WRITING NET IONIC EQUATIONS

9.2. WRITING NET IONIC EQUATIONS

When a substance made up of ions is dissolved in water, the dissolved ions behave independently. That is, they undergo their own characteristic reactions regardless of what other ions may be present. For example, silver ions in solution, Ag + , always react with chloride ions in solution, Cl − , to form an insoluble ionic compound, AgCl(s), no matter what other ions are present in the solution. If a solution of sodium chloride, NaCl, and a

solution of silver nitrate, AgNO 3 , are mixed, a white solid, silver chloride, is produced. The solid can be separated from the solution by filtration, and the resulting solution contains sodium nitrate, just as it would if solid NaNO 3 were added to water. In other words, when the two solutions are mixed, the following reaction occurs:

AgNO 3 + NaCl −→ AgCl(s) + NaNO 3

or

Ag + + NO − 3 + Na + + Cl − −→ AgCl(s) + Na + + NO − 3

Written in the latter manner, the equation shows that, in effect, the sodium ions and the nitrate ions have not changed. They began as ions in solution and wound up as those same ions in solution. They are called spectator ions. Since they have not reacted, it is really not necessary to include them in the equation. If they are left out, a net ionic equation results:

Ag + + Cl − −→ AgCl(s)

CHAP. 9]

NET IONIC EQUATIONS

This equation may be interpreted to mean that any soluble silver salt will react with any soluble ionic chloride to produce (insoluble) silver chloride.

EXAMPLE 9.1. Write three equations that the preceding net ionic equation can represent. Ans.

The following equations represent three of many possible equations:

AgNO 3 + NaCl −→ NaNO 3 + AgCl(s) AgClO 3 + KCl −→ KClO 3 + AgCl(s) AgC 2 H 3 O 2 + NH 4 Cl −→ NH 4 C 2 H 3 O 2 + AgCl(s)

Obviously, it is easier to remember the net ionic equation than the many possible overall equations that it represents. (See Problem 9.12.)

Net ionic equations may be written whenever reactions occur in solution in which some of the ions originally present are removed from solution or when ions not originally present are formed. Usually, ions are removed from solution by one or more of the following processes:

1. Formation of an insoluble ionic compound (Table 8-2)

2. Formation of molecules containing only covalent bonds

3. Formation of new ionic species

4. Formation of a gas (a corollary of 2) Examples of these processes include

1. AgClO

3 + NaCl −→ AgCl(s) + NaClO

3 − Ag + Cl −→ AgCl(s)

2. HI + NaOH −→ H O + NaI

H 2 + + OH − −→ H 2 O

3. Cu + 2 AgNO 3 −→ 2 Ag + Cu(NO 3 ) 2 Cu + 2 Ag + −→ Cu 2+ + 2 Ag

CO 2− 3 + 2H + −→ CO 2 + H 2 O The question arises how the student can tell whether a compound is ionic or covalent. The following

4. NH 4 CO 3 + 2 HCl −→ CO 2 + H 2 O + 2 NH 4 Cl

generalizations will be of some help in deciding:

1. Binary compounds of two nonmetals are covalently bonded. However, strong acids (Table 8-3) in water form ions completely.

2. Binary compounds of a metal and nonmetal are usually ionic.

3. Ternary compounds are usually ionic, at least in part, except if they contain no metal atoms or ammonium ion.

EXAMPLE 9.2. Predict which of the following will contain ionic bonds: (a) NiCl 2 , (b) SO 2 , (c) Al 2 O 3 , (d) NH 4 NO 3 ,

(e) H 2 SO 4 , (f ) HCl, and (g) NCl 3 .

Ans. (a) NiCl 2 , (c) Al 2 O 3 , and (d) NH 4 NO 3 contain ionic bonds. NH 4 NO 3 also has covalent bonds within each ion. (e) H 2 SO 4 and (f ) HCl would form ions if allowed to react with water.

When do we write compounds as separate ions, and when do we write them as complete compounds? Ions can act independently in solution, and so we write ionic compounds as separate ions only when they are soluble. We write compounds together when they are not ionic or when they are not in solution.

EXAMPLE 9.3. Write each of the following compounds as it should be written in an ionic equation. (a) KCl, (b) BaSO 4 ,

(c) SO 2 , and (d) Ca(HCO 3 ) 2 .

NET IONIC EQUATIONS

[ CHAP. 9

Ans.

(a) K + + Cl − (b) BaSO (insoluble) (c) SO (not ionic) (d) Ca 2+ + 2 HCO − 4 2 3

The insoluble compound is written as one compound even though it is ionic. The covalent compound is written together because it is not ionic.

EXAMPLE 9.4. Each of the following reactions produces 56 kilojoules (kJ) per mole of water produced. Is this just a coincidence? If not, explain why the same value is obtained each time.

KOH + HCl −→ KCl + H 2 O

LiOH + HBr −→ LiBr + H 2 O

RbOH + HI −→ RbI + H 2 O Ans.

NaOH + HNO 3 −→ NaNO 3 + H 2 O

The same quantity of heat is generated per mole of water formed in each reaction because it is really the same reaction in each case:

OH − + H + −→ H 2 O

It does not matter whether it is a K + ion in solution that undergoes no reaction or an Na + ion in solution that undergoes no reaction. As long as the spectator ions undergo no reaction, they do not contribute anything to the heat of the reaction.

EXAMPLE 9.5. Write a net ionic equation for the reaction of aqueous Ba(OH) 2 with aqueous HNO 3 . Ans.

The overall equation is

Ba(OH) 2 ( aq) + 2 HNO 3 −→ Ba(NO 3 ) 2 + 2H 2 O

In ionic form: Ba 2+ + 2 OH − + 2H + + 2 NO − 3 −→ Ba 2+ + 2 NO − 3 + 2H 2 O Leaving out the spectator ions yields

2 OH − + 2H + −→ 2H 2 O

Dividing each side by 2 yields the net ionic equation

OH − + H + −→ H 2 O

The net ionic equation is the same as that in Example 9.4. Writing net ionic equations does not imply that any solution can contain only positive ions or only negative

ions. For example, the net ionic equation

Ba 2+ + SO 2− 4 −→ BaSO 4 ( s)

does not imply that there is any solution containing Ba 2+ ions with no negative ions, or any solution containing SO 2− 4 ions with no positive ions. It merely implies that whatever negative ion is present with the barium ion and whatever positive ion is present with the sulfate ion, these unspecified ions do not make any difference to the reaction that will occur.

Net ionic equations must always have the same net charge on each side of the equation. (The same number of each type of spectator ion must be omitted from both sides of the equation.) For example, the equation

Cu + Ag + −→ Cu 2+ + Ag ( unbalanced)

has the same number of each type of atom on its two sides, but it is still not balanced. (One cannot add just one nitrate ion to the left side of an equation and two to the right.) The net charge must also be balanced:

Cu + 2 Ag + −→ Cu 2+ + 2 Ag

There is a net 2+ charge on each side of the balanced equation.

CHAP. 9]

NET IONIC EQUATIONS