A Molarity of Ions in Solution

APPENDIX
A
Molarity of Ions in Solution
Often it is necessary to calculate not only the concentration (in molarity) of a compound
in aqueous solution but also the concentration of each ion in aqueous solution. The
coefficients from the balanced dissolution equation are used in this type of calculation.
Example
Calculate the concentrations of Na1 and SO422 in an aqueous solution of 2.0 M Na2SO4.
Step 1
Write the balanced dissolution equation (i.e., equation showing how the strong electrolyte dissolves in water).
H2O
Na2SO4(s) → 2 Na1(aq) 1 SO422(aq)
Step 2
Use the coefficients from the balanced dissolution equation to calculate the concentration of each ion in aqueous solution.
Molarity of Na1:
2.0 mol Na 2SO4
4.0 mol Na +
2 mol Na +
× = OR 4.0 M Na +
L Soln
1 L Soln
1 mol Na 2SO4
↑
↑
molarity of soln coefficients from the
balanced dissolution eqn.
Molarity of SO422:
2.0 mol SO4–2
2.0 mol Na 2SO4
1 mol SO4–2
OR 2.0 M SO4−2
× = L Soln
1 L Soln
1 mol Na 2SO4
↑
↑
molarity of soln coefficients from the
balanced dissolution eqn.
153
Chemistry 116 General Chemistry
Example
Calculate the concentrations of Mg12 and Cl2 in an aqueous solution prepared by dissolving 19 g MgCl2 in enough water to prepare 250. mL of solution.
Step 1
Calculate the molarity of the MgCl2 solution.
19 g MgCl 2
1 mol MgCl 2
1000 mL soln
= 0.80 M MgCl 2
× × 1 L soln
250 mL soln 95.21 g MgCl 2
Step 2
Write the balanced dissolution equation (i.e., equation showing how the strong electrolyte dissolves in water).
H2O
MgCl2(s) → Mg12(aq) 1 2 Cl2(aq)
Step 3
Use the coefficients from the balanced dissolution equation to calculate the concentration of each ion in aqueous solution.
Molarity of Mg12:
0.80 mol Mg +2
0.80 mol MgCl 2
1 mol Mg +2
× = OR 0.80 M Mg +2
L soln
1 L soln
1 mol MgCl 2
↔
coefficients from the
balanced dissolution eqn.
↔
Molarity of Cl2:
0.80 mol MgCl 2
1.6 mol Cl –
2 mol Cl –
× = OR 1.6 M Cl –
L soln
1 L soln
1 mol MgCl 2
154
Molarity of Ions in Solution Appendix A
APPENDIX
A
Name
ID No.
Worksheet
Instructor
Course/Section
Date (of Lab Meeting)
Questions
1. A 0.25 M aqueous solution of CaI2 is ________ M in Ca12 and ________ M in I2.
2. A 0.75 M aqueous solution of Fe(ClO4)3 is ________ M in Fe13 and ________ M
in ClO42.
3. A 2.5 M aqueous solution of Fe2(SO4)3 is ________ M in Fe13 and ________M in
SO422.
4. An aqueous solution of Al2(SO4)3 is 0.36 M in SO422. Calculate the concentration
(in M) of Al2(SO4)3 in this solution.
155
Chemistry 116 General Chemistry
5. Calculate the concentrations of K1 and NO32 in an aqueous solution prepared by
dissolving 30.3 g KNO3 in enough water to make 300. mL of solution.
6. Calculate the concentrations of Al13 and SO422 in an aqueous solution prepared by
dissolving 17.1 g Al2(SO4)3 in enough water to make 400. mL of solution.
7. Calculate the concentrations of Na1 and SO422 in an aqueous solution prepared by
dissolving 852 g Na2SO4 in enough water to make 4.00 L of solution.
8. What mass (in g) of CaCl2 is needed to prepare 100. mL of an aqueous solution
that is 0.25 M in Cl2?
9. If 100. mL of 18.0 M H2SO4 solution are diluted to 10.0 L, what are the concentrations of H1 and SO422 in the diluted solution?
