1. Chapter 4 Chemical Quantities and Aqueous Reactions 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro

2. Tro, Chemistry: A Molecular Approach2 Reaction Stoichiometry the numerical relationships between chemical amounts in a reaction is called stoichiometry the coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) 2 molecules of C 8 H 18 react with 25 molecules of O 2 to form 16 molecules of CO 2 and 18 molecules of H 2 O 2 moles of C 8 H 18 react with 25 moles of O 2 to form 16 moles of CO 2 and 18 moles of H 2 O 2 mol C 8 H 18 : 25 mol O 2 : 16 mol CO 2 : 18 mol H 2 O

3. Tro, Chemistry: A Molecular Approach3 Predicting Amounts from Stoichiometry the amounts of any other substance in a chemical reaction can be determined from the amount of just one substance How much CO 2 can be made from 22.0 moles of C 8 H 18 in the combustion of C 8 H 18 ? 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) 2 moles C 8 H 18 : 16 moles CO 2

4. Tro, Chemistry: A Molecular Approach4 Example – Estimate the mass of CO 2 produced in 2004 by the combustion of 3.4 x 10 15 g gasoline assuming that gasoline is octane, C 8 H 18, the equation for the reaction is: 2 C 8 H 18 (l) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g) the equation for the reaction gives the mole relationship between amount of C 8 H 18 and CO 2, but we need to know the mass relationship, so the Concept Plan will be: g C 8 H 18 mol CO 2 g CO 2 mol C 8 H 18

5. Example – Estimate the mass of CO 2 produced in 2004 by the combustion of 3.4 x 10 15 g gasoline since 8x moles of CO 2 as C 8 H 18, but the molar mass of C 8 H 18 is 3x CO 2, the number makes sense 1 mol C 8 H 18 = 114.22g, 1 mol CO 2 = 44.01g, 2 mol C 8 H 18 = 16 mol CO 2 3.4 x 10 15 g C 8 H 18 g CO 2 Check: Solution: Concept Plan: Relationships: Given: Find: g C 8 H 18 mol CO 2 g CO 2 mol C 8 H 18

6. Tro, Chemistry: A Molecular Approach6 Practice According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose? C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O(l) 1.convert 9.0 g of glucose into moles (MM 180) 2.convert moles of glucose into moles of water 3.convert moles of water into grams (MM 18.02) 4.convert grams of water into mL a)How? what is the relationship between mass and volume? density of water = 1.00 g/mL

7. Tro, Chemistry: A Molecular Approach7 Practice According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose? C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O (l)

8. Tro, Chemistry: A Molecular Approach8 Limiting Reactant for reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others when this reactant is used up, the reaction stops and no more product is made the reactant that limits the amount of product is called the limiting reactant sometimes called the limiting reagent the limiting reactant gets completely consumed reactants not completely consumed are called excess reactants the amount of product that can be made from the limiting reactant is called the theoretical yield

9. Tro, Chemistry: A Molecular Approach9 Things Don’t Always Go as Planned! many things can happen during the course of an experiment that cause the loss of product the amount of product that is made in a reaction is called the actual yield generally less than the theoretical yield, never more! the efficiency of product recovery is generally given as the percent yield

10. Tro, Chemistry: A Molecular Approach10 Limiting and Excess Reactants in the Combustion of Methane CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) Our balanced equation for the combustion of methane implies that every 1 molecule of CH 4 reacts with 2 molecules of O 2 H H C H H + O O C + OO OO + O HH O HH +

11. 11 Limiting and Excess Reactants in the Combustion of Methane If we have 5 molecules of CH 4 and 8 molecules of O 2, which is the limiting reactant? H H C H H + OO OO OO OO OO OO OO OO ? H H C H H H H C H H H H C H H H H C H H CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)

12. 12 Limiting and Excess Reactants in the Combustion of Methane H H C H H H H C H H + OO OO OO OO OO OO OO OO H H C H H H H C H H H H C H H CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) since less CO 2 can be made from the O 2 than the CH 4, the O 2 is the limiting reactant

13. Example 4.4 Finding Limiting Reactant, Theoretical Yield, and Percent Yield

14. Tro, Chemistry: A Molecular Approach14 Example: When 28.6 kg of C are allowed to react with 88.2 kg of TiO 2 in the reaction below, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield.

