Q14P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: How many total moles of ions are released when each of the following samples dissolves completely in water? (a) 0.32 mol of   (b) 25.4 g of  (c)   formula units of LiCl

Step-by-Step Solution

Answer

Answer 

 

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

(a) 0.63 mol of ions

(b) 0.242  mol of ions

(c)   $\mathbf{1}\mathbf{.}\mathbf{79}\mathbf{\times }\mathbf{ }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$mol of ions

Step 1: Determine the solubility of Ionic compounds in water

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

Most ionic compounds dissociate and dissolve in water easily.

Step 2: Calculate total moles of ions released when dissolve in water

 

The dissociation reaction of $\mathit{N}{\mathit{H}}_{\mathbf{4}}\mathit{C}\mathit{l}$  is:

                      $\mathit{N}{\mathit{H}}_{\mathbf{4}}\mathit{C}\mathit{l}\mathbf{\to }\mathit{N}{\mathit{H}}^{\mathbf{4}\mathbf{+}}\mathbf{+}\mathit{C}{\mathit{I}}^{\mathbf{-}}$                              

Since each mole of $\mathit{N}{\mathit{H}}_{\mathbf{4}}\mathit{C}\mathit{l}$  when dissolve in water give 1 mole of $\mathit{N}{\mathit{H}}^{\mathbf{4}\mathbf{+}}$  ions and 1 mole of   $\mathit{C}{\mathit{I}}^{\mathbf{-}}$ions therefore total of 2 moles of ions are produced.

 $\left(0.32\times N{H}_{4}CI\right)\mathbf{x}\mathbf{(}\frac{\mathbf{2}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{ }\mathbf{i}\mathbf{o}\mathbf{n}}{\left(1 mol N{H}_{4}CI\right)}\mathbf{1}\mathbf{.}\mathbf{62}\mathit{m}\mathit{o}\mathit{l}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{i}\mathit{o}\mathit{n}\mathit{s}$

Step 3: Calculate total moles of ions released when dissolve in water

The dissociation reaction of $\mathit{B}\mathit{a}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{8}{\mathit{H}}_{\mathbf{2}}\mathit{O}$  is:

          $\mathit{B}\mathit{a}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\to }\mathit{B}{\mathit{a}}^{\mathbf{+}\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{O}{\mathit{H}}^{\mathbf{-}}$                                         

Each mole of   $\mathit{B}\mathit{a}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{8}{\mathit{H}}_{\mathbf{2}}\mathit{O}$ 1 mol of $\mathit{B}{\mathit{a}}^{\mathbf{+}\mathbf{2}}$of   ions and 2 moles of $\mathit{O}{\mathit{H}}^{\mathbf{-}}$  ions

therefore, a total of 3 moles of ions are produced.

$$\begin{gathered} \mathbf{=}\mathbf{ }\mathbf{25}\mathbf{.}\mathbf{4}\mathbf{ }\mathit{g}\mathit{m}\mathit{B}\mathit{a}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{8}{\mathit{H}}_{\mathbf{2}}\mathit{O}\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ofBa}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}{\mathbf{315}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{gmBa}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{3}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ions}}{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ofBa}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\mathbf{.}\mathbf{8}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{)} \\ \mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{24179}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{i}\mathit{o}\mathit{n} \\ \mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{242}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{i}\mathit{o}\mathit{n} \end{gathered}$$

Step 4: Calculate total moles of ions released when LiCl dissolve in water

 

The dissociation reaction of LiCl is:

                                           $\mathbf{ }\mathit{L}\mathit{i}\mathit{C}\mathit{I}\mathbf{\to }\mathit{L}{\mathit{i}}^{\mathbf{+}}\mathbf{+}\mathit{C}{\mathit{I}}^{\mathbf{-}}$   

Each mole of LiCl forms 1 mol of  $\mathit{L}{\mathit{i}}^{\mathbf{+}}$ and 1 mole of  $\mathit{C}{\mathit{I}}^{\mathbf{-}}$ ions therefore a total of 2 moles of ions are produced.

$$\begin{gathered} \mathbf{=}\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{55}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathbf{ }\mathit{g}\mathit{m}\mathbf{ }\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}\mathbf{ }\mathit{u}\mathit{n}\mathit{i}\mathit{t}\mathbf{ }\mathit{L}\mathit{i}\mathit{C}\mathit{I}\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ofLiCI}}{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{ }\mathbf{F}\mathbf{.}\mathbf{U}\mathbf{.}\mathbf{LiCI}}\mathbf{\left)}\mathbf{\right(}\frac{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ions}}{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{LiCl}}\mathbf{)} \\ \mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{79}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{i}\mathit{o}\mathit{n}\mathit{s} \end{gathered}$$