Precipitation Reactions Total Ionic Equations

The net ionic equation is the same as the total ionic equation. This is a precipitation and a neutralization reaction. There are no spectator ions because all the ions are used to form the two products. 

Section 4.1 polar molecule (109) solvated (110) electrolyte (110) nonelectrolyte (112) Section 4.2 molecular equation (113) total ionic equation (114) spectator ion (114) net ionic equation (114) Section 4.3 precipitation reaction (115) precipitate (115) metathesis reaction (116) Section 4.4 acid-base reaction (117) neutralization reaction (117) acid (117) base (118) salt (119) titration (11 9) equivalence point (120) end point (120) Section 4.5 oxidation-reduction (redox) reaction (123) oxidation (124) reduction (124) oxidizing agent (124) reducing agent (124) oxidation number (O.N.) (or oxidation state) (124) Section 4.6 activity series of the metals (130)... 

Now, what happens if we substitute a solution of lead(II) nitrate, Pb(N03)2, for the KNO3 solution The reactant ions are Na+, I", Pb +, and N03. In addition to the two soluble reactants, Nal and Pb(N03)2, the other two possible cation-anion combinations are NaN03 and Pbl2. According to Table 4.1, NaN03 is soluble (Rule 1), but Pbl2 is not (Rule 3). The total ionic equation shows the reaction that occurs as Pb + and I" ions collide and form a precipitate ... 

When nickel nitrate and sodium carbonate solutions are combined, solid nickel carbonate precipitates, leaving a solution of sodium nitrate. Write the conventional equation, total ionic equation, and net ionic equation for this reaction. 

This time both new combinations of ions precipitate. There are no spectators, so the total ionic equation is the net ionic equation. Actually, two separate reactions are taking place at the same time ... 

The reaction of an acid often leads to an ion combination that yields a molecular product instead of a precipitate. Except for the difference in the product, the equations are written in exactly the same way. Just as you had to recognize an insoluble product and not break it up in total ionic equations, you must now recognize a molecular product and not break it into ions. Water and weak acids are the two kinds of molecular products you will find. 

Just as you can write the net ionic equation for a precipitation reaction without the conventional and total ionic equations, so you can write the net ionic equation for a reaction that forms a molecule. Again, you must recognize the product from the formulas of the reactants. The acid will contribute a hydrogen ion to the molecular product. It will form a molecule with the anion from the other reactant. If the molecule is water or a weak acid, you have the reactants and product of the net ionic equation. 

The simplest balanced chemical equation for a precipitation reaction is a net ionic equation that has ions as the reactants and a neutral solid as the product. In a precipitation reaction, reactant ions combine to form a neutral ionic solid. One reactant carries positive charge and the other carries negative charge, but the product is electrically neutral. Because electrical charge always is conserved, the total positive charge of the reacting cations... 

Will a precipitate form when aqueous solutions of Ca(NOj)2 and NaCl are mixed in reasonable concentrations Write balanced formula unit, total ionic, and net ionic equations for any reaction. 

Exercises 69 and 70 describe precipitation reactions in aqueous solutions. For each, write balanced (i) formula unit, (ii) total ionic, and (hi) net ionic equations. Refer to the solubility guidelines as necessary. 

When each of the following pairs of aqueous solutions is mixed, does a precipitation reaction occur If so, write balanced molecular, total ionic, and net ionic equations ... 

Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate, (a) Write balanced total and net ionic equations for its reaction with aqueous NaOH. (b) What mass of precipitate forms when 185.5 mL of 0.533 M NaOH is added to 627 mL of a solution that contains 15.8 g of aluminum sulfate per liter ... 

The previous findings, however, cannof be generalized to the precipitant species or species other than the hardness and its associated ionic species. Eor example, in Equation (10.16), if the above findings were applied to the precipitant Ca(OH)2, its equivalent mass would be Ca(OH)2/2 however, this is not correct— the equivalent mass of Ca(OH)2 in this equation is 2Ca(OH)2/2. To conclude, the equivalent mass of a precipitant species or species other than the hardness and its associated species cannot be generalized as molecular mass divided by the total number of valences of the species but must be deduced from the chemical reaction. 

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