15. Tro, Chemistry: A Molecular Approach15 Example: When 28.6 kg of C reacts with 88.2 kg of TiO 2, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Write down the given quantity and its units. Given:28.6 kg C 88.2 kg TiO 2 42.8 kg Ti produced

16. Tro, Chemistry: A Molecular Approach16 Write down the quantity to find and/or its units. Find: limiting reactant theoretical yield percent yield Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti

17. 17 Write a Concept Plan : Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) kg TiO 2 kg C mol C mol TiO 2 mol Ti mol Ti } smallest amount is from limiting reactant g TiO 2 gCgC smallest mol Ti g Ti% Yield kg Ti T.Y.

18. 18 Collect Needed Relationships: 1000 g = 1 kg Molar Mass TiO 2 = 79.87 g/mol Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol 1 mole TiO 2 : 1 mol Ti (from the chem. equation) 2 mole C  1 mol Ti (from the chem. equation) Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g)

19. Tro, Chemistry: A Molecular Approach19 Apply the Concept Plan: Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g) smallest moles of Ti Limiting Reactant

20. Tro, Chemistry: A Molecular Approach20 Apply the Concept Plan: Theoretical Yield Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g)

21. Tro, Chemistry: A Molecular Approach21 Apply the Concept Plan: Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g)

22. Tro, Chemistry: A Molecular Approach22 Check the Solutions: Limiting Reactant = TiO 2 Theoretical Yield = 52.9 kg Percent Yield = 80.9% Since Ti has lower molar mass than TiO 2, the T.Y. makes sense The Percent Yield makes sense as it is less than 100%. Information Given:28.6 kg C, 88.2 kg TiO 2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld. CP: kg rct  g rct  mol rct  mol Ti pick smallest mol Ti  TY kg Ti  %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO 2 = 79.87g; 1000g = 1 kg; 1 mol TiO 2 : 1 mol Ti; 2 mol C : 1 mol Ti Example: Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO 2 (s) + 2 C(s)  Ti(s) + 2 CO(g)

23. Tro, Chemistry: A Molecular Approach23 Practice – How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l)

24. 1 mol NH 3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N 2 = 28.02 g 2 mol NH 3 = 1 mol N 2, 3 mol CuO = 1 mol N 2 9.05 g NH 3, 45.2 g CuO g N 2 Concept Plan: Relationships: Given: Find: g NH 3 mol N 2 mol NH 3 g CuOmol N 2 mol CuOg N 2 smallest moles N 2

25. Practice – How many grams of N 2 (g) can be made from 9.05 g of NH 3 reacting with 45.2 g of CuO? 2 NH 3 (g) + 3 CuO(s) → N 2 (g) + 3 Cu(s) + 3 H 2 O(l) units are correct, and since there are fewer moles of N 2 than CuO in the reaction and N 2 has a smaller mass, the number makes sense Check: Solution:

26. Tro, Chemistry: A Molecular Approach26 Solutions when table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can tell from the taste, or simply boiling away the water homogeneous mixtures are called solutions the component of the solution that changes state is called the solute the component that keeps its state is called the solvent if both components start in the same state, the major component is the solvent

27. Tro, Chemistry: A Molecular Approach27 Describing Solutions since solutions are mixtures, the composition can vary from one sample to another pure substances have constant composition salt water samples from different seas or lakes have different amounts of salt so to describe solutions accurately, we must describe how much of each component is present we saw that with pure substances, we can describe them with a single name because all samples identical

