is given below.

$\begin{array}{l}\text{=}\left({\text{3.85 x 10}}^{\text{-3}}\text{ g mol Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)\text{ x }\left(\frac{\text{1 mol Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{164.10gm Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\text{ x }\left(\frac{\text{3mol ions}}{\text{1mol Ca }{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\\ {\text{= 7.038 x 10}}^{\text{-5}}\text{ mol of ions }\end{array}$

Calculate total moles of ions released when dissolve in water

The dissociation reaction of $\mathrm{Sr}{\left({\mathrm{HCO}}_3\right)}_2$  is:

      $Sr{\left(HC{O}_3\right)}_2 \to S{r}^{2+}+2HC{O_3}^{-}$                               

Since each mole of $Sr{\left(HC{O}_3\right)}_2$ when dissolved in water gives 1 mole of $S{r}^{2+}$  ions and 2 moles of ${{\mathrm{HCO}}_3}^{-}$  ions, therefore a total of 3 moles of ions are produced.

Number of moles of ions formed when  $4.03\times {10}^{19}\text{\hspace{0.17em}g}$ of $Sr{\left(HC{O}_3\right)}_2$  is dissolved in water is given below.

$\begin{array}{l}=4.03x{10}^{19}formula units of Sr{\left(HC{O}_3\right)}_{2}x\frac{1 mol Sr{\left(HC{O}_3\right)}_2}{6.022x{10}^{23}formula units of Sr{\left(HC{O}_3\right)}_2}x\frac{3molions}{1 mol Sr{\left(HC{O}_3\right)}_2}\\ =2.008x{10}^{-4}mol of ions\end{array}$