Q4.37P - Chemistry: Molecular Nature Of Matter And Change

Question

With ions shown as spheres and solvent molecules omitted for clarity, the circle (right) illustrates the solid formed when a solution containing ${\mathbf{\text{K}}}^{\mathbf{\text{+}}}{\mathbf{\text{,\hspace{0.17em}Mg}}}^{\mathbf{\text{2+}}}{\mathbf{\text{,\hspace{0.17em}Ag}}}^{\mathbf{\text{+}}}{\mathbf{\text{,\hspace{0.17em}or\hspace{0.17em}Pb}}}^{\mathbf{\text{2+}}}\mathbf{\left(}\mathbf{\text{blue}}\mathbf{\right)}$  is mixed with one containing  ${\mathbf{\text{ClO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{-}}}{\mathbf{\text{,\hspace{0.17em}NO}}}_{\mathbf{\text{3}}}^{\mathbf{\text{-}}}{\mathbf{\text{,\hspace{0.17em}or\hspace{0.17em}SO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2-}}}\mathbf{\left(}\mathbf{\text{yellow}}\mathbf{\right)}\mathbf{\text{.}}$

(a) Identify the solid.

(b) Write a balanced net ionic equation for the reaction. 

(c) If each sphere represents ${\mathbf{\text{5.0×10}}}^{\mathbf{\text{4-}}}\mathbf{\text{\hspace{0.17em}mol}}$ of ion, what mass of product forms?

Step-by-Step Solution

Answer

1. The solid is  ${\mathbf{\text{PbSO}}}_{\mathbf{\text{4}}}\mathbf{.}$
2.  ${\mathbf{\text{Pb}}}^{\mathbf{\text{2+}}}{\mathbf{\text{(aq) + SO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2-}}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{PbSO}}}_{\mathbf{\text{4}}}\mathbf{\text{(s)}}$
3. The mass of product forms is 1.516 gm.

Step 1: Rules to determine the charge on the ions

The ions present in the reactants for each pair of solutions and write the cation-anion combinations if any of them are insoluble or soluble. For the molecular equation, we predict the products:

1. All common metal hydroxides are insoluble, except those of Group 1A (1) and the larger members of Group 2A (2) (beginning with ${\mathrm{Ca}}^{2+}$  ).

2. All common carbonates  $\left({\mathrm{CO}}_3^{2-}\right)$ and phosphates  $\left({\mathrm{PO}}_4^{3-}\right)$ are insoluble, except those of Group 1A (1) and  ${\mathrm{NH}}_4^{+}.$

3. All common sulfides are insoluble except those of Group 1A (1), Group 2A (2), and ${\mathrm{NH}}_4^{+}.$

4. All common chlorides $\left({\mathrm{Cl}}^{-}\right),$   bromides $\left({\mathrm{Br}}^{-}\right),$  and iodides  $\left({\mathrm{I}}^{-}\right),$ are soluble, except those of  ${\mathrm{Ag}}^{+},{\mathrm{Pb}}^{2+},{\mathrm{Cu}}^{+},\mathrm{and}{\mathrm{Hg}}_2^{2+}.$ 

Step 2: Possibilities to form salt solution are:

The different possibilities to form salts are: ${\mathrm{KNO}}_3,{\mathrm{KClO}}_4,{\mathrm{K}}_2{\mathrm{SO}}_4,{\mathrm{AgClO}}_4,{\mathrm{AgNO}}_3,{\mathrm{PbClO}}_4,\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$  and  ${\mathrm{PbSO}}_4.$

But according to previous knowledge, we know that these all salts do not form precipitate except  ${\mathbf{\text{PbSO}}}_{\mathbf{\text{4}}}\mathbf{.}$

Therefore, all are soluble in the solvent and remain as ions in the solution except   ${\mathrm{PbSO}}_4.$

Hence, the given solid in the figure is  ${\mathrm{PbSO}}_4.$

Step 3: Write the balanced net ionic equation of the reaction

The balanced net ionic equation for the reaction is:

 ${\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)\to {\mathrm{PbSO}}_4\left(\mathrm{s}\right)$

Step 4: Calculate the mass of the product form

Number of spheres of ${\text{Pb}}^{\text{+2}}$   = Number of spheres of  ${\text{SO}}_{\text{4}}^{\text{-2}}$  =  $10$ 

Since each sphere contains $5.0\times {10}^{-4}$  mol ions,

Number of mole of ions of  ${\text{Pb}}^{\text{+2}}$= Number of mol of ions of ${\text{SO}}_{\text{4}}^{\text{-2}}$ = $\begin{array}{l}=10\times 5.0\times {10}^{-4}\\ =5.0\times {10}^{-3}\end{array}$ 

1 mole of  ${\text{Pb}}^{\text{+2}}$ combines with 1 mol of  ${\text{SO}}_{\text{4}}^{\text{-2}}$ to give 1 mole of  ${\text{PbSO}}_{\text{4}}$.

Therefore,

Total number of moles of   ${\text{PbSO}}_{\text{4}}$ formed =  $5.0\times {10}^{-3}$

Mass of 1 mole of  ${\text{PbSO}}_{\text{4}}$ =  $303.26\text{\hspace{0.17em}g}$ 

Mass of  $5.0\times {10}^{-3}$ moles of    ${\text{PbSO}}_{\text{4}}$$\begin{array}{c}=3{\text{03.26 \hspace{0.17em}g/mol × 5.0×10}}^{\text{-3}}\text{\hspace{0.17em}moles}\\ \text{=1.51\hspace{0.17em}g}\end{array}$

Therefore, mass of the product formed is 1.51 g.