Workshop Writing and Balancing Reaction Equations in Solution

Changes in matter are shown using chemical reaction equations. Different branches of chemistry use slightly different methods to write chemical reactions. In this workshop we will explore writing, completing, and balancing three types of reaction equations in solutions.

For every reaction equation, the left side of the equation contains the reactants, what is mixed together, the arrow indicates a reaction and the right side shows the products, the result of the chemical reaction. In inorganic and organic chemistry the reactants and products are typically shown using their chemical formulas, in biochemistry the names of the compounds are usually used.

Remember that matter is neither created nor destroyed in a chemical reaction. Elements and even subatomic particles (specifically electrons) are conserved. To illustrate this, the same number of each element must appear on both sides of a reaction equation.

Before you can balance the chemical reaction all of the reactants and products must be written with the correct formula. For ionic compounds they must be written as electrically neutral.

International Union of Pure and Applied Chemists Nomenclature (IUPAC) and chemical formulas for ionic compounds

Compounds made of simple ions:

Metals form positive ions, which are also called cations, and nonmetals, anions, form negative ions. Chemical formulas are written in the order cation anion.

For main groups elements, which have an A group number, the positive charge equals the roman numeral or group number. A periodic table is shown below with the typical charges an element may have in a stable ion. For nonmetals, the negative charge is equal to the A group number minus 8.

These elements are typically +3, but may also be +1

Examples: Write the formulas for stable compounds between the following elements.

Sodium and chloride:

Sodium has the symbol Na, it would form a +1 charge. Chloride is Cl and will have a -1 charge. The ratio between them should be 1:1, the correct formula is NaCl.

Calcium and Fluoride

Calcium has the symbol Ca, it would form a +2 ion. Fluoride is F and will have a -1 charge. The ratio needs to be 2 F for every one Ca, the correct formulas is CaF₂.

Table 1

Table 1 contains information to make all of the simple ionic compounds. The names of ionic compounds are always given as the cation is listed first, followed by the anion. Cations are typically metals, and they are named the same as the element. If an element has more than one possible cation the charge is given in the name with Roman numerals in parenthesis. This replaces the older stock system. Stock names are still occasionally seen.

Table 2 Transition Metal Ion Names

For simple anions the name is changed to have an “ide” ending. These are given in Table 3.

Table 3: Common Anions and Their Names

Example 3: What is the formula for the ionic compound between iron (III) and Oxygen.

Notice that I need to specify the charge on iron because more than one ionic charge is stable.

Fe⁺³ and O^-2 would form a compound with the formula Fe₂O₃.

The name of this compound is Iron (III) oxide.

There is a trick to finding the formula for more complicated ionic compounds, take the charge of cation and that is the number of anions need and the charge of the anion is number of cations needed:

Fe⁺³ O^-2 Fe₂O₃

Do not forget, if the charge is the same on the cation and the anion, they must be in a 1:1 ratio.

Practice problem:

Form and name the compounds made with the following elements

a. Ca and Cl

b. Mg and O

c. K and O

d. Pb⁺⁴ and F

e. Al and O

f. Fe⁺³ and S

g. Na and C

h. Li and S

Compound Ions

Ions may also contain more than one element, which are covalently bonded together. These are compound ions. Many of the common compound ions are deprotonated oxoacids. Table 4 lists many of the common oxoacids.

Table 4 Common Oxoacids

Table 5 shows some of the more common compound ions. Many of these are deprotonated oxoacids. If the oxoacid has an “ic acid” ending, the ending of the anion is “ate.” If the oxoacid has an “ous acid” ending, the ending of the anion is “ite.” The prefixes, “per” and “hypo” are unchanged.

Table 5 Common Compound Ions

When writing formulas containing compound ions, the compound ion is treated as a unit. If more than one of a compound ion is needed to maintain electric neutrality, the compound ion is placed in parenthesis.

Example: Write the formula for aluminum hydroxide:

Al⁺³ and OH^–. Three hydroxides are needed to maintain electric neutrality, therefore the formula should be written as: Al(OH)₃.

Practice problem 2

a. Ca and hydroxide

b. ammonium and Cl

c. Cu⁺² and sulfate

d. K and sulfite

e. Mg and dihydrogenphosphate

f. Na and carbonate

g. Ba and nitrate

h. Mg and phosphate

Writing and Balancing Reaction Equations

Remember that matter is neither created nor destroyed in a chemical reaction. Elements and even subatomic particles (specifically electrons) are conserved. Therefore, the same number of each element must appear on both sides of a reaction equation. The ratios of each substance in a reaction equation are called the reaction stoichiometry. Once the correct formulas for the reactants and products are known, they remain unchanged. The number of each of substances is changed to make each element appear in the same number on both sides of the reaction equation.

