Molarity of Ions in Solution

APPENDIX A Molarity of Ions in Solution ften 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 Na 1 and in an aqueous solution of.0 M Na. Step 1 Write the balanced dissolution equation (i.e., equation showing how the strong electrolyte dissolves in water). H Na (s) Na 1 1 Step Use the coefficients from the balanced dissolution equation to calculate the concentration of each ion in aqueous solution. Molarity of Na 1 :.0 mol Na S 1 L Soln 4 molarity of soln + mol Na 1 mol Na S 4 = coefficients from the balanced dissolution eqn. 4.0 mol Na L Soln + R 4.0 M Na + Molarity of :.0 mol Na S 1 L Soln molarity of soln 1 mol S 1 mol Na S 4 4 coefficients from the balanced dissolution eqn. 4.0 mol S4 = R.0 M L Soln 153

Chemistry 116 General Chemistry Example Calculate the concentrations of Mg 1 and Cl in an aqueous solution prepared by dissolving 19 g MgCl in enough water to prepare 50. ml of solution. Step 1 Calculate the molarity of the MgCl solution. 19 g MgCl 50 ml soln 1 mol MgCl 95.1 g MgCl 1000 ml soln = 0.80 M MgCl 1 L soln Step Write the balanced dissolution equation (i.e., equation showing how the strong electrolyte dissolves in water). H MgCl (s) Mg 1 1 Cl Step 3 Use the coefficients from the balanced dissolution equation to calculate the concentration of each ion in aqueous solution. Molarity of Mg 1 : 0.80 mol MgCl 1 L soln + 1 mol Mg = 1 mol MgCl 0.80 mol Mg L soln + R 0.80 M Mg + coefficients from the balanced dissolution eqn. Molarity of Cl : 0.80 mol MgCl 1 L soln mol Cl = 1 mol MgCl 1.6 mol Cl L soln R 1.6 M Cl 154

Molarity of Ions in Solution Appendix A APPENDIX A Name ID No. Worksheet Instructor Date (of Lab Meeting) Course/Section Questions 1. A 0.5 M aqueous solution of CaI is M in Ca 1 and M in I.. A 0.75 M aqueous solution of Fe(Cl 4 ) 3 is M in Fe 13 and M in Cl 4. 3. A.5 M aqueous solution of Fe ( ) 3 is M in Fe 13 and M in. 4. An aqueous solution of Al ( ) 3 is 0.36 M in. Calculate the concentration (in M) of Al ( ) 3 in this solution. 155

Chemistry 116 General Chemistry 5. Calculate the concentrations of K 1 and N 3 in an aqueous solution prepared by dissolving 30.3 g KN 3 in enough water to make 300. ml of solution. 6. Calculate the concentrations of Al 13 and in an aqueous solution prepared by dissolving 17.1 g Al ( ) 3 in enough water to make 400. ml of solution. 7. Calculate the concentrations of Na 1 and in an aqueous solution prepared by dissolving 85 g Na in enough water to make 4.00 L of solution. 8. What mass (in g) of CaCl is needed to prepare 100. ml of an aqueous solution that is 0.5 M in Cl? 9. If 100. ml of 18.0 M H solution are diluted to 10.0 L, what are the concentrations of H 1 and 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 really exists in solution as dissociated and separated ions of Na 1 and Cl. 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, Strong Acids HCl HCl 4 H HBr HCl 3 HN 3 HI an aqueous solution of the strong acid perchloric acid, HCl 4, actually consists of dissociated and separated ions of H 1 and Cl 4. Neutral molecules of HCl 4 do not exist dissolved in solution. Likewise, an aqueous solution of the strong acid hydrochloric acid, HCl, really consists of H 1 157

Chemistry 116 General Chemistry and Cl. Neutral molecules of HCl do not exist dissolved in solution. The dissolution equations for HCl 4 and HCl are represented as shown below. When dissolved in HCl 4 H H 1 1 Cl 4 100% 0% dissolved as 100% dissociated and dissolved neutral molecules as separate ions H HCl H 1 1 Cl 100% 0% dissolved as 100% dissociated and dissolved neutral molecules as separate ions water, any other acid, e.g., HC, H C 3, HF, H S, 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, HC, readily dissolves in water. However, the majority of this acid (~99%) dissolves as neutral molecules of HC. nly a small amount (~1%) dissociates and exists as separate ions of H 1 and C. The dissolution equation for HC is represented as shown below. HC H H 1 1 C ~1% ~99% dissolved as ~1% dissociated and dissolved neutral molecules 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 IIA LiH RbH Ca(H) NaH CsH Sr(H) KH Ba(H) hydroxide, Ca(H), really consists of dissociated and separated ions of Ca 1 and H. Neutral formula units of Ca(H) do not exist dissolved in solution. However, if the solution is saturated with Ca 1 and H ions, solid undissolved Ca(H) may be present as a precipitate at the bottom of the vessel. The dissolution equation 158

