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?

Short 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-by-step solution

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 ${\mathrm{Ag}}^{+},{\mathrm{Pb}}^{2+},{\mathrm{Cu}}^{+},\mathrm{and}{\mathrm{Hg}}_2^{2+}.$

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 ${\mathrm{PbSO}}_4.$

Write the balanced net ionic equation of the reaction 

The balanced net ionic equation for the reaction is:

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

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.