1. 4-1 Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Seventh Edition Martin S. Silberberg and Patricia G. Amateis Copyright  McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

2. 4-2 Chapter 4 Three Major Classes of Chemical Reactions

3. 4-3 Learning Objectives LO 2.15 The student is able to explain observations regarding the solubility of ionic solids and molecules in water and other solvents on the basis of particle views that include intermolecular interactions and entropic effects. LO 3.4 The student is able to relate quantities (measured mass of substances, volumes of solutions, or volumes and pressures of gases) to identify stoichiometric relationships for a reaction, including situations involving limiting reactants and situations in which the reaction has not gone to completion.

4. 4-4 Precipitation Reactions In a precipitation reaction two soluble ionic compounds react to give an insoluble product, called a precipitate. The precipitate forms through the net removal of ions from solution. It is possible for more than one precipitate to form in such a reaction.

5. 4-5 Figure 4.8The precipitation of calcium fluoride. Ca 2+ (aq) + 2F – (aq) → CaF 2 (s) 2Na + (aq) + 2F – (aq) + Ca 2+ (aq) + 2Cl – (aq) → CaF 2 (s) + 2Na + (aq) + 2Cl – (aq) 2NaF (aq) + CaCl 2 (aq) → CaF 2 (s) + 2NaCl (aq)

6. 4-6 Predicting Whether a Precipitate Will Form Note the ions present in the reactants. Consider all possible cation-anion combinations. Use the solubility rules to decide whether any of the ion combinations is insoluble. –Any insoluble combination identifies a precipitate that will form.

7. 4-7

8. 4-8 Figure 4.9 The precipitation of PbI 2, a metathesis reaction. Precipitation reactions are also called double displacement reactions or metathesis reactions. 2NaI (aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + NaNO 3 (aq) 2Na + (aq) + 2I – (aq) + Pb 2+ (aq) + 2NO 3 – (aq) → PbI 2 (s) + 2Na + (aq) + 2NO 3 - (aq) 2NaI (aq) + Pb(NO 3 ) 2 (aq) → PbI 2 (s) + 2NaNO 3 (aq) Pb 2+ (aq) + 2I - (aq) → PbI 2 (s) Ions exchange partners and a precipitate forms, so there is an exchange of bonds between reacting species.

9. 4-9 Sample Problem 4.8 Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations PROBLEM:Predict whether or not a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) potassium fluoride (aq) + strontium nitrate (aq) → (b) ammonium perchlorate (aq) + sodium bromide (aq) → Note reactant ions, write the possible cation-anion combinations, and use Table 4.1 to decide if the combinations are insoluble. Write the appropriate equations for the process. PLAN:

10. 4-10 SOLUTION: K + and NO 3 - are spectator ions Sample Problem 4.8 Molecular equation: 2KF (aq) + Sr(NO 3 ) 2 (aq) → 2KNO 3 (aq) + SrF 2 (s) (a) The reactants are KF and Sr(NO 3 ) 2. The possible products are KNO 3 and SrF 2. KNO 3 is soluble, but SrF 2 is an insoluble combination. Net ionic equation: Sr 2+ (aq) + 2F - (aq) → SrF 2 (s) Total ionic equation: 2K + (aq) + 2F - (aq) + Sr 2+ (aq) + 2NO 3 - (aq) → 2K + (aq) + 2NO 3 - (aq) + SrF 2 (s)

11. 4-11 SOLUTION: Sample Problem 4.8 Molecular equation: NH 4 ClO 4 (aq) + NaBr (aq) → NH 4 Br (aq) + NaClO 4 (aq) (b) The reactants are NH 4 ClO 4 and NaBr. The possible products are NH 4 Br and NaClO 4. Both are soluble, so no precipitate forms. All ions are spectator ions and there is no net ionic equation. The compounds remain in solution as solvated ions. Total ionic equation: NH 4 + (aq) + ClO 4 - (aq) + Na + (aq) + Br - (aq) → NH 4 + (aq) + Br - (aq) + Na + (aq) + ClO 4 - (aq)

