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10/7/2014 Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 4 Chemical Quantities and Aqueous Reactions Global Warming • Scientists have measured an average 0.6 °C rise in atmospheric temperature since 1860 • During the same period atmospheric CO2 levels • have risen 25% Are the two trends causal? Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 2 Copyright © 2011 Pearson Education, Inc. The Sources of Increased CO2 • One source of CO2 is the combustion reactions of fossil fuels we use to get energy • Another source of CO2 is volcanic action • How can we judge whether global warming is natural or due to our use of fossil fuels? Quantities in Chemical Reactions • The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction Law of Conservation of Mass Balancing equations by balancing atoms • The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry Tro: Chemistry: A Molecular Approach, 2/e 3 Copyright © 2011 Pearson Education, Inc. Reaction Stoichiometry • The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 2 molecules of C8H18 react with 25 molecules of O2 to form 16 molecules of CO2 and 18 molecules of H2O 2 moles of C8H18 react with 25 moles of O2 to form 16 moles of CO2 and 18 moles of H2O 2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O Tro: Chemistry: A Molecular Approach, 2/e 5 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 4 Copyright © 2011 Pearson Education, Inc. Making Pizza • The number of pizzas you can make depends on the amount of the ingredients you use 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza • This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza • If you want to make more or less than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make assuming you have enough crusts and tomato sauce Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright © 2011 Pearson Education, Inc. 1 10/7/2014 Predicting Amounts from Stoichiometry • The amounts of any other substance in a • Practice • According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C6H12O6 + 6 O2 6 CO2 + 6 H2O chemical reaction can be determined from the amount of just one substance How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18? 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 2 moles C8H18 : 16 moles CO2 Tro: Chemistry: A Molecular Approach, 2/e 7 Copyright © 2011 Pearson Education, Inc. Practice How many moles of water are made in the combustion of 0.10 moles of glucose? Given: Find: 0.10 moles C6H12O6 moles H2O Conceptual Plan: Relationships: mol C6H12O6 Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasolne • Assuming that gasoline is octane, C8H18, the equation for the reaction is mol H2O 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) C6H12O6 + 6 O2 → 6 CO2 + 6 H2O 1 mol C6H12O6 : 6 mol H2O Solution: • The equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the conceptual plan will be g C8H18 0.6 mol H2O = 0.60 mol H2O Check: 8 mol C8H18 mol CO2 g CO2 because 6x moles of H2O as C6H12O6, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 9 Copyright © 2011 Pearson Education, Inc. Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasoline 1015 Given: 3.4 x Find: g CO2 Conceptual g C8H18 Plan: g C8H18 mol C8H18 mol CO2 Tro: Chemistry: A Molecular Approach, 2/e Solution: produced 1.1 x 1016 g of CO2 just from petroleum combustion in 2007 1.1 x 1013 kg CO2 • Estimates of volcanic CO2 production are • Check: because 8x moles of CO as C H , but the molar mass of 2 8 18 C8H18 is 3x CO2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 11 Copyright © 2011 Pearson Education, Inc. Copyright © 2011 Pearson Education, Inc. Which Produces More CO2; Volcanoes or Fossil Fuel Combustion? • Our calculation just showed that the world g CO2 Relationships:1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2 10 2 x 1011 kg/year This means that volcanoes produce less than 2% of the CO2 added to the air annually 2.0 1011 kg yr 1.1 1013 kg yr Tro: Chemistry: A Molecular Approach, 2/e 100% 1.8% 12 Copyright © 2011 Pearson Education, Inc. 2 10/7/2014 Example 4.1: How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis? Given: Find: 37.8 g CO2, 6 CO2 + 6 H2O C6H12O6+ 6 O2 g C6H12O6 Practice — How many grams of O2 can be made from the decomposition of 100.0 g of PbO2? 