28. Tro, Chemistry: A Molecular Approach28 Solution Concentration qualitatively, solutions are often described as dilute or concentrated dilute solutions have a small amount of solute compared to solvent concentrated solutions have a large amount of solute compared to solvent quantitatively, the relative amount of solute in the solution is called the concentration

29. Tro, Chemistry: A Molecular Approach29 Solution Concentration Molarity moles of solute per 1 liter of solution used because it describes how many molecules of solute in each liter of solution

30. Tro, Chemistry: A Molecular Approach30 Preparing 1 L of a 1.00 M NaCl Solution

31. Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution since most solutions are between 0 and 18 M, the answer makes sense 1 mol KBr = 119.00 g, M = moles/L 25.5 g KBr, 1.75 L solution Molarity, M Check: Check Solution: Follow the Concept Plan to Solve the problem Concept Plan: Relationships: Strategize Given: Find: Sort Information g KBrmol KBr L sol’n M

32. Tro, Chemistry: A Molecular Approach32 Using Molarity in Calculations molarity shows the relationship between the moles of solute and liters of solution If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar 2 liters = 4.0 moles sugar 0.5 liters = 1.0 mole sugar 1 L solution : 2 moles sugar

33. Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH? since each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L 0.125 mol NaOH = 1 L solution 0.125 M NaOH, 0.255 mol NaOH liters, L Check: Check Solution: Follow the Concept Plan to Solve the problem Concept Plan: Relationships: Strategize Given: Find: Sort Information mol NaOHL sol’n

34. Tro, Chemistry: A Molecular Approach34 Dilution often, solutions are stored as concentrated stock solutions to make solutions of lower concentrations from these stock solutions, more solvent is added the amount of solute doesn’t change, just the volume of solution moles solute in solution 1 = moles solute in solution 2 the concentrations and volumes of the stock and new solutions are inversely proportional M 1 ∙V 1 = M 2 ∙V 2

35. Example 4.7 – To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH? since the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does M 1 V 1 = M 2 V 2 V 1 = 0.200L, M 1 = 15.0 M, M 2 = 3.00 M V 2, L Check: Check Solution: Follow the Concept Plan to Solve the problem Concept Plan: Relationships: Strategize Given: Find: Sort Information V 1, M 1, M 2 V2V2

36. Tro, Chemistry: A Molecular Approach36 Solution Stoichiometry since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

37. Example 4.8 – What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO 3 ) 2 in the reaction 2 KCl(aq) + Pb(NO 3 ) 2 (aq)  PbCl 2 (s) + 2 KNO 3 (aq) since need 2x moles of KCl as Pb(NO 3 ) 2, and the molarity of Pb(NO 3 ) 2 > KCl, the volume of KCl should be more than 2x volume Pb(NO 3 ) 2 1 L Pb(NO 3 ) 2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO 3 ) 2 = 2 mol KCl 0.150 M KCl, 0.150 L of 0.175 M Pb(NO 3 ) 2 L KCl Check: Check Solution: Follow the Concept Plan to Solve the problem Concept Plan: Relationships: Strategize Given: Find: Sort Information L Pb(NO 3 ) 2 mol KClL KClmol Pb(NO 3 ) 2

38. 38 What Happens When a Solute Dissolves? there are attractive forces between the solute particles holding them together there are also attractive forces between the solvent molecules when we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules if the attractions between solute and solvent are strong enough, the solute will dissolve

39. Tro, Chemistry: A Molecular Approach39 Table Salt Dissolving in Water Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity

40. Tro, Chemistry: A Molecular Approach40 Electrolytes and Nonelectrolytes materials that dissolve in water to form a solution that will conduct electricity are called electrolytes materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes

41. Tro, Chemistry: A Molecular Approach41 Molecular View of Electrolytes and Nonelectrolytes in order to conduct electricity, a material must have charged particles that are able to flow electrolyte solutions all contain ions dissolved in the water ionic compounds are electrolytes because they all dissociate into their ions when they dissolve nonelectrolyte solutions contain whole molecules dissolved in the water generally, molecular compounds do not ionize when they dissolve in water  the notable exception being molecular acids