Net Ionic Equations Appendix B for Ca(H) 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(H) (s) H Ca 1 1 H 100% 0% dissolved as 100% dissociated and dissolved neutral molecules as separate ions base ammonia, N, readily dissolves in water. However, the majority of this base (~99%) dissolves as neutral molecules of N. A small amount (~1%) reacts with water to form separated ions of NH 41 and H, as shown below in the dissolution equation. N 1 H (l) H NH 41 1 H ~1% ~99% dissolved as ~1% dissolved as separate ions neutral molecules 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, K and Ba. The fourth solubility rule specifies that K is soluble, whereas Ba is insoluble. Therefore, of the two sulfates, only the K is a strong electrolyte. Consequently, an aqueous solution of potassium sulfate, K, really consists of dissociated and separated ions of K 1 and. Neutral formula units of K do not exist dissolved in solution. However, if the solution is saturated with K 1 and ions, solid undissolved K may be present as a precipitate at the bottom of the vessel. The dissolution equation for K is represented as shown below. Conversely, the insoluble Ba does not dissolve to any significant extent and will remain as a precipitate at the bottom of the vessel. 1 K (s) H K 1 1 100% 0% dissolved as 100% dissociated and dissolved as neutral formula units separate ions 1 The small amount of an insoluble ionic compound that does dissolve is dissociated 100% into ions. For example, insoluble Ba has a molar solubility of 1310 5 M. This small amount of Ba that dissolves, dissociates 100% to give relatively low Ba + and concentrations of 1310 5 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. calcium chloride = CaCl sodium carbonate = Na C 3 Reactants B. If not given, predict products by allowing reactants to trade partners. Ca 1 Cl CaC 3 Products Na 1 C 3 NaC1 C. Write the balanced molecular equation. Balanced Molecular Equation: CaCl 1 Na C 3 CaC 3 1 NaCl Step 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. CaCl : Na C 3 : CaC 3 : 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 160

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: Ca 1 1 Cl 1 Na 1 1 C 3 CaC 3 (s) 1 Na1 1 Cl 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: Na 1 and Cl These ions are present on both sides of the equation in the same form (dissolved in solution) and with the same charge. Reacting Ions: Ca 1 and C 3 These ions are changing form. n 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: Ca 1 1 C 3 CaC 3 (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 NH 4 Cl magnesium hydroxide 5 Mg(H) Reactants magnesium chloride 5 MgCl ammonia 5 N water 5 H Products 161

Chemistry 116 General Chemistry B. Write the balanced molecular equation. Step A. Identify strong electrolytes. Balanced Molecular Equation: NH 4 Cl 1 Mg(H) MgCl 1 N 1 H NH 4 Cl: ionic compound soluble by Rule #1 strong electrolyte Mg(H) : ionic compound insoluble by Rule #6 solid MgCl : ionic compound soluble by Rule #3 strong electrolyte N : molecular compound weak base weak electrolyte H : 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. Step 3 A. Identify spectator ions. Balanced Full Ionic Equation: NH 4 1 1 Cl 1 Mg(H) (s) Mg 1 1 Cl 1 N 1 H Spectator Ion: Cl B. Write the net ionic equation by removing spectator ions. Balanced Net Ionic Equation: NH 4 1 1 Mg(H) (s) Mg 1 1 N 1 H 16

Net Ionic Equations Appendix B APPENDIX B Name ID No. Worksheet Instructor Date (of Lab Meeting) Course/Section 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: BaS 3 is insoluble.) 1. Zn(C ) 1 Na S. Pb(N 3 ) 1 NH 4 Cl 3. Na 3 P 4 1 MgCl 4. Cu(N 3 ) 1 NaH 163

Chemistry 116 General Chemistry 5. Zn 1 BaS 6. Na C 3 1 HCl (HINT: H C 3 normally decomposes to C (g) and H (l)) 7. NiS 1 HCl 8. BaCl 1 Na S 3 9. Zn 1 HCl (HINT: This is a redox reaction. Zn is oxidized to Zn 1 and H 1 is reduced to H (g)) 10. Hg(N 3 ) 1 (NH 4 ) 164

Net Ionic Equations Appendix B 11. barium acetate 1 ammonium sulfate 1. 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: NH 4 H does not exist and decomposes to N (g or aq) and H (l)) 0. calcium carbonate 1 sulfuric acid (HINT: H C 3 normally decomposes to C (g) and H (l)) 166