2 PbO2(s) → 2 PbO(s) + O2(g) (PbO2 = 239.2, O2 = 32.00) Conceptual g CO2 mol CO2 mol C6H12O6 g C6H12O6 1mol 180.2 g Plan: 1mol C6H12O6 44.01 g Relationships: 1 mol 6 mol CO2 1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6 : 6 mol CO2 Solution: 1 mol C6H12O6 180.2 g C6H12O6 1 mol CO2 37.8 g CO2 44.01 g CO2 6 mol CO2 1 mol C6H12O6 25.8 g C6H12O6 Check: because 6x moles of CO as C H O , but the molar mass 2 6 12 6 of C6H12O6 is 4x CO2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 13 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e Practice — How many grams of O2 can be made from the decomposition of 100.0 g of PbO2? 2 PbO2(s) → 2 PbO(s) + O2(g) Conceptual g PbO2 Plan: Relationships: mol PbO2 mol O2 g O2 1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 : 2 mol PbO2 Solution: aAbB Solution 100.0 g PbO2, 2 PbO2 → 2 PbO + O2 g O2 Copyright © 2011 Pearson Education, Inc. Stoichiometry Road Map Pure Substance Given: Find: 14 % A(aq) ppm A(aq) M A(aq) M B(aq) % B(aq) ppm B(aq) M = moles L Moles A MM mass A density equation 22.4 L Moles B MM 22.4 L Volume A(g) equation mass B Volume B(g) volume A (l) density volume B(l) Check: because ½ moles of O as PbO , and the molar mass of 2 2 PbO2 is 7x O2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 15 Copyright © 2011 Pearson Education, Inc. More Making Pizzas • We know that 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza • But what would happen if we had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese? Tro: Chemistry: A Molecular Approach, 2/e 17 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 16 Copyright © 2011 Pearson Education, Inc. More Making Pizzas, Continued • Each ingredient could potentially make a different number of pizzas • But all the ingredients have to work together! • We only have enough tomato sauce to make three pizzas, so once we make three pizzas, the tomato sauce runs out no matter how much of the other ingredients we have. Tro: Chemistry: A Molecular Approach, 2/e 18 Copyright © 2011 Pearson Education, Inc. 3 10/7/2014 The Limiting Reactant More Making Pizzas, Continued • The tomato sauce limits the amount of pizzas we can make. In chemical reactions we call this the limiting reactant. also known as the limiting reagent • The maximum number of pizzas we can make depends on this ingredient. In chemical reactions, we call this the theoretical yield. it also determines the amounts of the other ingredients we will use! • 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 Tro: Chemistry: A Molecular Approach, 2/e 19 Copyright © 2011 Pearson Education, Inc. Limiting and Excess Reactants in the Combustion of Methane • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Our balanced equation for the combustion of methane implies that every one molecule of CH4 reacts with two molecules of O2 Tro: Chemistry: A Molecular Approach, 2/e 21 Copyright © 2011 Pearson Education, Inc. Practice — How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2 Si3N4? Tro: Chemistry: A Molecular Approach, 2/e 20 Copyright © 2011 Pearson Education, Inc. Limiting and Excess Reactants in the Combustion of Methane CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) • If we have five molecules of CH4 and eight molecules since less CO2 can be made from the O2 than the CH4, so the O2 is the limiting reactant of O2, which is the limiting reactant? Tro: Chemistry: A Molecular Approach, 2/e 22 Copyright © 2011 Pearson Education, Inc. Practice — How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2 Si3N4? Given: Find: Conceptual Plan: 1.20 mol Si, 1.00 mol N2 mol Si3N4 mol Si mol Si3N4 mol N2 mol Si3N4 Pick least amount Relationships: Limiting reactant and theoretical yield 2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N4 Solution: Limiting reactant Theoretical yield Tro: Chemistry: A Molecular Approach, 2/e 23 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 24 Copyright © 2011 Pearson Education, Inc. 4 10/7/2014 More Making Pizzas • Let’s now assume that as we are making • pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield. Tro: Chemistry: A Molecular Approach, 2/e 25 Copyright © 2011 Pearson Education, Inc. Example 