It is also customary to indicate the phase of the reactants and products, this is done by placing one of the four abbreviations in parenthesis after the chemical formula.

Table 6 Phase Abbreviations

For reactions in solution there are three major types of reactions.

1. Precipitation reactions:

In a precipitation reaction ions which are soluble in combination with other ions, when mixed form an insoluble compound. Whether or not this type of reaction will occur can be determined by referring to the solubility rules.

Solubility Rules for Ionic compounds in water

Soluble compounds

1. All compounds containing alkali metals (Group IA) are soluble.

2. All salts containing NH₄⁺, NO₃^–, ClO₄^–, ClO₃^–, and C₂H₃O₂^– are soluble.

3. All chlorides, Cl-, bromides, Br-, and iodides, I-, are soluble, except with Ag⁺, Pb⁺², and Hg₂⁺².

4. All sulfates, SO₄^-2, are soluble, except with Pb⁺², Ag⁺ and Hg₂⁺², Ba⁺², Ca⁺², and Sr⁺².

Insoluble compounds

1. All hydroxides, OH^–, and metal oxides, O^-2, are insoluble except group IA, and Ba⁺², Ca⁺², and Sr⁺² are soluble.

2. All compounds that contain PO₄^-3, CO₃^-2, SO₃^-2,and S^-2 are insoluble except those of Group IA and NH₄⁺.

Example 1

An aqueous solution of lead (II) nitrate is mixed with an aqueous solution of potassium iodide.

Writing the reactants in their correct form on the correct side of the reaction equation yields:

Pb(NO₃)₂(aq) + KI(aq)

Now swap the cations and anions Pb⁺² is combined with I^–, and K⁺ is combined with NO₃^–. Write the compounds between these elements on the product side of the reaction equation.

Pb(NO₃)₂(aq) + KI(aq) KNO₃ + PbI₂

Are these new compounds soluble or insoluble? A check of the rules shows that PbI₂ is insoluble, and KNO₃ is soluble. This means the KNO₃ stays dissolved in water and the PbI₂ will form a precipitate.

Pb(NO₃)₂(aq) + KI(aq) KNO₃(aq) + PbI₂(s)

Everything is in the correct form, but the reaction is clearly not balanced. Usually in this type of reaction it is best to think of the compound ion, in this case NO₃^–, as one group. Then it is clear that we need 2KNO₃. To balance K and I we also need 2KI. The balanced equation is:

Pb(NO₃)₂(aq) + 2KI(aq) 2KNO₃(aq) + PbI₂(s)

Example 2:

An aqueous solution of barium nitrate is mixed with an aqueous solution of sodium sulfate.

Writing the reactants in their correct form on the correct side of the reaction equation yields:

Ba(NO₃)₂(aq) + Na₂SO₄ (aq)

Now swap the cations and anions Ba⁺² is combined with SO₄^-2, and Na⁺ is combined with NO₃^–. Write the compounds between these elements on the product side of the reaction equation.

Ba(NO₃)₂(aq) + Na₂SO₄ (aq) NaNO₃ + BaSO₄

Are these new compounds soluble or insoluble? A check of the rules shows that BaSO₄ is insoluble, and NaNO₃ is soluble. This means the NaNO₃ stays dissolved in water and the BaSO₄ will form a precipitate.

Ba(NO₃)₂(aq) + Na₂SO₄ (aq) NaNO₃ (aq) + BaSO₄(s)

Everything is in the correct form, but the reaction is clearly not balanced. Usually in this type of reaction it is best to think of the compound ion, in this case NO₃^–, as one group. Then it is clear that we need 2NaNO₃. The balanced equation is:

Ba(NO₃)₂(aq) + Na₂SO₄ (aq) 2NaNO₃ (aq) + BaSO₄(s)

2. Acid-Base Reactions

There is more than one definition for acids or bases, but in aqueous solutions the most useful definition of an acid is that is a proton (H⁺) donor, and a base is an H⁺ acceptor. An acid-base reaction is a proton, H⁺, transfer reaction.

Example 1: Strong Acid/Strong Base

What happens when a solution of sulfuric acid is mixed with a solution of potassium hydroxide?