42. Tro, Chemistry: A Molecular Approach42 Salt vs. Sugar Dissolved in Water ionic compounds dissociate into ions when they dissolve molecular compounds do not dissociate when they dissolve

43. Tro, Chemistry: A Molecular Approach43 Acids acids are molecular compounds that ionize when they dissolve in water the molecules are pulled apart by their attraction for the water when acids ionize, they form H + cations and anions the percentage of molecules that ionize varies from one acid to another acids that ionize virtually 100% are called strong acids HCl(aq)  H + (aq) + Cl - (aq) acids that only ionize a small percentage are called weak acids HF(aq)  H + (aq) + F - (aq)

44. Tro, Chemistry: A Molecular Approach44 Strong and Weak Electrolytes strong electrolytes are materials that dissolve completely as ions ionic compounds and strong acids their solutions conduct electricity well weak electrolytes are materials that dissolve mostly as molecules, but partially as ions weak acids their solutions conduct electricity, but not well when compounds containing a polyatomic ion dissolve, the polyatomic ion stays together Na 2 SO 4 (aq)  2 Na + (aq) + SO 4 2- (aq) HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 - (aq)

45. Tro, Chemistry: A Molecular Approach45 Classes of Dissolved Materials

46. Tro, Chemistry: A Molecular Approach46 Solubility of Ionic Compounds some ionic compounds, like NaCl, dissolve very well in water at room temperature other ionic compounds, like AgCl, dissolve hardly at all in water at room temperature compounds that dissolve in a solvent are said to be soluble, while those that do not are said to be insoluble NaCl is soluble in water, AgCl is insoluble in water the degree of solubility depends on the temperature even insoluble compounds dissolve, just not enough to be meaningful

47. Tro, Chemistry: A Molecular Approach47 When Will a Salt Dissolve? Predicting whether a compound will dissolve in water is not easy The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results we call this method the empirical method

48. Tro, Chemistry: A Molecular Approach48 Compounds Containing the Following Ions are Generally Soluble Exceptions (when combined with ions on the left the compound is insoluble) Li +, Na +, K +, NH 4 + none NO 3 –, C 2 H 3 O 2 – none Cl –, Br –, I – Ag +, Hg 2 2+, Pb 2+ SO 4 2– Ag +, Ca 2+, Sr 2+, Ba 2+, Pb 2+ Solubility Rules Compounds that Are Generally Soluble in Water

49. Tro, Chemistry: A Molecular Approach49 Compounds Containing the Following Ions are Generally Insoluble Exceptions (when combined with ions on the left the compound is soluble or slightly soluble) OH – Li +, Na +, K +, NH 4 +, Ca 2+, Sr 2+, Ba 2+ S 2– Li +, Na +, K +, NH 4 +, Ca 2+, Sr 2+, Ba 2+ CO 3 2–, PO 4 3– Li +, Na +, K +, NH 4 + Solubility Rules Compounds that Are Generally Insoluble

50. Tro, Chemistry: A Molecular Approach50 Precipitation Reactions reactions between aqueous solutions of ionic compounds that produce an ionic compound that is insoluble in water are called precipitation reactions and the insoluble product is called a precipitate

51. 51 2 KI(aq) + Pb(NO 3 ) 2 (aq)  PbI 2 (s) + 2 KNO 3 (aq)

52. Tro, Chemistry: A Molecular Approach52 No Precipitate Formation = No Reaction KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq) all ions still present,  no reaction

53. Tro, Chemistry: A Molecular Approach53 Process for Predicting the Products of a Precipitation Reaction 1. Determine what ions each aqueous reactant has 2. Determine formulas of possible products Exchange ions  (+) ion from one reactant with (-) ion from other Balance charges of combined ions to get formula of each product 3. Determine Solubility of Each Product in Water Use the solubility rules If product is insoluble or slightly soluble, it will precipitate 4. If neither product will precipitate, write no reaction after the arrow