156
APPENDIX
B
Net Ionic Equations
Introduction
For reactions which occur in aqueous solution, the net ionic equation is particularly useful since only those species which participate in the reaction, i.e.,
reacting species, are included. The principal step in writing the net ionic equation
is to determine which molecular species or salts are strong electrolytes in aqueous
solution. Therefore, it is necessary to have a complete understanding of strong
electrolytes before proceeding to the writing of net ionic equations.
Strong Electrolytes
In aqueous solution, strong electrolytes dissolve and dissociate (separate) 100%
to form ions. Thus, an aqueous solution of a strong electrolyte contains a relatively high concentration of ions. For example, when dissolved in water, the
ionic compound NaCl is classified as a strong electrolyte. As a result, NaCl(aq)
really exists in solution as dissociated and separated ions of Na1(aq) and Cl2(aq).
Neutral formula units of NaCl do not exist dissolved in solution. Strong acids,
strong bases, and soluble ionic compounds are classified as strong electrolytes in
aqueous solution.
Most general chemistry books agree that the seven compounds given below are
strong acids (ergo strong electrolytes) when dissolved in water. Consequently,
HCl(aq)
HBr(aq)
HI(aq)
Strong Acids
HClO4(aq)
HClO3(aq)
H2SO4(aq)
HNO3(aq)
an aqueous solution of the strong acid perchloric acid, HClO4(aq), actually
consists of dissociated and separated ions of H1(aq) and ClO42(aq). Neutral
molecules of HClO4 do not exist dissolved in solution. Likewise, an aqueous
solution of the strong acid hydrochloric acid, HCl(aq), really consists of H1(aq)
157
Chemistry 116 General Chemistry
and Cl2(aq). Neutral molecules of HCl do not exist dissolved in solution. The dissolution
equations for HClO4 and HCl are represented as shown below. When dissolved in
HClO4
H2O
→
H1(aq) 1 ClO42(aq) 100%
0% dissolved as
neutral molecules
HCl 100% dissociated and dissolved
as separate ions
H2O
→
H1(aq) 1 Cl2(aq) 100%
0% dissolved as
neutral molecules
100% dissociated and dissolved
as separate ions
water, any other acid, e.g., HC2H3O2, H2CO3, HF, H2S, etc., is a weak acid and therefore
a weak electrolyte. Weak electrolytes dissolve but dissociate less than 100% to form ions.
Thus, an aqueous solution of a weak electrolyte contains a relatively low concentration
of ions. For example, the weak acid acetic acid, HC2H3O2, readily dissolves in water.
However, the majority of this acid (~99%) dissolves as neutral molecules of HC2H3O2.
Only a small amount (~1%) dissociates and exists as separate ions of H1(aq) and
C2H3O22(aq). The dissolution equation for HC2H3O2 is represented as shown below.
HC2H3O2(aq) H2O
~1%
~99% dissolved as
neutral molecules
H1(aq) 1 C2H3O22(aq) ~1% dissociated and dissolved
as separate ions
Most general chemistry books agree that the Group IA metal hydroxides and the heavier
Group IIA metal hydroxides given below are strong bases (ergo strong electrolytes)
when dissolved in water. Consequently, an aqueous solution of the strong base calcium
Strong Bases
IA LiOH(aq) RbOH(aq) NaOH(aq) CsOH(aq)
KOH(aq) IIA
Ca(OH)2(aq)
Sr(OH)2(aq)
Ba(OH)2(aq)
hydroxide, Ca(OH)2(aq), really consists of dissociated and separated ions of Ca12(aq) and
OH2(aq). Neutral formula units of Ca(OH)2 do not exist dissolved in solution. However, if the solution is saturated with Ca12 and OH2 ions, solid undissolved Ca(OH)2
may be present as a precipitate at the bottom of the vessel. The dissolution equation
158
Net Ionic Equations
Appendix B
for Ca(OH)2 is represented as shown below. When dissolved in water, weak bases, such
as ammonia and its derivatives, are classified as weak electrolytes. For example, the weak
Ca(OH)2(s)
H2O
→
100%
0% dissolved as
neutral molecules
Ca12(aq) 1 2 OH2(aq) 100% dissociated and dissolved
as separate ions
base ammonia, NH3, readily dissolves in water. However, the majority of this base (~99%)
dissolves as neutral molecules of NH3. A small amount (~1%) reacts with water to form
separated ions of NH41(aq) and OH2(aq), as shown below in the dissolution equation.
