This is "Unit 8", section 8.4 from the book General Chemistry (v. 1.0).

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8.4 Ionic Equations

Learning Objective

  1. To understand what information is obtained by each type of ionic equation.

The chemical equations we have discussed so far have shown the identities and the number of formula units of the reactants and the products, but they told us very little about what was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent.

Let’s consider the reaction of silver nitrate with potassium dichromate. When aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall chemical equationA chemical equation that shows all the reactants and products as undissociated, electrically neutral compounds. for the reaction shows each reactant and product as undissociated, electrically neutral compounds:

Equation 8.4(eq1)

2AgNO3(aq) + K2Cr2O7(aq) → Ag2Cr2O7(s) + 2KNO3(aq)

Although Equation 8.4(eq1) gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as AgNO3 and K2Cr2O7 are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag2Cr2O7 is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equationA chemical equation that shows which ions and molecules are hydrated and which are present in other forms and phases., showing which ions and molecules are hydrated and which are present in other forms and phases:

Equation 8.4(eq2)

2Ag+(aq) + 2NO3(aq) + 2K+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) + 2K+(aq) + 2NO3(aq)

Note that K+(aq) and NO3(aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectator ionsIons that do not participate in the actual reaction. because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equationA chemical equation that shows only those species that participate in the chemical reaction., which shows only those species that participate in the chemical reaction:

Equation 8.4(eq3)

2Ag+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s)

Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation 8.4(eq3), the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag2Cr2O7 formula unit.

By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows:

Equation 8.4(eq4)

2AgF(aq) + (NH4)2Cr2O7(aq) → Ag2Cr2O7(s) + 2NH4F(aq)

The complete ionic equation for this reaction is as follows:

Equation 8.4(eq5)

2Ag+(aq) + 2F(aq) + 2NH4+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s) + 2NH4+(aq) + 2F(aq)

Because two NH4+(aq) and two F(aq) ions appear on both sides of Equation 8.4(eq5), they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation 8.4(eq6)), which is identical to Equation 8.4(eq3):

Equation 8.4(eq6)

2Ag+(aq) + Cr2O72−(aq) → Ag2Cr2O7(s)

If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.

Example 8.4-1

Write the net ionic equation for each chemical reaction.

  1. K+(aq) + Br-(aq) + Ag+(aq) + C2H3O2-(aq) → K+(aq) + C2H3O2-(aq) + AgBr(s)
  2. Mg2+(aq) + SO42-(aq) + Ba2+(aq) + 2NO3-(aq) → Mg2+(aq) + 2NO3-(aq) + BaSO4(s)

Solution

  1. In the first equation, the K+(aq) and C2H3O2-(aq) ions are spectator ions, so they are canceled:

    K+(aq)+Br-(aq)+Ag+(aq)+C2H3O2-(aq) K+(aq)+C2H3O2-(aq)+AgBr(s)

    The net ionic equation is

    Br-(aq) + Ag+(aq) → AgBr(s)
  2. In the second equation, the Mg2+(aq) and NO3-(aq) ions are spectator ions, so they are canceled:

    Mg2+(aq)+SO42-(aq)+Ba2+(aq)+ 2NO3-(aq)Mg2+(aq)+2NO3-(aq)+BaSO4(s)

    The net ionic equation is

    SO42-(aq) + Ba2+(aq) → BaSO4(s)

Test Yourself

Write the net ionic equation for

CaCl2(aq) + Pb(NO3)2(aq) → Ca(NO3)2(aq) + PbCl2(s)

Answer

Pb2+(aq) + 2Cl-(aq) → PbCl2(s)

Example 8.4-2

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate.

Given: reactants and products

Asked for: overall, complete ionic, and net ionic equations

Strategy:

Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.

Solution:

From the information given, we can write the unbalanced chemical equation for the reaction:

Ba(NO3)2(aq) + Na3PO4(aq) → Ba3(PO4)2(s) + NaNO3(aq)

Because the product is Ba3(PO4)2, which contains three Ba2+ ions and two PO43− ions per formula unit, we can balance the equation by inspection:

3Ba(NO3)2(aq) + 2Na3PO4(aq) → Ba3(PO4)2(s) + 6NaNO3(aq)

This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form:

3Ba2+(aq) + 6NO3(aq) + 6Na+(aq) + 2PO43−(aq) → Ba3(PO4)2(s) + 6Na+(aq) + 6NO3(aq)

The six NO3(aq) ions and the six Na+(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:

3Ba2+(aq) + 2PO43−(aq) → Ba3(PO4)2(s)

Exercise

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride.

