Precipitation Reactions Total Molecular Equations

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)... 

U.24 Complete the following precipitation reactions with balanced molecular, total ionic, and net ionic equations ... 

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 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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