4.4: Finding limiting reactant, theoretical yield, and percent yield Theoretical and Actual Yield • As we did with the pizzas, in order to determine • the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make The theoretical yield will always be the least possible amount of product the theoretical yield will always come from the limiting reactant • Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield Tro: Chemistry: A Molecular Approach, 2/e 26 Copyright © 2011 Pearson Education, Inc. Example: • When 28.6 kg of C are allowed to react with 88.2 kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2 (s) 2 C(s) Ti(s) 2 CO(g) Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. • Write down the given quantity and its units Given: 28.6 kg C 88.2 kg TiO2 42.8 kg Ti produced 29 28 Copyright © 2011 Pearson Education, Inc. Information Example: Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) Example: When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) Tro: Chemistry: A Molecular Approach, 2/e Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. • Write down the quantity to find and/or its units Find: limiting reactant theoretical yield percent yield Tro: Chemistry: A Molecular Approach, 2/e 30 Copyright © 2011 Pearson Education, Inc. 5 10/7/2014 Information Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: lim. rct., theor. yld., % yld. Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) • Write a conceptual plan • Collect needed relationships kg C } kg TiO2 smallest amount is from limiting reactant smallest mol Ti Tro: Chemistry: A Molecular Approach, 2/e Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) Information Given: 28.6 kg C, 88.2 kg TiO2, 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 TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) 31 Copyright © 2011 Pearson Education, Inc. Information Given: 28.6 kg C, 88.2 kg TiO2, 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 TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti • Apply the conceptual plan 1000 g = 1 kg Molar Mass TiO2 = 79.87 g/mol Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol 1 mole TiO2 : 1 mol Ti (from the chem. equation) 2 mole C : 1 mol Ti (from the chem. equation) Tro: Chemistry: A Molecular Approach, 2/e Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) 32 Copyright © 2011 Pearson Education, Inc. Information Given: 28.6 kg C, 88.2 kg TiO2, 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 TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti • Apply the conceptual plan theoretical yield limiting reactant Tro: Chemistry: A Molecular Approach, 2/e Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) smallest moles of Ti 33 Copyright © 2011 Pearson Education, Inc. Information Given: 28.6 kg C, 88.2 kg TiO2, 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 TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti • Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) 34 Copyright © 2011 Pearson Education, Inc. Information Given: 28.6 kg C, 88.2 kg TiO2, 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 TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti • Check the solutions limiting reactant = TiO2 theoretical yield = 52.9 kg percent yield = 80.9% Because Ti has lower molar mass than TiO2, the T.Y. makes sense and the percent yield makes sense as it is less than 100% Tro: Chemistry: A Molecular Approach, 2/e 35 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 36 Copyright © 2011 Pearson Education, Inc. 6 10/7/2014 Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? Given: Find: 9.05 g NH3, 45.2 g CuO g N2 Conceptual g NH 3 Plan: mol NH3 mol N2 Choose smallest g CuO mol CuO g N2 mol N2 Relationships: 1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g 2 mol NH3 : 1 mol N2, 3 mol CuO : 1 mol N2 Tro: Chemistry: A Molecular Approach, 2/e 37 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? 