The precipitation reaction between 25.0 mL of a solution containing a cation (purple) and 35.0 mL of a solution containing an anion (green) is depicted below (with ions shown as spheres and solvent molecules omitted for clarity).

(a) Given the following choices of reactants, write balanced total ionic and net ionic equations that best represent the reaction:

(1) KNO₃(aq) + CuCl₂(aq)

(2) NaClO₄(aq) + CaCl₂(aq)

(3) Li2SO₄(aq) + AgNO₃(aq)

(4) NH₄Br(aq) + Pb (CH₃COO)₂(aq)

(b) If each sphere represents 2.5 x 10^-3 mol of ion, find the total number of ions present.

(c) What is the mass of solid formed?

We note 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 Ca²⁺).

2. All common carbonates (CO3^2-) and phosphates (PO4 ^3-) are insoluble, except those of Group 1A (1) and NH4 ⁺.

All common sulfides are insoluble except those of Group 1A (1), Group 2A (2), and NH4⁺

4. All common chlorides (Cl^-), bromides (Br^-), and iodides (I^-) are soluble, except those of Ag⁺, Pb²⁺, Cu+, and Hg2 ²⁺

(a) From the given fig we conclude that we have 9 cations and 7 anions which gives 4 anionic spheres to react with 8 cationic spheres which means 2 moles of cations are combined with the 1 mole of the anion and form 4 moles of solid.

Hence, from the given choices (C) is depicted as the best choice for the given figure, which gives PbSO₄ as a product.

The balanced total ionic reaction is:

${\mathbf{Li}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{Ag}\mathbf{\left(}{\mathbf{NO}}_{\mathbf{3}}\mathbf{\right)}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{Li}}^{\mathbf{+}}\mathbf{+}{\mathbf{NO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Ag}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

Total ionic reaction:

$2{\mathrm{Li}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{Li}}^{+}\left(\mathrm{aq}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+{\mathrm{Ag}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)$

Net ionic reaction:
$2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)\to {\mathrm{Ag}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)$

(b) If each sphere represents 2.5 x 10^-3 mol of ion

Thus, the moles of cation = 9 x 2.5 x 10^-3 = 22.5 x 10^-3

the moles of anion = 7 x 2.5 x 10^-3 = 17.5 x 10^-3

the total moles of ions = 4 x 10^-1

therefore, the total number of ions present:

$\mathrm{total}\mathrm{no}.\mathrm{of}\mathrm{ions}\mathrm{present}=4\mathrm{x}{10}^{-1}\mathrm{moles}\mathrm{of}\mathrm{ion}\mathrm{present}\mathrm{x}\frac{6.022\mathrm{x}{10}^{23}\mathrm{ions}}{1\mathrm{mol}\mathrm{ions}}\mathrm{x}\frac{1\mathrm{mol}{\mathrm{Ag}}_2{\mathrm{SO}}_4}{1\mathrm{mol}\mathrm{ions}}$

= 2.41 x 10²² ions