54. Tro, Chemistry: A Molecular Approach54 Process for Predicting the Products of a Precipitation Reaction 5. If either product is insoluble, write the formulas for the products after the arrow – writing (s) after the product that is insoluble and will precipitate, and (aq) after products that are soluble and will not precipitate 6. Balance the equation

55. Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 1. Write the formulas of the reactants K 2 CO 3 (aq) + NiCl 2 (aq)  2. Determine the possible products a)Determine the ions present (K + + CO 3 2- ) + (Ni 2+ + Cl - )  b)Exchange the Ions (K + + CO 3 2- ) + (Ni 2+ + Cl - )  (K + + Cl - ) + (Ni 2+ + CO 3 2- ) c)Write the formulas of the products  cross charges and reduce K 2 CO 3 (aq) + NiCl 2 (aq)  KCl + NiCO 3

56. Tro, Chemistry: A Molecular Approach56 3. Determine the solubility of each product KCl is soluble NiCO 3 is insoluble 4. If both products soluble, write no reaction does not apply since NiCO 3 is insoluble Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride

57. Tro, Chemistry: A Molecular Approach57 5. Write (aq) next to soluble products and (s) next to insoluble products K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s) 6. Balance the Equation K 2 CO 3 (aq) + NiCl 2 (aq)  KCl(aq) + NiCO 3 (s) Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride

58. Tro, Chemistry: A Molecular Approach58 Ionic Equations equations which describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) + Mg(NO 3 ) 2 (aq)  2 KNO 3 (aq) + Mg(OH) 2 (s) equations which describe the actual dissolved species are called complete ionic equations aqueous strong electrolytes are written as ions  soluble salts, strong acids, strong bases insoluble substances, weak electrolytes, and nonelectrolytes written in molecule form  solids, liquids, and gases are not dissolved, therefore molecule form 2K +1 (aq) + 2OH -1 (aq) + Mg +2 (aq) + 2NO 3 -1 (aq)  K +1 (aq) + 2NO 3 -1 (aq) + Mg(OH) 2(s)

59. Tro, Chemistry: A Molecular Approach59 Ionic Equations ions that are both reactants and products are called spectator ions 2K +1 (aq) + 2OH -1 (aq) + Mg +2 (aq) + 2NO 3 -1 (aq)  K +1 (aq) + 2NO 3 -1 (aq) + Mg(OH) 2(s) an ionic equation in which the spectator ions are removed is called a net ionic equation 2OH -1 (aq) + Mg +2 (aq)  Mg(OH) 2(s)

60. Tro, Chemistry: A Molecular Approach60 Acid-Base Reactions also called neutralization reactions because the acid and base neutralize each other’s properties 2 HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) the net ionic equation for an acid-base reaction is H + (aq) + OH  (aq)  H 2 O(l) as long as the salt that forms is soluble in water

61. Tro, Chemistry: A Molecular Approach61 Acids and Bases in Solution acids ionize in water to form H + ions more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H 3 O +  most chemists use H + and H 3 O + interchangeably bases dissociate in water to form OH  ions bases, like NH 3, that do not contain OH  ions, produce OH  by pulling H off water molecules in the reaction of an acid with a base, the H + from the acid combines with the OH  from the base to make water the cation from the base combines with the anion from the acid to make the salt acid + base  salt + water