38 Copyright © 2011 Pearson Education, Inc. Solutions • When table salt is mixed with water, it seems to Solution: 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 Theoretical yield • 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 Check: because the percent yield is less than 100, the answer makes sense Tro: Chemistry: A Molecular Approach, 2/e 39 Copyright © 2011 Pearson Education, Inc. Describing Solutions • Because solutions are mixtures, the composition solvent if both components start in the same state, the major component is the solvent Tro: Chemistry: A Molecular Approach, 2/e • 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 are identical Tro: Chemistry: A Molecular Approach, 2/e 41 Copyright © 2011 Pearson Education, Inc. Copyright © 2011 Pearson Education, Inc. Solution Concentration • Qualitatively, solutions are can vary from one sample to another pure substances have constant composition saltwater samples from different seas or lakes have different amounts of salt 40 • • 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 Tro: Chemistry: A Molecular Approach, 2/e 42 Copyright © 2011 Pearson Education, Inc. 7 10/7/2014 Concentrations—Quantitative Descriptions of Solutions • A more precise method for describing a • solution is to quantify the amount of solute in a given amount of solution Concentration = amount of solute in a given amount of solution 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 occasionally amount of solvent Tro: Chemistry: A Molecular Approach, 2/e 43 Copyright © 2011 Pearson Education, Inc. Preparing 1 L of a 1.00 M NaCl Solution Tro: Chemistry: A Molecular Approach, 2/e 44 Copyright © 2011 Pearson Education, Inc. Example 4.5: Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution Given: Find: Conceptual Plan: 25.5 g KBr, 1.75 L solution molarity, M g KBr mol KBr M L sol’n Relationships: 1 mol KBr = 119.00 g, M = moles/L Solution: Check: Tro: Chemistry: A Molecular Approach, 2/e 45 Copyright © 2011 Pearson Education, Inc. Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in 200.0 mL of solution? because most solutions are between 0 and 18 M, the answer makes sense Tro: Chemistry: A Molecular Approach, 2/e 46 Copyright © 2011 Pearson Education, Inc. Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in 200.0 mL of solution? Given: 3.4 0.20gmol NH3NH , 200.0 mL solution L solution 3, 0.2000 Find: M Conceptual g NH3 mol NH3 Plan: mL sol’n L sol’n M Relationships: M = mol/L, 1 mol NH3 = 17.03 g, 1 mL = 0.001 L Solve: Check: the unit is correct, the number is reasonable because the fraction of moles is less than the fraction of liters Tro: Chemistry: A Molecular Approach, 2/e 47 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 48 Copyright © 2011 Pearson Education, Inc. 8 10/7/2014 Example 4.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH? 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 Given: 0.125 M NaOH, 0.255 mol NaOH Find: liters, L Conceptual Plan: L sol’n mol NaOH Relationships: 0.125 mol NaOH = 1 L solution Solution: • 1 L solution : 2 moles sugar Check: Tro: Chemistry: A Molecular Approach, 2/e 49 Copyright © 2011 Pearson Education, Inc. Practice — Determine the mass of CaCl2 (MM = 110.98) in 1.75 L of 1.50 M solution because each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should require a little more than 2 L Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. 50 Practice — Determine the mass of CaCl2 (MM = 110.98) in 1.75 L of 1.50 M solution Given: 1.50 M CaCl2, 1.75 L Find: mass CaCl2, g Conceptual Plan: L sol’n mol CaCl2 g CaCl2 Relationships: 1.50 mol CaCl2 = 1 L solution; 110.98 g CaCl2 = 1 mol Solution: Check: Tro: Chemistry: A Molecular Approach, 2/e 51 Copyright © 2011 Pearson Education, Inc. Example: How would you prepare 250.0 mL of a 1.00 M solution CuSO45 H2O(MM 249.69)? Given: 250.0 mL solution Find: mass CuSO4 5 H2O, g mol CuSO4 Conceptual mL sol’n L sol’n Plan: Relationships: because each L has 1.50 mol CaCl2, it makes sense that 1.75 L should have almost 3 moles Tro: Chemistry: A Molecular Approach, 2/e 52 Copyright © 2011 Pearson Education, Inc. Practice – How would you prepare 250.0 mL of 0.150 M CaCl2 (MM = 110.98)? g CuSO4 1.00 L sol’n = 1.00 mol; 1 mL = 0.001 L; 1 mol = 249.69 g Solution: Dissolve 62.4 g of CuSO4∙5H2O in enough water to total 250.0 mL Check: the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter Tro: Chemistry: A Molecular Approach, 2/e 53 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 54 Copyright © 2011 Pearson Education, Inc. 9 10/7/2014 Practice – How would you