91Ó°ÊÓ

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 ${\mathrm{Ag}}^{+},\text{ }{\mathrm{Pb}}^{2+},\text{ }{\mathrm{Cu}}^{+},\text{ }\mathrm{and}\text{ â¶Ä‰}{\mathrm{Hg}}_2^{2+}.$

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

The balanced net ionic equation for the reaction is:

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

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

Since each sphere contains $5.0×{10}^{−4}$ mol ions,

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

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 = $5.0×{10}^{−3}$

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

Mass of $5.0×{10}^{−3}$moles of ${\text{PbSO}}_{\text{4}}$$\begin{array}{c}=3{\text{03.26 g/³¾´Ç±ô×5.0×10}}^{\text{-3}}\text{″¾´Ç±ôes}\\ \text{=1.51 g}\end{array}$

Therefore, mass of the product formed is 1.51 g.