62. Common Acids

63. Tro, Chemistry: A Molecular Approach63 Common Bases

64. Tro, Chemistry: A Molecular Approach64 HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)

65. Tro, Chemistry: A Molecular Approach65 Example - Write the molecular, ionic, and net- ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 1. Write the formulas of the reactants HNO 3 (aq) + Ca(OH) 2 (aq)  2. Determine the possible products a)Determine the ions present when each reactant dissociates (H + + NO 3 - ) + (Ca +2 + OH - )  b)Exchange the ions, H +1 combines with OH -1 to make H 2 O(l) (H + + NO 3 - ) + (Ca +2 + OH - )  (Ca +2 + NO 3 - ) + H 2 O(l) c)Write the formula of the salt cross the charges (H + + NO 3 - ) + (Ca +2 + OH - )  Ca(NO 3 ) 2 + H 2 O(l)

66. Tro, Chemistry: A Molecular Approach66 3. Determine the solubility of the salt Ca(NO 3 ) 2 is soluble 4. Write an (s) after the insoluble products and a (aq) after the soluble products HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + H 2 O(l) 5. Balance the equation 2 HNO 3 (aq) + Ca(OH) 2 (aq)  Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) Example - Write the molecular, ionic, and net- ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide

67. Tro, Chemistry: A Molecular Approach67 Example - Write the molecular, ionic, and net- ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 6. Dissociate all aqueous strong electrolytes to get complete ionic equation not H 2 O 2 H + (aq) + 2 NO 3 - (aq) + Ca +2 (aq) + 2 OH - (aq)  Ca +2 (aq) + 2 NO 3 - (aq) + H 2 O(l) 7. Eliminate spectator ions to get net-ionic equation 2 H +1 (aq) + 2 OH -1 (aq)  H 2 O(l) H +1 (aq) + OH -1 (aq)  H 2 O(l)

68. Tro, Chemistry: A Molecular Approach68 Titration often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio the unknown solution is added slowly from an instrument called a burette  a long glass tube with precise volume markings that allows small additions of solution

69. Tro, Chemistry: A Molecular Approach69 Acid-Base Titrations the difficulty is determining when there has been just enough titrant added to complete the reaction the titrant is the solution in the burette in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity the chemical is called an indicator at the endpoint of an acid-base titration, the number of moles of H + equals the number of moles of OH  aka the equivalence point

70. Tro, Chemistry: A Molecular Approach70 Titration

71. Tro, Chemistry: A Molecular Approach71 Titration The base solution is the titrant in the burette. As the base is added to the acid, the H + reacts with the OH – to form water. But there is still excess acid present so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color.

72. Tro, Chemistry: A Molecular Approach72 Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Write down the given quantity and its units. Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH

73. Tro, Chemistry: A Molecular Approach73 Write down the quantity to find, and/or its units. Find: concentration HCl, M Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

74. Tro, Chemistry: A Molecular Approach74 Collect Needed Equations and Conversion Factors: HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l)  1 mole HCl = 1 mole NaOH 0.100 M NaOH  0.100 mol NaOH  1 L sol’n Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH Find:M HCl Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

75. Tro, Chemistry: A Molecular Approach75 Write a Concept Plan: mL NaOH L NaOH mol NaOH mol HCl Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH Find:M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Example 4.14: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? mL HCl L HCl

76. Tro, Chemistry: A Molecular Approach76 Apply the Solution Map: = 1.25 x 10 -3 mol HCl Information Given:10.00 mL HCl 12.54 mL of 0.100 M NaOH Find:M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

77. Tro, Chemistry: A Molecular Approach77 Apply the Concept Plan: Information Given:10.00 mL HCl 12.54 mL NaOH Find:M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

78. Tro, Chemistry: A Molecular Approach78 Check the Solution: HCl solution = 0.125 M The units of the answer, M, are correct. The magnitude of the answer makes sense since the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated. Information Given:10.00 mL HCl 12.54 mL NaOH Find:M HCl CF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Example: The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

79. Tro, Chemistry: A Molecular Approach79 Gas Evolving Reactions Some reactions form a gas directly from the ion exchange K 2 S(aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + H 2 S(g) Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K 2 SO 3 (aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + H 2 SO 3 (aq) H 2 SO 3  H 2 O(l) + SO 2 (g)