prepare 250.0 mL of 0.150 M CaCl2? Given: 250.0 mL solution Find: mass CaCl2, g Conceptual mL sol’n L sol’n Plan: • Often, solutions are stored as concentrated stock solutions mol CaCl2 g CaCl2 Relationships: 1.00 L sol’n = 0.150 mol; 1 mL = 0.001L; 1 mol = 110.98 g Solution: Dissolve 4.16 g of CaCl2 in enough water to total 250.0 mL Check: the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter Tro: Chemistry: A Molecular Approach, 2/e 55 Copyright © 2011 Pearson Education, Inc. Example 4.7: To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH? Given: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M Find: V2, L Conceptual Plan: Relationships: V1, M1, M2 Dilution • 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 M1∙V1 = M2∙V2 Tro: Chemistry: A Molecular Approach, 2/e 56 Copyright © 2011 Pearson Education, Inc. Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to 135.0 mL? V2 M1V1 = M2V2 Solution: Check: because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does Tro: Chemistry: A Molecular Approach, 2/e 57 Copyright © 2011 Pearson Education, Inc. Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to 135.0 mL? Given: V1 = 45.0 mL, M1 = 8.25 M, V2 = 135.0 mL Find: M2, L Conceptual Plan: Relationships: V1, M1, V2 Tro: Chemistry: A Molecular Approach, 2/e 58 Copyright © 2011 Pearson Education, Inc. Practice – How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution? M2 M1V1 = M2V2 Solution: Check: because the solution is diluted by a factor of 3, the molarity should decrease by a factor of 3, and it does Tro: Chemistry: A Molecular Approach, 2/e 59 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 60 Copyright © 2011 Pearson Education, Inc. 10 10/7/2014 Practice – How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution? Given: M1 = 2.0 M, M2 = 0.25 M, V2 = 200.0 mL Find: V1, L Conceptual Plan: Relationships: M1, M2, V2 V1 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 M1V1 = M2V2 Solution: Dilute 25 mL of 2.0 M solution up to 200.0 mL Check: because the solution is diluted by a factor of 8, the volume should increase by a factor of 8, and it does Tro: Chemistry: A Molecular Approach, 2/e 61 Copyright © 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 63 Copyright © 2011 Pearson Education, Inc. When Will a Salt Dissolve? Tro: Chemistry: A Molecular Approach, 2/e 62 Copyright © 2011 Pearson Education, Inc. Solubility of Ionic Compounds • Some ionic compounds, such as NaCl, dissolve very well in water at room temperature • Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature • Compounds that dissolve in a solvent are said to be soluble, where as 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 Tro: Chemistry: A Molecular Approach, 2/e 64 Copyright © 2011 Pearson Education, Inc. Solubility Rules Compounds that Are Generally Soluble in Water • 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 Tro: Chemistry: A Molecular Approach, 2/e 65 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 66 Copyright © 2011 Pearson Education, Inc. 11 10/7/2014 Solubility Rules Compounds that Are Generally Insoluble in Water Tro: Chemistry: A Molecular Approach, 2/e 67 Copyright © 2011 Pearson Education, Inc. Electrolytes and Nonelectrolytes • Materials that dissolve • Practice – Determine if each of the following is soluble in water KOH KOH is soluble because it contains K+ AgBr AgBr is insoluble; most bromides are soluble, but AgBr is an exception CaCl2 CaCl2 is soluble; most chlorides are soluble, and CaCl2 is not an exception Pb(NO3)2 Pb(NO3)2 is soluble because it contains NO3− PbSO4 PbSO4 is insoluble; most sulfates are soluble, but PbSO4 is an exception Tro: Chemistry: A Molecular Approach, 2/e 68 Copyright © 2011 Pearson Education, Inc. Molecular View of Electrolytes and Nonelectrolytes • To conduct electricity, a material must have 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 charged particles that are able to flow • Electrolyte solutions all contain ions dissolved in the water ionic compounds are electrolytes because they 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 Tro: Chemistry: A Molecular Approach, 2/e 69 Copyright © 2011 Pearson Education, Inc. Salt vs. Sugar Dissolved in Water Tro: Chemistry: A Molecular Approach, 2/e 70 Copyright © 2011 Pearson Education, Inc. 