12. 4-12 Sample Problem 4.9 Using Molecular Depictions in Precipitation Reactions PROBLEM: The following molecular views show reactant solutions for a precipitation reaction (with H 2 O molecules omitted for clarity). (a)Which compound is dissolved in beaker A: KCl, Na 2 SO 4, MgBr 2, or Ag 2 SO 4 ? (b)Which compound is dissolved in beaker B: NH 4 NO 3, MgSO 4, Ba(NO 3 ) 2, or CaF 2 ? Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

13. 4-13 Sample Problem 4.9 PLAN:Note the number and charge of each kind of ion and use Table 4.1 to determine the ion combinations that are soluble. SOLUTION: (a)Beaker A contains two 1+ ions for each 2- ion. Of the choices given, only Na 2 SO 4 and Ag 2 SO 4 are possible. Na 2 SO 4 is soluble while Ag 2 SO 4 is not. Beaker A therefore contains Na 2 SO 4. (b)Beaker B contains two 1- ions for each 2+ ion. Of the choices given, only CaF 2 and Ba(NO 3 ) 2 match this description. CaF 2 is not soluble while Ba(NO 3 ) 2 is soluble. Beaker B therefore contains Ba(NO 3 ) 2.

14. 4-14 PROBLEM: (c)Name the precipitate and spectator ions when solutions A and B are mixed, and write balanced molecular, total ionic, and net ionic equations for this process. (d)If each particle represents 0.010 mol of ions, what is the maximum mass (g) of precipitate that can form (assuming complete reaction)? Sample Problem 4.9 PLAN: (c) Consider the cation-anion combinations from the two solutions and use Table 4.1 to decide if either of these is insoluble. SOLUTION:The reactants are Ba(NO 3 ) 2 and Na 2 SO 4. The possible products are BaSO 4 and NaNO 3. BaSO 4 is insoluble while NaNO 3 is soluble.

15. 4-15 Sample Problem 4.9 Na + and NO 3 - are spectator ions Molecular equation: Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) → 2NaNO 3 (aq) + BaSO 4 (s) Net ionic equation: Ba 2+ (aq) + SO 4 2- (aq) → BaSO 4 (s) Total ionic equation: Ba 2+ (aq) + 2NO 3 - (aq) + 2Na + (aq) + SO 4 2- (aq) → 2Na + (aq) + 2NO 3 - (aq) + BaSO 4 (s)

16. 4-16 Sample Problem 4.9 PLAN: (d) Count the number of each kind of ion that combines to form the solid. Multiply the number of each reactant ion by 0.010 mol and calculate the mol of product formed from each. Decide which ion is the limiting reactant and use this information to calculate the mass of product formed. SOLUTION:There are 4 Ba 2+ particles and 5 SO 4 2- particles depicted. 4 Ba 2+ particles x 0.010 mol Ba 2+ 1 particle x 1 mol BaSO 4 1 mol Ba 2+ = 0.040 mol BaSO 4 4 SO 4 2- particles x 0.010 mol SO 4 2- 1 particle x 1 mol BaSO 4 1 mol SO 4 2- = 0.050 mol BaSO 4

17. 4-17 Sample Problem 4.9 233.4 g BaSO 4 1 mol BaSO 4 0.040 mol BaSO 4 x = 9.3 g BaSO 4 Ba 2+ ion is the limiting reactant, since the given amount yields less BaSO 4.

18. 4-18 Sample Problem 4.10 Calculating Amounts of Reactants and Products in a Precipitation Reaction PROBLEM:Magnesium is the second most abundant metal in sea water after sodium. The first step in its industrial extraction involves the reaction of Mg 2+ with Ca(OH) 2 to precipitate Mg(OH) 2. What mass of Mg(OH) 2 is formed when 0.180 L of 0.0155 M MgCl 2 reacts with excess Ca(OH) 2 ? PLAN: We are given the molarity (0.0155 M) and the volume (0.180 L) of MgCl 2 solution that reacts with excess Ca(OH) 2, and we must calculate the mass of the precipitate, Mg(OH) 2. (i)Write the balanced equation. (ii)Find the amount (mol) of MgCl 2 from its molarity and volume. (iii)Use molar (stoichiometric) ratios to calculate the amount (mol) of Mg(OH) 2 precipitated. (iv)Use molar mass of Mg(OH) 2 to convert amount (mol) to mass (g).

19. 4-19 Sample Problem 4.10 multiply by M (mol/L) 1 L MgCl 2 solution = 0.0155 mol MgCl 2 Volume (L) of MgCl 2 solution Amount (mol) of MgCl 2 Mass (g) of Mg(OH) 2 multiply by M (g/mol) Amount (mol) of Mg(OH) 2 molar ratio as conversion factor Road Map

20. 4-20 Sample Problem 4.10 = 0.00279 mol MgCl 2 0.0155 mol MgCl 2 1 L MgCl 2 SOLUTION: (i)Write the balanced equation. MgCl 2 (aq) + Ca(OH) 2 (aq) → Mg(OH) 2 (s) + CaCl 2 (aq) (ii) Find the amount (mol) of MgCl 2 from its molarity and volume. Amount (mol) of MgCl 2 = 0.180 L MgCl 2 x (iii) Use molar (stoichiometric) ratios to calculate the amount (mol) of Mg(OH) 2 precipitated. Amount (mol) of Mg(OH) 2 = 0.00279 mol MgCl 2 x 1 mol MgCl 2 1 mol Mg(OH) 2 = 0.00279 mol Mg(OH) 2

21. 4-21 Sample Problem 4.10 SOLUTION: (iv) Use molar mass of Mg(OH) 2 to convert amount (mol) to mass (g). Mass (g) of Mg(OH) 2 = 0.00279 mol Mg(OH) 2 x CHECK: The answer seems reasonable; rounding to check the math shows that we have (0.2 L) x (0.02 M) = 0.004 mol MgCl 2 ; due to the 1:1 molar ratio, 0.004 mol Mg(OH) 2 is produced. 58.33 g Mg(OH) 2 1 mol Mg(OH) 2 = 0.163 g Mg(OH) 2

22. 4-22 Figure 4.10 Summary of amount-mass-number relationships for a chemical reaction in solution.

23. 4-23 Sample Problem 4.11 Solving Limiting-Reactant Problems for Precipitation Reactions PROBLEM:In a simulation mercury removal from industrial wastewater, 0.050 L of 0.010 M mercury(II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams of mercury(II) sulfide form? Write a reaction table for this process. PLAN:Write a balanced chemical reaction. Determine limiting reactant. Calculate the grams of mercury(II) sulfide product. multiply by M mole ratio Mass of HgS multiply by M mole ratio Volume of Hg(NO 3 ) 2 soln Amount (mol) of Hg(NO 3 ) 2 Amount (mol) of HgS Volume of Na 2 S soln Amount (mol) of Na 2 S Amount (mol) of HgS select lower number of moles of HgS multiply by M

24. 4-24 = 5.0 x 10 -4 mol HgS = 2.0 x 10 -3 mol HgS Hg(NO 3 ) 2 is the limiting reactant because the given amount yields less HgS. 5.0 x 10 -4 mol HgS x 232.7 g HgS 1 mol HgS = 0.12 g HgS Sample Problem 4.11 SOLUTION:Hg(NO 3 ) 2 (aq) + Na 2 S (aq) → HgS (s) + 2NaNO 3 (aq) 0.050 L Hg(NO 3 ) 2 x 1 mol HgS 1 mol Hg(NO 3 ) 2 0.010 mol Hg(NO 3 ) 2 1 L Hg(NO 3 ) 2 x0.020 L Na 2 S x 1 mol HgS 1 mol Na 2 S 0. 10 mol Na 2 S 1 L Na 2 S x

25. 4-25 AmountHg(NO 3 ) 2 (aq) +Na 2 S (aq) →HgS (s) +2NaNO 3 (aq) Initial Change 5.0 x 10 -4 -5.0 x 10 -4 2.0 x 10 -3 -5.0 x 10 -4 0 +5.0 x 10 -4 0 +1.0 x 10 -3 Final 0 1.5 x 10 -3 5.0 x 10 -4 +1.0 x 10 -3 Sample Problem 4.11 The reaction table is constructed using the amount of Hg(NO 3 ) 2 to determine the changes, since it is the limiting reactant.