80. Tro, Chemistry: A Molecular Approach80 NaHCO 3 (aq) + HCl(aq)  NaCl(aq) + CO 2 (g) + H 2 O(l)

81. Tro, Chemistry: A Molecular Approach81 Compounds that Undergo Gas Evolving Reactions Reactant Type Reacting With Ion Exchange Product Decom- pose? Gas Formed Example metal n S, metal HS acidH2SH2SnoH2SH2S K 2 S(aq) + 2HCl(aq)  2KCl(aq) + H 2 S(g) metal n CO 3, metal HCO 3 acidH 2 CO 3 yesCO 2 K 2 CO 3 (aq) + 2HCl(aq)  2KCl(aq) + CO 2 (g) + H 2 O(l) metal n SO 3 metal HSO 3 acidH 2 SO 3 yesSO 2 K 2 SO 3 (aq) + 2HCl(aq)  2KCl(aq) + SO 2 (g) + H 2 O(l) (NH 4 ) n anionbaseNH 4 OHyesNH 3 KOH(aq) + NH 4 Cl(aq)  KCl(aq) + NH 3 (g) + H 2 O(l)

82. Tro, Chemistry: A Molecular Approach82 Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 1. Write the formulas of the reactants Na 2 CO 3 (aq) + HNO 3 (aq)  2. Determine the possible products a)Determine the ions present when each reactant dissociates (Na +1 + CO 3 -2 ) + (H +1 + NO 3 -1 )  b)Exchange the anions (Na +1 + CO 3 -2 ) + (H +1 + NO 3 -1 )  (Na +1 + NO 3 -1 ) + (H +1 + CO 3 -2 ) c)Write the formula of compounds cross the charges Na 2 CO 3 (aq) + HNO 3 (aq)  NaNO 3 + H 2 CO 3

83. Tro, Chemistry: A Molecular Approach83 3. Check to see either product H 2 S - No 4. Check to see if either product decomposes – Yes H 2 CO 3 decomposes into CO 2 (g) + H 2 O(l) Na 2 CO 3 (aq) + HNO 3 (aq)  NaNO 3 + CO 2 (g) + H 2 O(l) Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves

84. Tro, Chemistry: A Molecular Approach84 5. Determine the solubility of other product NaNO 3 is soluble 6. Write an (s) after the insoluble products and a (aq) after the soluble products Na 2 CO 3 (aq) + 2 HNO 3 (aq)  NaNO 3 (aq) + CO 2 (g) + H 2 O(l) 7. Balance the equation Na 2 CO 3 (aq) + 2 HNO 3 (aq)  NaNO 3 + CO 2 (g) + H 2 O(l) Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves

85. Tro, Chemistry: A Molecular Approach85 Other Patterns in Reactions the precipitation, acid-base, and gas evolving reactions all involved exchanging the ions in the solution other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions also known as redox reactions many involve the reaction of a substance with O 2 (g) 4 Fe(s) + 3 O 2 (g)  2 Fe 2 O 3 (s)

86. Tro, Chemistry: A Molecular Approach86 Combustion as Redox 2 H 2 (g) + O 2 (g)  2 H 2 O(g)

87. Tro, Chemistry: A Molecular Approach87 Redox without Combustion 2 Na(s) + Cl 2 (g)  2 NaCl(s) 2 Na  2 Na + + 2 e  Cl 2 + 2 e   2 Cl 

88. Tro, Chemistry: A Molecular Approach88 Reactions of Metals with Nonmetals consider the following reactions: 4 Na(s) + O 2 (g) → 2 Na 2 O(s) 2 Na(s) + Cl 2 (g) → 2 NaCl(s) the reaction involves a metal reacting with a nonmetal in addition, both reactions involve the conversion of free elements into ions 4 Na(s) + O 2 (g) → 2 Na + 2 O – (s) 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s)