Write the reactants, sulfuric acid is H₂SO₄, and potassium hydroxide is KOH. To find the products of this reaction, we need to exchange the ions. Sulfuric acid dissolved in water forms 2H⁺ and one SO₄^-2. KOH will form in solution K⁺ and OH^–. Again swap the cations and the two products will be K₂SO₄ and HOH or H₂O. The final balanced reaction equation is:

H₂SO₄(aq) + 2KOH(aq) K₂SO₄(aq) + 2 H₂O(l)

Example 2: Strong acid/weak base

What happens when a solution of hydrochloric acid (HCl) is mixed with the weak base ammonia, NH₃. The weak base ammonia will accept the H⁺, forming NH₄⁺. The reaction equation should be:

HCl(aq) + NH₃(aq) NH₄⁺(aq) + Cl^–(aq)

The two ions can be combined to complete the equation in molecular form.

HCl(aq) + NH₃(aq) NH₄Cl(aq)

Example 3: Weak acid/strong base

What happens when a solution of acetic acid, C₂H₃OOH is mixed with a solution of NaOH?

Acetic acid has only one protic hydrogen. Normally we expect to see the acid hydrogen in the cation position, but for acetic acid it is the hydrogen connected to an oxygen. Therefore the complete reaction equation is:

C₂H₃OOH(aq) + NaOH(aq) C₂H₃OONa (aq) + H₂O(l)

Example 4: Weak acid/weak base

What happens when an aqueous solution of acetic acid is mixed with an aqueous solution of ammonia?

Based on our previous examples, the reaction equation is:

C₂H₃OOH(aq) + NH₃ (aq) C₂H₃OONH₄ (aq)

Ionic and Net ionic reaction equations:

The previous reaction equations were all written in the molecular form. This illustrates the compounds mixed together and which compounds can be recovered as products. Two other methods are used to better show what is present in solution and the net change. To show what is present in solution the ionic method is used. In this method, each strong electrolyte is split into its ions. In the net ionic form, only substances, which differ on both sides of the reaction, equation are shown.

Examples of reaction equations written in all three forms:

Molecular: Pb(NO₃)₂(aq) + 2KI(aq) 2KNO₃(aq) + PbI₂(s)

Ionic: Pb⁺²(aq) + 2NO₃^– (aq) + 2K⁺(aq) + 2I^– (aq) 2K⁺(aq) + 2NO₃^– (aq) + PbI₂(s)

Net ionic: Pb⁺²(aq) + 2I^– (aq) PbI₂(s)

Molecular: H₂SO₄(aq) + 2KOH(aq) K₂SO₄(aq) + 2 H₂O(l)

Ionic: 2H⁺(aq)+ SO₄^-2(aq) + 2K⁺(aq) + 2OH^–(aq) 2 H₂O(l) + SO₄^-2(aq) + 2K⁺(aq)

Net ionic:H⁺(aq) + OH^–(aq) H₂O(l) This simplified by dividing everything by 2.

Molecular: HCl(aq) + NH₃(aq) NH₄Cl(aq)

Ionic: H⁺(aq) + Cl^– (aq) + NH₃(aq) NH₄⁺ + Cl^– (aq)

Net ionic: H⁺(aq) + NH₃(aq) NH₄⁺ (aq)

Note in the next example the weak electrolyte is not separated into ions.

Molecular: C₂H₃OOH(aq) + NaOH(aq) C₂H₃OONa (aq) + H₂O(l)

Ionic: C₂H₃OOH(aq) + Na⁺ (aq) + OH^– (aq) C₂H₃OO^– (aq) + Na⁺ (aq) + H₂O(l)

Net ionic: C₂H₃OOH(aq) + OH^– (aq) C₂H₃OO^– (aq) + H₂O(l)

Molecular: C₂H₃OOH(aq) + NH₃ (aq) C₂H₃OONH₄ (aq)

Ionic: C₂H₃OOH(aq) + NH₃ (aq) C₂H₃OO^–(aq) + NH₄⁺ (aq)

Net ionic: C₂H₃OOH(aq) + NH₃ (aq) C₂H₃OO^–(aq) + NH₄⁺ (aq)

In the final example the ionic and net ionic are the same.

Practice Problem #3. Complete and Balance the following reaction equations in molecular, ionic and net ionic form.

a. NaCl(aq) + Pb(NO₃)₂(aq)

b. CaBr₂ (aq) + Na₂CO₃

c. HCl(aq) + Ba(OH)₂(aq)

d. AgNO₃(aq) + NaCl(aq)

e. Mg(OH)₂(aq) + HCl(aq)

f. CuSO₄(aq) + NaOH(aq)