Answer:

overall chemical equation: 3AgF(aq) + Na3PO4(aq) → Ag3PO4(s) + 3NaF(aq)

complete ionic equation: 3Ag+(aq) + 3F(aq) + 3Na+(aq) + PO43−(aq) → Ag3PO4(s) + 3Na+(aq) + 3F(aq)

net ionic equation: 3Ag+(aq) + PO43−(aq) → Ag3PO4(s)

Acid-Base Reactions

If we write the complete ionic equation for the reaction in Equation 8.4(eq7), we see that Na+(aq) and Br(aq) are spectator ions and are not involved in the reaction:

Equation 8.4(eq7)

H + (aq) + Br (aq) + Na + (aq) + OH (aq) H 2 O(l) + Na + (aq) + Br (aq)

The overall reaction is therefore simply the combination of H+(aq) and OH(aq) to produce H2O, as shown in the net ionic equation:

Equation 8.4(eq8)

H+(aq) + OH(aq) → H2O(l)

The net ionic equation for the reaction of any strong acid with any strong base is identical to Equation 8.4(eq8).

In the case of the reaction of a weak acid with a strong base, the ionic equation needs to reflect the fact that the weak acid stays mostly undissociated. For this reason, the formula of the weak acid is left whole and not written as ions as was the case for the strong acid reactions. Consider the reaction of nitrous acid with sodium hydroxide shown in Equation 8.4(eq9)

Equation 8.4(eq9)

HNO2(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) +NO2-(aq)

The overall reaction is showing the abstaction of an H+(aq) from nitrous acid by OH(aq) to produce H2O. And the net ionic equation omits only the sodium ion, as a spectator, from the total ionic equation:

Equation 8.4(eq10)

HNO2(aq) + OH-(aq) → H2O(l) + NO2-(aq)

Summary

The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist in solution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction.

Key Takeaway

  • A complete ionic equation consists of the net ionic equation and spectator ions.

Exercises

  1. Write a chemical equation that represents NaBr(s) dissociating in water.

  2. Write a chemical equation that represents SrCl2(s) dissociating in water.

  3. Write a chemical equation that represents (NH4)3PO4(s) dissociating in water.

  4. Write a chemical equation that represents Fe(C2H3O2)3(s) dissociating in water.

  5. Write the complete ionic equation for the reaction of FeCl2(aq) and AgNO3(aq). You may have to consult the solubility rules.

  6. Write the complete ionic equation for the reaction of BaCl2(aq) and Na2SO4(aq). You may have to consult the solubility rules.

  7. Write the complete ionic equation for the reaction of KCl(aq) and NaC2H3O2(aq). You may have to consult the solubility rules.

  8. Write the complete ionic equation for the reaction of Fe2(SO4)3(aq) and Sr(NO3)2(aq). You may have to consult the solubility rules.

  9. Write the net ionic equation for the reaction of FeCl2(aq) and AgNO3(aq). You may have to consult the solubility rules.

  10. Write the net ionic equation for the reaction of BaCl2(aq) and Na2SO4(aq). You may have to consult the solubility rules.

  11. Write the net ionic equation for the reaction of KCl(aq) and NaC2H3O2(aq). You may have to consult the solubility rules.

  12. Write the net ionic equation for the reaction of Fe2(SO4)3(aq) and Sr(NO3)2(aq). You may have to consult the solubility rules.

  13. Identify the spectator ions in Exercises 9 and 10.

  14. Identify the spectator ions in Exercises 11 and 12.

Answers

  1. NaBr(s)H2O Na+(aq) + Br-(aq)

  2. (NH4)3PO4(s) H2O 3NH4+(aq) + PO43-(aq)

  3. Fe2+(aq) + 2Cl-(aq) + 2Ag+(aq) + 2NO3-(aq) → Fe2+(aq) + 2NO3-(aq) + 2AgCl(s)

  4. K+(aq) + Cl-(aq) + Na+(aq) + C2H3O2-(aq) → Na+(aq) + Cl-(aq) + K+(aq) + C2H3O2-(aq)

  5. 2Cl-(aq) + 2Ag+(aq) → 2AgCl(s)

  6. There is no overall reaction.

  7. In Exercise 9, Fe2+(aq) and NO3-(aq) are spectator ions; in Exercise 10, Na+(aq) and Cl-(aq) are spectator ions.