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 also 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) ionic compounds dissociate into ions when they dissolve Tro: Chemistry: A Molecular Approach, 2/e molecular compounds do not dissociate when they dissolve 71 Copyright © 2011 Pearson Education, Inc. • Acids that only ionize a small percentage are called weak acids HF(aq) H+(aq) + F−(aq) Tro: Chemistry: A Molecular Approach, 2/e 72 Copyright © 2011 Pearson Education, Inc. 12 10/7/2014 Dissociation and Ionization 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 HC2H3O2(aq) H+(aq) + C2H3O2−(aq) Tro: Chemistry: A Molecular Approach, 2/e 73 Copyright © 2011 Pearson Education, Inc. Practice – Write the equation for the process that occurs when the following strong electrolytes dissolve in water • When ionic compounds dissolve in water, the anions and cations are separated from each other. This is called dissociation. Na2S(aq) 2 Na+(aq) + S2-(aq) When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion Na2SO4(aq) 2 Na+(aq) + SO42−(aq) When strong acids dissolve in water, the molecule ionizes into H+ and anions H2SO4(aq) H+(aq) + HSO4−(aq) HSO4−(aq) H+(aq) + SO42− (aq) Tro: Chemistry: A Molecular Approach, 2/e 74 Copyright © 2011 Pearson Education, Inc. Precipitation Reactions • Precipitation reactions are CaCl2 CaCl2(aq) Ca2+(aq) + 2 Cl−(aq) HNO3 HNO3(aq) H+(aq) + NO3−(aq) (NH4)2CO3 (NH4)2CO3(aq) 2 NH4+(aq) + CO32−(aq) Tro: Chemistry: A Molecular Approach, 2/e 75 Copyright © 2011 Pearson Education, Inc. 2 KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2 KNO3(aq) reactions in which a solid forms when we mix two solutions reactions between aqueous solutions of ionic compounds produce an ionic compound that is insoluble in water the insoluble product is called a precipitate Tro: Chemistry: A Molecular Approach, 2/e 76 Copyright © 2011 Pearson Education, Inc. No Precipitate Formation = No Reaction KI(aq) + NaCl(aq) KCl(aq) + NaI(aq) all ions still present, no reaction Tro: Chemistry: A Molecular Approach, 2/e 77 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 78 Copyright © 2011 Pearson Education, Inc. 13 10/7/2014 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 Process for Predicting the Products of a Precipitation Reaction 5. If any of the possible products are insoluble, write their formulas as the products of the reaction using (s) after the formula to indicate solid. Write any soluble products with (aq) after the formula to indicate aqueous. 6. Balance the equation remember to only change coefficients, not subscripts 4. If neither product will precipitate, write no reaction after the arrow Tro: Chemistry: A Molecular Approach, 2/e 79 Copyright © 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 80 Copyright © 2011 Pearson Education, Inc. 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 K2CO3(aq) + NiCl2(aq) 2. Determine the possible products 3. Determine the solubility of each product KCl is soluble NiCO3 is insoluble a) determine the ions present (K+ + CO32−) + (Ni2+ + Cl−) b) exchange the Ions (K+ + CO32−) + (Ni2+ + Cl−) (K+ + Cl−) + (Ni2+ + CO32−) c) write the formulas of the products 4. If both products are soluble, write no reaction does not apply because NiCO3 is insoluble balance charges K2CO3(aq) + NiCl2(aq) KCl + NiCO3 Tro: Chemistry: A Molecular Approach, 2/e 81 Copyright © 2011 Pearson Education, Inc. 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 5. Write (aq) next to soluble products and (s) next to insoluble products K2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s) 6. Balance the equation K2CO3(aq) + NiCl2(aq) 2 KCl(aq) + NiCO3(s) Tro: Chemistry: A Molecular Approach, 2/e 83 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 82 Copyright © 2011 Pearson Education, Inc. Practice – Predict the products and balance the equation KCl(aq) + AgNO3(aq) (K+ + Cl−) + (Ag+ + NO3−) → (K+ + NO3−) + (Ag+ + Cl−) KCl(aq) + AgNO3(aq) → KNO3 + AgCl KCl(aq) + AgNO3(aq) KNO3(aq) + AgCl(s) Na2S(aq) + CaCl2(aq) (Na+ + S2−) + (Ca2+ + Cl−) → (Na+ + Cl−) + (Ca2+ + S2−) Na2S(aq) + CaCl2(aq) → NaCl + CaS Na2S(aq) + CaCl2(aq) NaCl(aq) + CaS(aq) No reaction Tro: Chemistry: A Molecular Approach, 2/e 