NH3(aq) 1 H2O(l) H2O
~1%
~99% dissolved as
neutral molecules
NH41(aq) 1 OH2(aq) ~1% dissolved as separate ions
Soluble ionic compounds are also classified as strong electrolytes when dissolved in
water. The solubility rules, given in Appendix C, are used to determine whether an ionic
compound is soluble (ergo a strong electrolyte) or insoluble (ergo not a strong electrolyte). For example, consider the two ionic compounds, K2SO4 and BaSO4. The fourth
solubility rule specifies that K2SO4 is soluble, whereas BaSO4 is insoluble. Therefore, of
the two sulfates, only the K2SO4 is a strong electrolyte. Consequently, an aqueous solution of potassium sulfate, K2SO4(aq), really consists of dissociated and separated ions of
K1(aq) and SO422(aq). Neutral formula units of K2SO4 do not exist dissolved in solution.
However, if the solution is saturated with K1 and SO422 ions, solid undissolved K2SO4
may be present as a precipitate at the bottom of the vessel. The dissolution equation for
K2SO4 is represented as shown below. Conversely, the insoluble BaSO4 does not dissolve
to any significant extent and will remain as a precipitate at the bottom of the vessel.1
K2SO4(s)
H2O
→
100%
0% dissolved as
neutral formula units 2 K1(aq) 1 SO422(aq) 100% dissociated and dissolved as
separate ions
1
he small amount of an insoluble ionic compound that does dissolve is dissociated 100% into ions. For example,
T
insoluble BaSO4 has a molar solubility of 131025 M. This small amount of BaSO4 that dissolves, dissociates
100% to give relatively low Ba+2 and SO422 concentrations of 131025 M.
159
Chemistry 116 General Chemistry
Net Ionic Equations
For chemical reactions taking place in aqueous solution, three different balanced
equations can be written, the molecular equation, the full ionic equation, and the net ionic
equation. The molecular equation includes full chemical formulas for all reactants and
products. The full ionic equation shows all reactants and products as they actually exist
in solution, i.e., strong electrolytes are shown dissociated into ions. The net ionic equation includes only the reacting species, i.e., ions or compounds that change in some way
during the course of the reaction. Systematic and correct writing of the three equations
in the order molecular, full ionic, and net ionic will result in correct derivation of the
net ionic equation. This process is outlined in the following two examples.
Example
Aqueous solutions of calcium chloride and sodium carbonate are mixed. Write the net
ionic equation.
Step 1
A. Write correct chemical formulas for each compound given.

 Reactants
sodium carbonate = Na 2CO3 

calcium chloride = CaCl 2
B. If not given, predict products by allowing reactants to trade partners.
Ca12
Cl2
Na1
CO322
CaCO3 
➡ NaC1
 Products

C. Write the balanced molecular equation.
Balanced Molecular Equation:
CaCl2 1 Na2CO3 → CaCO3 1 2 NaCl
Step 2
A. Identify strong electrolytes. Remember…strong electrolytes are strong acids, strong
bases, and soluble ionic compounds. Use Appendix C to determine whether an ionic
compound is soluble or insoluble.
160
CaCl2:
Na2CO3:
CaCO3:
NaCl:
ionic compound → soluble by Rule #3 → strong electrolyte
ionic compound → soluble by Rule #1 → strong electrolyte
ionic compound → insoluble by Rule #6 → precipitates
ionic compound → soluble by Rule #1 → strong electrolyte
Net Ionic Equations
Appendix B
B. Write the full ionic equation by showing all species as they really exist in solution,
i.e., strong electrolytes are shown dissociated into ions.
Balanced Full Ionic Equation:
Ca12 1 2 Cl2 1 2 Na1 1 CO322 → CaCO3(s) 1 2 Na1 1 2 Cl2
Step 3
A. Identify spectator ions. Spectator ions are ions that appear on both sides of the equation in exactly the same form.
Spectator Ions: Na1 and Cl2
These ions are present on both sides of the equation
in the same form (dissolved in solution) and with
the same charge.
