Cation spectator ions

The cationic spectator ion is K, and the anionic spectator ion is S04 . The Fe ion will need to occur twice in the product Fc2( 804)3, so there must be an even number of Fe atoms. So the net ionic equation is multiphed by two. It now becomes... 

Most anions behave this way, except those derived from strong acids (Cl-, Br-, I-, N03 , C104-), which show little or no tendency to react with water to harm OH- ions. Like the cations listed in the left column of Table 13.5, they act as spectator ions as far as pH is concerned. 

A note on good practice The oxides and hydroxides of the alkali and alkaline earth metals are not Bronsted bases the oxide and hydroxide ions they contain are the bases (the cations are spectator ions). However, for convenience, chemists often refer to the compounds themselves as bases. 

FIGURE 11.15 If the concentration of one of the ions of a slightly soluble salt is increased, the concentration of the other decreases to maintain a constant value of Ksp. (a) The cations (pink) and anions (green) in solution, (b) When more anions are added (together with their accompanying spectator ions, which are not shown), the concentration of cations decreases. In other words, the solubility of the original compound is reduced by the presence of a common ion. In the insets, the blue background represents the solvent (water). 

Mixing the two solutions will produce 2.50 X 10 mol of Fe (0H)3 precipitate, which is 2.67 g. The mixed solution contains Na cations and Cl anions, too, but we can ignore these spectator ions in our calculations. Notice that this precipitation reaction is treated just like other limiting reactant problems. Examples and further illustrate the application of general stoichiometric principles to precipitation reactions. 

The formula unit equation is obtained by recognizing that there is no net charge in a solution, so all the cations are paired with anions to neutralize the charge. In this case the Cu2+ and H+ are paired with N03. This does mean that some N03 did not react but remained as spectator ions. 

Chemistry is often conducted in aqueous solutions. Soluble ionic compounds dissolve into their component ions, and these ions can react to form new products. In these kinds of reactions, sometimes only the cation or anion of a dissolved compound reacts. The other ion merely watches the whole affair, twiddling its charged thumbs in electrostatic boredom. These uninvolved ions cire called spectator ions. 

Leaving the spectator ions out of a net ionic equation does not imply that their presence is irrelevant. Certainly, if a reaction occurs by mixing a solution of Pb2+ ions with a solution of I- ions, then those solutions must also contain additional ions to balance the charge in each The Pb2+ solution must contain an anion, and the 1 solution must contain a cation. Leaving these other ions out of the net ionic equation merely implies that the specific identity of the spectator ions is not important any nonreactive ions could fill the same role. 

Two general styles are used when writing oxidation-reduction reactions. The first is to include all species that are in the reaction—that is to say that there is nothing left out, even if it is a spectator. In this style, acids and bases are written into the equation. The second style leaves out spectators. Because the anion in acids and the cation in bases are often spectators (ions common with one or more of the compounds in the reaction), the... 

Just like sodium ions, chloride ions are spectator ions in acid-base chemistry. Their job is to provide a charge balance to the cations in solution. So, in calculating the pH of lidocaine hydrochloride we ignore the chloride ion. Now we could draw out the structure or write the molecular formula of lidocaine and its conjugate acid, but it is tedious to do so. Let s do what most chemists do, and postulate a temporary abbreviation for these species. How about using L for lidocaine, and HL+ for its conjugate acid Now, we can write an equation for the acid ionization equilibrium reaction. 

Mixing a solution that contains silver ions with a solution that contains chloride ions produces a white precipitate of silver chloride. There must have been other ions present in each solution, as well. You know this because it is impossible to have a solution of just a cation or just an anion. Perhaps the solution that contained silver ions was prepared using silver nitrate or silver acetate. Similarly, the solution that contained chloride ions might have been prepared by dissolving NaCl in water, or perhaps NH4C1 or another soluble chloride. Any solution that contains Ag+(aq) will react with any other solution that contains Cl (aq) to form a precipitate of AgCl(s). The other ions in the solutions are not important to the net result. These ions are like passive onlookers. They are called spectator ions. 

Note that the cations are not eliminated because they are not the same on each side. There has been a change from iron(III) to iron(II). The uncharged metal atom has also changed and cannot be eliminated as a spectator ion. The net ionic equation indicates that iron metal will react with any soluble iron(III) compound as long as the corresponding iron(II) compound is soluble. 

Charged species such as OH and NH2 are used as salts, with cations such as LE, Na", or to balance the negative charge. These cations are called counterions or spectator ions, and their identity is usually incoiLsequential. For this reason, the counterion is often omitted. 

All reactions in this problem follow the same pattern the only difference is the nucleophile (" Nuc). Only the nucleophile is listed below. (Cations like Na or K accompany the nucleophile but are simply spectator ions and do not take part in the reaction they are not shown here.)... 

Plan For each pair of solutions, we note the ions present in the reactants, write the cation-anion combinations, and refer to Table 4.1 to see if any are insoluble. For the molecular equation, we predict the products. For the total ionic equation, we write the soluble compounds as separate ions. For the net ionic equation, we eliminate the spectator ions. Solution (a) In addition to the reactants, the two other ion combinations are strontium sulfate and sodium nitrate. Table 4.1 shows that strontium sulfate is insoluble, so a reaction does occur. Writing the molecular equation ... 

We can generalize this type of substitution for any nucleophile and substrate. To simplify things we have ignored the cation for the time being, as it doesn t participate in the reaction. The general equation for this type of substitution at a saturated carbon atom can be written as follows, ignoring the spectator ions... 

ACID-BASE PROPERTIES OF SALTS (SECTION 16.9) The acid-base properties of salts can be ascribed to the behavior of their respective cations and anions. The reaction of ions with water, with a resultant change in pH, is called hydrolysis. The cations of the alkali metals and the alkaline earth metals as well as the anions of strong acids, such as Cl , Br , F, and NO3 , do not undergo hydrolysis. They are always spectator ions in acid-base chemistry. A cation that is the conjugate acid of a weak base produces H upon hydrolysis. 

Figure 7.15 Anions are above and to the right oftheline cations are below and to the left of the line. Figure 7.16 No. The Na" and N03 ions will simply be spectator ions. The ions of an acid are needed in order to dissolve NiO. Figure 7.17 No. As seen in the photo, sulfur crumbles as it is hit with a hammer, typical of a solid non-metal. Figure 7.21 Because Rb is below K in the periodic table, we expect Rb to be more reactive with water than K. Figure 7.23 LUac (see Figure 7.22). Figure 7.25 The bubbles are due to H2(g).

The preceding equation is an example of an ionic equation, which shows dissolved species as free ions. To see whether a precipitate might form from this solution, we first combine the cation and anion from different compounds that is, Pbl2 and KNO3. Referring to Table 4.2, we see that PW2 is an insoluble compound and KNO3 is soluble. Therefore, the dissolved KNO3 remains in solution as separate and NO3 ions, which are called spectator ions, or ions that are not involved in the overall reaction. Because spectator ions appear on both sides of an equation, they can be eliminated from the ionic equation... 

The white insoluble solid that s formed is silver chloride. You can drop out the potassium cation and nitrate anion spectator ions, because they don t change during the reaction and are found on both sides of the equation in an identical form. (If you re totally confused about all those plus and minus symbols in the equations, or don t know what a cation or an anion is, just flip to Chapter 6. It tells all you need to know about this stuff.)... 

If it s necessary to add spectator ions to one side of the equation, add the same number to the other side of the equation. For example, there are 8 on the left side of the equation. In the original equation, the H was in the molecular form of HNO3. You need to add the NOs spectator ions back to it. You already have 2 on the left, so you simply add 6 more. You then add 6 N03 to the right-hand side to keep things b anced. Those are the spectator ions that you need for the Cu cation to convert it back to the molecular form that you want. 

Analyze and Plan We are asked to predict whether a solution of Na2pIP04 will be acidic or basic. Because Na2pIP04 is an ionic compound, we divide it into its component ions, Na and HP04, and consider whether each is acidic or basic. Because Na is the cation of a strong base, NaOH, we know that Na has no influence on pH. It is merely a spectator ion in acid-base chemistry. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HP04 ion. We need to con-... 

Cations such as Na+ or K+ are also present in this solution as spectator Ions necessary to balance charge. 

Start by writing the acid and base reactants and the salt product. Recall that the salt consists of cations from the base and anions from the acid. Note also that water is frequently a product if it is, you will not be able to balance the equation without it. Once you have the balanced molecular equation, write the complete ionic equation. Do that by representing any strong electrolytes by the formulas of their ions. Finally, write the net ionic equation by canceling any spectator ions, and from it note the proton transfer. 

The precipitate forms as a result of the very strong attractive forces between the Pb + cations and the 1 anions. The other product is the water-soluble salt potassium nitrate, KNO3. The potassium and nitrate ions do not take part in the reaction. They remain in solution as aqueous ions and therefore are often referred to as spectator ions. The guidelines that help identify which ions form a precipitate and which ions remain in solution are developed in a later chapter on ions in aqueous solutions. 

---