84 Copyright © 2011 Pearson Education, Inc. 14 10/7/2014 Practice – Write an equation for the reaction that takes place when an aqueous solution of (NH4)2SO4 is mixed with an aqueous solution of Pb(C2H3O2)2. (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) (NH4+ + SO42−) + (Pb2+ + C2H3O2−) → (NH4+ + C2H3O2−) + (Pb2+ + SO42−) (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) NH4C2H3O2 + PbSO4 Ionic Equations • Equations that describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) + Mg(NO3)2(aq) 2 KNO3(aq) + Mg(OH)2(s) • Equations that describe the material’s structure when dissolved are called complete ionic equations aqueous strong electrolytes are written as ions (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) NH4C2H3O2(aq) + PbSO4(s) (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) 2 NH4C2H3O2(aq) + PbSO4(s) Tro: Chemistry: A Molecular Approach, 2/e 85 Copyright © 2011 Pearson Education, Inc. soluble salts, strong acids, strong bases insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form solids, liquids, and gases are not dissolved, therefore molecule form 2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s) Tro: Chemistry: A Molecular Approach, 2/e • Ions that are both reactants and products are called 2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s) An ionic equation in which the spectator ions are removed is called a net ionic equation 2 OH−(aq) + Mg2+(aq) Mg(OH)2(s) Copyright © 2011 Pearson Education, Inc. Practice – Write the ionic and net ionic equation for each Ionic Equations spectator ions 86 K2SO4(aq) + 2 AgNO3(aq) 2 KNO3(aq) + Ag2SO4(s) 2K+(aq) + SO42−(aq) + 2Ag+(aq) + 2NO3−(aq) 2K+(aq) + 2NO3−(aq) + Ag2SO4(s) 2 Ag+(aq) + SO42−(aq) Ag2SO4(s) Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l) 2Na+(aq) + CO32−(aq) + 2H+(aq) + 2Cl−(aq) 2Na+(aq) + 2Cl−(aq) + CO2(g) + H2O(l) CO32−(aq) + 2 H+(aq) CO2(g) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 87 Copyright © 2011 Pearson Education, Inc. Acid-Base Reactions • Also called neutralization reactions because the Tro: Chemistry: A Molecular Approach, 2/e • The net ionic equation for an acid-base reaction is H+(aq) + OH(aq) H2O(l) as long as the salt that forms is soluble in water Copyright © 2011 Pearson Education, Inc. 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, H3O+ acid and base neutralize each other’s properties 2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l) 88 most chemists use H+ and H3O+ interchangeably • Bases dissociate in water to form OH ions bases, such as NH3, 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 Tro: Chemistry: A Molecular Approach, 2/e 89 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 90 Copyright © 2011 Pearson Education, Inc. 15 10/7/2014 Common Acids Tro: Chemistry: A Molecular Approach, 2/e 91 Copyright © 2011 Pearson Education, Inc. Common Bases Tro: Chemistry: A Molecular Approach, 2/e HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 92 Copyright © 2011 Pearson Education, Inc. Example: Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 1. Write the formulas of the reactants HNO3(aq) + Ca(OH)2(aq) 2. Determine the possible products a) determine the ions present when each reactant dissociates or ionizes (H+ + NO3−) + (Ca2+ + OH−) b) exchange the ions, H+ combines with OH− to make H2O(l) (H+ + NO3−) + (Ca2+ + OH−) (Ca2+ + NO3−) + H2O(l) c write the formula of the salt (H+ + NO3−) + (Ca2+ + OH−) Ca(NO3)2 + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 93 Copyright © 2011 Pearson Education, Inc. Example: Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 3. Determine the solubility of the salt Ca(NO3)2 is soluble 4. Write an (s) after the insoluble products and an (aq) after the soluble products HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + H2O(l) 5. Balance the equation 2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 95 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 94 Copyright © 2011 Pearson Education, Inc. Example: Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 6. Dissociate all aqueous strong electrolytes to get complete ionic equation not H2O 2 H+(aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq) Ca2+(aq) + 2 NO3−(aq) + H2O(l) 7. Eliminate spectator ions to get net-ionic equation 2 H+(aq) + 2 OH−(aq) 2 H2O(l) H+(aq) + OH−(aq) H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 96 Copyright © 2011 Pearson Education, Inc. 16 10/7/2014 Practice – Predict the products and balance the equation HCl(aq) + Ba(OH)2(aq) Gas-Evolving Reactions • Some reactions form a gas directly from the ion (H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−) HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2 2 HCl(aq) + Ba(OH)2(aq) 2 H2O(l) + BaCl2(aq) H2SO4(aq) + Sr(OH)2(aq) (H+ + SO42−) + (Sr2+ + OH−) → (H+ + OH−) + (Sr2+ + SO42−) H2SO4(aq) + Sr(OH)2(aq) → H2O(l) + SrSO4 • exchange K2S(aq) + H2SO4(aq) K2SO4(aq) + H2S(g) Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq) H2SO3 H2O(l) + SO2(g) H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4 H2SO4(aq) + Sr(OH)2(aq) 2 H2O(l) + SrSO4(s) Tro: Chemistry: A Molecular Approach, 2/e 97 Copyright © 2011 Pearson Education, Inc. NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 99 Copyright © 2011 Pearson Education, Inc. Example 4.14: 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 Na2CO3(aq) + HNO3(aq) 2. Determine the possible products a) determine the ions present when each reactant dissociates or ionizes (Na+ + CO32−) + (H+ + NO3−) b) exchange the anions (Na+ + CO32−) + (H+ + NO3−) (Na+ + NO3−) + (H+ + CO32−) c) write the formula of compounds balance the charges Na2CO3(aq) + HNO3(aq) NaNO3 + H2CO3 Tro: Chemistry: A Molecular Approach, 2/e 101 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 98 Copyright © 2011 Pearson Education, Inc. Compounds that Undergo Gas-Evolving Reactions Tro: Chemistry: A Molecular Approach, 2/e 100 Copyright © 2011 Pearson Education, Inc. Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 3. Check to see if either product is H2S - No 4. Check to see if either product decomposes – Yes H2CO3 decomposes into CO2(g) + H2O(l) Na2CO3(aq) + HNO3(aq) NaNO3 + CO2(g) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 102 Copyright © 2011 Pearson Education, Inc. 17 10/7/2014 Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 5. Determine the solubility of other product NaNO3 is soluble 6. Write an (s) after the insoluble products and an (aq) after the soluble products Na2CO3(aq) + 2 HNO3(aq) 2 NaNO3(aq) + CO2(g) + H2O(l) 7. Balance the equation Na2CO3(aq) + 2 HNO3(aq) 2 NaNO3(aq) + CO2(g) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 103 Copyright © 2011 Pearson Education, Inc. Practice – Predict the products and balance the equation HCl(aq) + Na2SO3(aq) (H+ + Cl−) + (Na+ + SO32−) → (H+ + SO32−) + (Na+ + Cl−) HCl(aq) + Na2SO3(aq) → H2SO3 + NaCl HCl(aq) + Na2SO3(aq) → H2O(l) + SO2(g) + NaCl 2 HCl(aq) + Na2SO3(aq) H2O(l) + SO2(g) + 2 NaCl(aq) H2SO4(aq) + CaS(aq) (H+ + SO42−) + (Ca2+ + S2−) → (H+ + S2−) + (Ca2+ + SO42−) H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4 H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4(s) Tro: Chemistry: A Molecular Approach, 2/e 104 Copyright © 2011 Pearson Education, Inc. Combustion as Redox 2 H2(g) + O2(g) 2 H2O(g) Other Patterns in Reactions • The precipitation, acid-base, and gas-evolving • reactions all involve 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 O2(g) 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) Tro: Chemistry: A Molecular Approach, 2/e 105 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e Redox without Combustion 2 Na(s) + Cl2(g) 2 NaCl(s) 106 Copyright © 2011 Pearson Education, Inc. Reactions of Metals with Nonmetals • Consider the following reactions: 2 Na 2 Na+ + 2 e • • 4 Na(s) + O2(g) → 2 Na2O(s) 2 Na(s) + Cl2(g) → 2 NaCl(s) The reactions involve a metal reacting with a nonmetal In addition, both reactions involve the conversion of free elements into ions 4 Na(s) + O2(g) → 2 Na+2O2– (s) 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Cl2 + 2 e 2 Cl Tro: Chemistry: A Molecular Approach, 2/e 107 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 108 Copyright © 2011 Pearson Education, Inc. 18 10/7/2014 Oxidation and Reduction Electron Bookkeeping • To convert a free element into an ion, the • For reactions that are not metal + nonmetal, or 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 even though they look like them, oxidation states are not ion charges! