Volume 2
2.6. IONIC EQUATIONS

Neutralization Reaction Revisited. Complete and Net Ionic Equations for Reactions of Strong Acids with Strong Bases • Ionic Equations for Neutralization Reactions Involving Weak Acids and Bases • Ionic Equations for Exchange Reactions • Reactions That Go to Completion Revisited • Exercises
2.6.1. Neutralization Reaction Revisited. Complete and Net Ionic Equations for Reactions of Strong Acids with Strong Bases. Acids and bases are the two most fundamental classes of chemical compounds. The acidity, as represented by the H+ cation (proton), and alkalinity (basicity), as represented by the OH- anion, are like chemical antipodes. Once an OH- ion and an H+ ion collide in solution, they "annihilate" or, as chemists say, neutralize each other to give a molecule of water. If I had to name one most important chemical reaction ever, it is the neutralization that I would probably choose.

A simple example of acid-base neutralization is the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH).

NaOH + HCl = NaCl + H2O

This chemical equation is presented in its standard molecular form, as if the acid and base as well as the salt produced were all in a non-dissociated state. However, NaOH, HCl, and NaCl (but not H2O) are strong electrolytes that exist in the form of ions in solution. We therefore can rewrite the equation in its ionic form (Figure 2-78). To derive an ionic equation from a molecular chemical equation, we just break up the molecular formulas into the corresponding ions for all of the strong electrolytes among the reactants and products. Since water is a weak electrolyte, we keep its formula "as is", in the molecular form. The resultant equation (Figure 2-78) is called a complete ionic equation because it displays all of the ions and molecules involved in the transformation. Figure 2-78. Deriving a complete ionic equation from a regular (molecular) equation for the neutralization reaction of NaOH with HCl.

To see more clearly the essence of a chemical transformation, it is sometimes useful to transform a complete ionic equation into a net ionic equation (Figure 2-79). Imagine two aqueous solutions, one of HCl and the other of NaOH. Being strong electrolytes, NaOH and HCl exist in solution as ions. The ions in our HCl solution are H+ and Cl-. Those in the NaOH solution are Na+ and OH-. Now we mix the two. In the moment of mixing, all four ions are present in the combined solution: Na+, OH-, H+, and Cl-. For this set, four types of collision between oppositely charged ions are possible, as follows.

- Na+ and OH- collide and part ways because NaOH is a strong electrolyte that immediately dissociates back to Na+ and OH-.

- H+ and Cl- collide and separate because HCl is a strong electrolyte that dissociates back to H+ and Cl-.

- Na+ and Cl- collide and separate because, being a salt, NaCl is a strong electrolyte that dissociates back to the ions.

- H+ and OH- collide to form H2O, a weak electrolyte that does not dissociate.

So, the Na+ cations and the Cl- anions just float around in our solution, doing virtually nothing. Being just spectators rather than active players in the reaction, the Na+ and Cl- can be cancelled out on both sides of the equation (Figure 2-79, middle line). This gives us the net ionic equation (Figure 2-79, bottom). Figure 2-79. Converting a complete ionic equation to a net ionic equation.

The net ionic ionic equation at the bottom of Figure 2-79 is the essence of any neutralization reaction between a soluble metal hydroxide (base) and a soluble acid.

H+ + OH- = H2O

This is the net ionic equation that we should always end up getting when converting a regular chemical equation for any neutralization reaction between any soluble metal hydroxide with any soluble acid, regardless of their basicity. This point is illustrated by Figure 2-80 showing how a balanced regular chemical equation for the neutralization reaction of Ca(OH)2, a dibasic base, with HNO3, a monobasic acid, is converted to the complete and net ionic equations. Figure 2-80. Converting a regular chemical equation for neutralization of Ca(OH)2 with HNO3 to complete and net ionic equations.

Practice by writing regular, complete ionic, and net ionic equations for the neutralization of H2SO4 with KOH to produce K2SO4. If you do not get H+ + OH- = H2O for the net ionic equation, then look for errors.

2.6.2. Ionic Equations for Neutralization Reactions Involving Weak Acids and Bases. The neutralization reaction is characteristic of not just strong acids and bases. Weak acids and bases also undergo neutralization. Let us consider the reaction of Cu(OH)2, a weak base, with H2SO4, a strong acid. Water-insoluble blue Cu(OH)2 readily dissolves on agitation with aqueous H2SO4 to give a blue solution of CuSO4 (Figure 2-81). Figure 2-81. Neutralization of Cu(OH)2 with H2SO4.

Let us now convert the molecular equation shown in Figure 2-81 to an ionic equation. As we do that, we should keep in mind that Cu(OH)2 is a weak electrolyte and, being insoluble, is not in the solution. Therefore, we write the formula of copper hydroxide as Cu(OH)2, not Cu2+ + 2 OH- (Figure 2-82). Figure 2-82. Converting a regular chemical equation for neutralization of Cu(OH)2 with H2SO4 to complete and net ionic equations.

Note that the net ionic equation is: Cu(OH)2 + 2 H+ = Cu2+ + 2 H2O, not H+ + OH- = H2O, as in the case of neutralization of a soluble strong base with a soluble strong acid.

Ionic equations for reactions of weak acids with strong bases are written similarly (Figure 2-83). To write a correct ionic equation for the reaction of H2SiO3, a weak electrolyte, with NaOH, a strong base, we should keep in mind that water-insoluble silicic acid is not dissociated. Figure 2-83. Converting a regular chemical equation for neutralization of H2SiO3 with NaOH to complete and net ionic equations.

Are you curious enough to wonder how to write equations for neutralization reactions between a weak base and a weak acid? If you are, I am afraid I have to disappoint you. Such reactions are murky waters, especially for an introductory chemistry course like ours. Sometimes there is no reaction whatsoever (for example, Al(OH)3 + H2S), sometimes efficient neutralization is observed (for example, Cu(OH)2 + acetic acid), and sometimes a hydro or hydroxo salt is formed.

2.6.3. Ionic Equations for Exchange Reactions. Suppose we need to make BaSO4 from BaCl2 and Na2SO4. For that, we just mix aqueous solutions of BaCl2 and Na2SO4. The insoluble BaSO4 product precipitates out within the time of mixing (Figure 2-84), after which it can be separated by filtration, washed with water, and dried. Figure 2-84. Precipitation of BaSO4 upon mixing solutions of BaCl2 and Na2SO4.

Let us analyze this reaction from the perspective of electrolytic dissociation, the way we did it for the neutralization reaction between NaOH and HCl in the previous subsection. Both BaCl2 and Na2SO4 are salts and, consequently, strong electrolytes. In the moment of mixing aqueous BaCl2 and Na2SO4, the resulting solution contains the ions Ba2+, Cl-, Na+, and SO42-. The oppositely charged ions collide as follows.

- Na+ and SO42- come together and immediately part ways because Na2SO4 is a strong electrolyte that dissociates back to Na+ and SO42-.

- Ba2+ and Cl- come together and separate because BaCl2 is a strong electrolyte, dissociating back to Ba2+ and Cl-.

- Na+ and Cl- collide and split up because NaCl is a strong electrolyte that dissociates back to the ions.

- Ba2+ and SO42- collide to form BaSO4, which is insoluble in water and precipitates out.

Figure 2-85 shows how a regular molecular equation for this reaction is converted to complete and net ionic equations. One might think that our ionic equations imply that BaSO4 is a weak electrolyte, which is not true. Although BaSO4 is a strong electrolyte, it is insoluble in water and precipitates out as a solid. There are no dissociated ions in solid compounds, even ionic ones. Barium sulfate is no exception. Therefore, we present the formula of barium sulfate in ionic equations as BaSO4, not "Ba2+ + SO42-". Figure 2-85. Molecular and ionic equations for the reaction of BaCl2 with Na2SO4.

As follows from the net ionic equation (Figure 2-85, bottom line) any soluble sources of Ba2+ and SO42- should be suitable for making BaSO4. Indeed, BaBr2, Ba(NO3)2, and Ba(OH)2 are all soluble in water and therefore can be treated with also water-soluble Na2SO4, K2SO4, or H2SO4 in any combination to produce BaSO4.

Like barium, magnesium is an alkaline-earth metal. Both belong in Group 2 of the periodic and both are divalent. Can we make MgSO4 from MgCl2 and Na2SO4 in the same manner as BaSO4 is made from BaCl2 and Na2SO4 (Figure 2-85)? No, we cannot. Unlike BaSO4, MgSO4 is soluble in water and does not precipitate out from a solution containing Mg2+ and SO42-.

Both MgCl2 and Na2SO4 are salts and, consequently, strong electrolytes that exist in the form of ions in solution. Therefore, the solution produced upon mixing MgCl2 and Na2SO4 in water contains Mg2+, Cl-, Na+, and SO42-. The standard ion collision analysis for this case follows.

- Na+ and SO42- come together and dissociate immediately because Na2SO4 is a strong electrolyte.

- Mg2+ and Cl- collide and part ways because MgCl2 is a strong electrolyte.

- Na+ and Cl- collide and part ways because NaCl is a strong electrolyte.

- Mg2+ and SO42- collide and part ways because MgSO4 is a strong electrolyte.

Clearly, our resultant solution is just a cocktail of ions, a blend of Mg2+, Cl-, Na+, and SO42- in water (Figure 2-86). The cations and anions associate in all possible combinations to form ion pairs nonstop and dissociate back nonstop in a continuous dynamic process. We use the double arrow sign for the molecular equation (Figure 2-86, top) to indicate that the forward and backward processes are reversible. The complete ionic equation (Figure 2-86, middle line) is particularly illustrative, showing that the solution is just a mix of ions. As all of the ions can be cancelled out (Figure 2-86 bottom), there is no net ionic equation in this case. Figure 2-86. A solution of MgCl2 and Na2SO4 is just a cocktail of ions produced on dissociation of the two salts.

It would be incorrect to state that MgCl2 and Na2SO4 do not react. They do, but the products of their reaction, MgSO4 and NaCl, also react nonstop to give back the starting salts. In other words, there is exchange between MgCl2 and Na2SO4 to give MgSO4 and NaCl, but, being fully reversible, this reaction does not go to completion.

2.6.3. Reactions That Go to Completion Revisited. Some reactions go to completion and some do not. For instance, neutralization reactions go to completion and so does the formation of insoluble BaSO4 from soluble sources of Ba2+ and SO42-. In contrast, the reaction between MgCl2 and Na2SO4 does not go to completion. As in real life we need to make chemical and materials, it is critical to know what reactions go to completion and what reactions do not. A chemical reaction proceeds to completion if:

- Water (or another weak electrolyte) is produced in the reaction. Acid-base neutralization is a typical example of such a chemical transformation.

- A precipitate is formed in the reaction. The formation of insoluble BaSO4 (Figures 2-84 and 2-85) is a good example. Another typical example is the reaction of a soluble silver salt with a soluble chloride salt to give insoluble AgCl, for example:

AgNO3 + KCl = AgCl↓ + KNO3

Complete ionic equation: Ag+ + NO3- + K+ + Cl- = AgCl↓ + K+ + NO3-

Net ionic equation: Ag+ + Cl- = AgCl↓

- A gas is formed, which bubbles off the reaction mixture. For example, mixing sodium sulfide (Na2S) with hydrochloric acid (HCl) produces H2S, a gas that escapes from the reaction solution (video):

Na2S + 2 HCl = 2 NaCl + H2S↑

Complete ionic equation: 2 Na+ + S2- + 2 H+ + 2 Cl- = 2 Na+ + 2 Cl- + H2S↑

Net ionic equation: S2- + 2 H+ = H2S↑

2.6.4. Exercises.

1. Write molecular, complete ionic, and net ionic equations for the following neutralization reactions. Assume complete neutralization, i.e. involvement of all of the OH groups of the metal hydroxides and all of the protons on the acid molecules. For the neutralization of H3PO4 with NaOH write two sets of complete and net ionic equations: (A) assuming that H3PO4 is a weak electrolyte and (B) assuming that only the first proton of H3PO4 dissociates.

(a) KOH + HNO3

(b) Mg(OH)2 + HCl

(c) Al(OH)3 + H2SO4

(d) NaOH + H3PO4

(e) Cu(OH)2 + HBr

2. Write regular, complete ionic, and net ionic equations for the following exchange reactions.

(a) Ca(NO3)2 + Na2CO3 (CaCO3 precipitates)

(b) CuSO4 + KOH (Cu(OH)2 precipitates)

(c) AgNO3 + MgBr2 (AgBr precipitates)

(d) NaF and CaCl2 (CaF2 precipitates).

3. To a solution of 100 g of BaCl2 was added Na2SO4 in excess. The experiment was then repeated using 100 g of BaBr2 in place of the BaCl2. Which of the two reactions produced a larger quantity of BaSO4? Solve this problem mentally, without performing calculations. Answer

4. Heating solid NaCl with concentrated H2SO4 in the presence of a small amount of water produces HCl gas and NaHSO4. Write regular, complete ionic, and net ionic equations for this reaction. Could HCl gas be made using dilute rather than concentrated sulfuric acid? Answer

5. Of the chemical reactions presented below, identify those proceeding to completion and write both complete and net ionic equations for them. Mark precipitates and gases in the products with arrows.

(a) Na2CO3 + 2 HCl = 2 NaCl + CO2 + H2O

(b) KNO3 + NaCl = KCl + NaNO3

(c) Cu(NO3)2 + Ca(OH)2 = Cu(OH)2 + Ca(NO3)2

(d) Al2(SO4)3 + 2 Na3PO4 = 2 AlPO4 + 3 Na2SO4

(e) Ca(OH)2 + CO2 = CaCO3 + H2O

(f) MgBr2 + 2 KCl = MgCl2 + 2 KBr

(g) Ag2SO4 + BaCl2 = 2 AgCl + BaSO4

(h) FeS + 2 HCl = FeCl2 + H2S