Reacting Ions:
Ca12 and CO322
These ions are changing form. On the reactant
side they are dissolved in solution; whereas, on
the product side they are present in an insoluble
precipitate.
B. Write the net ionic equation by removing spectator ions from the full ionic equation and balance (if the molecular equation was not previously balanced in Step 1C).
Balanced Net Ionic Equation:
Ca12 1 CO322 → CaCO3(s)
Example
Aqueous solutions of ammonium chloride and magnesium hydroxide are mixed. Write
the net ionic equation if the products of this reaction are magnesium chloride, ammonia, and water.
Step 1
A. Write correct chemical formulas for each compound given.
ammonium chloride 5 NH4Cl
magnesium hydroxide 5 Mg(OH)2
Reactants
magnesium chloride 5 MgCl2
ammonia 5 NH3
water 5 H2O
Products
161
Chemistry 116 General Chemistry
B. Write the balanced molecular equation.
Balanced Molecular Equation:
2 NH4Cl 1 Mg(OH)2 → MgCl2 1 2 NH3 1 2 H2O
Step 2
A. Identify strong electrolytes.
NH4Cl:
Mg(OH)2:
MgCl2:
NH3:
H2O:
ionic compound → soluble by Rule #1 → strong electrolyte
ionic compound → insoluble by Rule #6 → solid
ionic compound → soluble by Rule #3 → strong electrolyte
molecular compound → weak base → weak electrolyte
molecular compound→ weak or non-electrolyte
B. Write the full ionic equation by showing all species as they really exist in solution,
i.e., strong electrolytes are shown dissociated into ions.
Balanced Full Ionic Equation:
2 NH41 1 2 Cl2 1 Mg(OH)2(s) → Mg12 1 2 Cl2 1 2 NH3 1 2 H2O
Step 3
A. Identify spectator ions.
Spectator Ion: Cl2
B. Write the net ionic equation by removing spectator ions.
Balanced Net Ionic Equation:
162
2 NH41 1 Mg(OH)2(s) → Mg12 1 2 NH3 1 2 H2O
Net Ionic Equations
Appendix B
APPENDIX
B
Name
ID No.
Worksheet
Instructor
Course/Section
Date (of Lab Meeting)
Questions
Predict products and write the balanced net ionic equation for each of the following
reactions. Show all work, including chemical formulae for products as well as intermediate full ionic equations. (Note: BaSO3 is insoluble.)
1. Zn(C2H3O2)2 1 Na2S →
2. Pb(NO3)2 1 NH4Cl →
3. Na3PO4 1 MgCl2 →
4. Cu(NO3)2 1 NaOH →
163
Chemistry 116 General Chemistry
5. ZnSO4 1 BaS →
6. Na2CO3 1 HCl →
(HINT: H2CO3(aq) normally decomposes to CO2(g) and H2O(l))
7. NiS 1 HCl →
8. BaCl2 1 Na2SO3 →
9. Zn 1 HCl →
(HINT: This is a redox reaction. Zn is oxidized to Zn12 and H1 is reduced to H2(g))
10. Hg(NO3)2 1 (NH4)2SO4 →
164
Net Ionic Equations
Appendix B
11. barium acetate 1 ammonium sulfate →
12. calcium hydroxide 1 sodium carbonate →
13. iron(III) nitrate 1 barium hydroxide →
14. barium hydroxide 1 hydrochloric acid →
15. silver nitrate 1 magnesium bromide →
16. acetic acid 1 potassium hydroxide →
165
Chemistry 116 General Chemistry
17. sodium chromate 1 silver nitrate →
18. calcium chlorate 1 sodium phosphate →
19. ammonium chloride 1 sodium hydroxide →
(HINT: NH4OH(aq) does not exist and decomposes to NH3(g or aq) and H2O(l))
20. calcium carbonate 1 sulfuric acid →
(HINT: H2CO3(aq) normally decomposes to CO2(g) and H2O(l))
166