17 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) 0.734 mol of ${\mathbf{\text{Na}}}_{\mathbf{\text{2}}}{\mathbf{\text{HPO}}}_{\mathbf{\text{4}}}$                                 

  (b) 3.86 g of ${\mathbf{\text{CuSO}}}_{\mathbf{\text{4}}}{\mathbf{\text{.5H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}$

(c) ${\mathbf{\text{8.66×10}}}^{\mathbf{\text{20}}}$  formula units of ${\mathbf{\text{NiCl}}}_{\mathbf{\text{2}}}$

Step-by-Step Solution

Answer

Total number of moles of ions  released when each of the following samples dissolves completely in the water is given below.

(a) 2.02 mol of ions

(b) $3.09\times {10}^{-2}$ mol of ions

(c) $4.31\times {10}^{-3}$ mol of ions 

Determine the solubility of Ionic compounds in water

The dissociation reaction of  ${\mathrm{Na}}_2{\mathrm{HPO}}_4$ is:                                        

      ${\mathrm{Na}}_2{\mathrm{HPO}}_4 \to 2{\mathrm{Na}}^{+}+{{\mathrm{HPO}}_4}^{2-}$                                   

Since each mole of ${\mathrm{Na}}_2{\mathrm{HPO}}_4$  when dissolved in water gives 2 moles of ${\mathrm{Na}}^{+}$  ions and 1 mole of  ${{\mathrm{HPO}}_4}^{2-}$ ions, therefore a total of 3 moles of ions are produced.

Total number of moles of ions produced when  $0.734\text{\hspace{0.17em}\hspace{0.17em}mol}$ of ${\mathrm{Na}}_2{\mathrm{HPO}}_4$ is dissolved in water.

$\left(0.734 mol N{a}_{2}HP{O}_4\right)x\frac{3 molion}{1 mol N{a}_{2}HP{O}_4}=2.202 mol of ions$

Calculate total moles of ions released when CuSO 4 . 5 H 2 O dissolve in water

The dissociation reaction of  ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ is:

             ${\mathrm{CuSO}}_4 \to {\mathrm{Cu}}^{2+}+{{\mathrm{SO}}_4}^{2-}$                                

Since each mole of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$  when dissolved in water gives 2 moles of ${\mathrm{Cu}}^{2+}$  ions and 1 mole of ${{\mathrm{SO}}_4}^{2-}$  ions, therefore a total of 2 moles of ions are produced.

Total number of moles of ions formed when $3.86\text{\hspace{0.17em}\hspace{0.17em}g}$  of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ is dissolved in water.

$\begin{array}{l}=\left(3.86 g CuS{O}_4.5H_{2}O\right)x\frac{1 mol CuS{O}_4.5H_{2}O}{249.72 gm CuS{O}_4.5H_{2}O}x\frac{2 mol ions}{1 mol CuS{O}_4.5H_{2}O}\\ =3.09x{10}^{-2} mol of ions\end{array}$

Calculate total moles of ions released when NiCl 2 dissolve in water

 

The dissociation reaction of ${\text{NiCl}}_{\text{2}}$  is:

   ${\mathrm{NiCl}}_2 \to {\mathrm{Ni}}^{+2}+2{\mathrm{Cl}}^{-}$                                             

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

Total number of moles of ions formed when $8.66\times {10}^{20}$  formula units of ${\text{NiCl}}_{\text{2}}$  is dissolved in water is given below.

$\begin{array}{l}=8.66x{10}^{20} formula unit NiC{l}_{2}x\frac{1molNiC{l}_2}{6.022x{10}^{23}formula units NiC{l}_2}x\frac{3molions}{1mol NiC{l}_2}\\ =4.31x{10}^{-3} mol of ions\end{array}$