89. Tro, Chemistry: A Molecular Approach89 Oxidation and Reduction in order to convert a free element into an ion, the atoms must gain or lose electrons of course, if one atom loses electrons, another must accept them reactions where electrons are transferred from one atom to another are redox reactions atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na → Na + + 1 e – oxidation Cl 2 + 2 e – → 2 Cl – reduction Leo Ger

90. Tro, Chemistry: A Molecular Approach90 Electron Bookkeeping for reactions that are not metal + nonmetal, or do not involve O 2, we need a method for determining how the electrons are transferred chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction even though they look like them, oxidation states are not ion charges!  oxidation states are imaginary charges assigned based on a set of rules  ion charges are real, measurable charges

91. Tro, Chemistry: A Molecular Approach91 Rules for Assigning Oxidation States rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = -1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

92. Tro, Chemistry: A Molecular Approach92 Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO 3 –, (+5) + 3(-2) = -1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl 2

93. Tro, Chemistry: A Molecular Approach93 Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority NonmetalOxidation StateExample FCF 4 H+1CH 4 O-2CO 2 Group 7ACCl 4 Group 6A-2CS 2 Group 5A-3NH 3

94. Tro, Chemistry: A Molecular Approach94 Practice – Assign an Oxidation State to Each Element in the following Br 2 K + LiF CO 2 SO 4 2- Na 2 O 2

95. Tro, Chemistry: A Molecular Approach95 Practice – Assign an Oxidation State to Each Element in the following Br 2 Br = 0, (Rule 1) K + K = +1, (Rule 2) LiFLi = +1, (Rule 4a) & F = -1, (Rule 5) CO 2 O = -2, (Rule 5) & C = +4, (Rule 3a) SO 4 2- O = -2, (Rule 5) & S = +6, (Rule 3b) Na 2 O 2 Na = +1, (Rule 4a) & O = -1, (Rule 3a)

96. Tro, Chemistry: A Molecular Approach96 Oxidation and Reduction Oxidation and Reduction Another Definition oxidation occurs when an atom’s oxidation state increases during a reaction reduction occurs when an atom’s oxidation state decreases during a reaction CH 4 + 2 O 2 → CO 2 + 2 H 2 O -4 +1 0 +4 –2 +1 -2 oxidation reduction

97. Tro, Chemistry: A Molecular Approach97 Oxidation–Reduction oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl 2 (g) → 2 Na + Cl – (s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl 2 is the oxidizing agent

98. Tro, Chemistry: A Molecular Approach98 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O

99. Tro, Chemistry: A Molecular Approach99 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O +1 -2 +5 -2 +1 0 +2 -2 +1 -2 ox agred ag +4 -2 +1 -1 +2 -1 0 +1 -2 oxidation reduction oxidation reduction red agox ag

100. Tro, Chemistry: A Molecular Approach100 Combustion Reactions Reactions in which O 2 (g) is a reactant are called combustion reactions Combustion reactions release lots of energy Combustion reactions are a subclass of oxidation- reduction reactions 2 C 8 H 18 (g) + 25 O 2 (g)  16 CO 2 (g) + 18 H 2 O(g)

101. Tro, Chemistry: A Molecular Approach101 Combustion Products to predict the products of a combustion reaction, combine each element in the other reactant with oxygen ReactantCombustion Product contains CCO 2 (g) contains HH 2 O(g) contains SSO 2 (g) contains NNO(g) or NO 2 (g) contains metalM2On(s)M2On(s)

102. Tro, Chemistry: A Molecular Approach102 Practice – Complete the Reactions combustion of C 3 H 7 OH(l) combustion of CH 3 NH 2 (g)

103. Tro, Chemistry: A Molecular Approach103 Practice – Complete the Reactions C 3 H 7 OH(l) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) CH 3 NH 2 (g) + 3 O 2 (g)  CO 2 (g) + 2 H 2 O(g) + NO 2 (g)