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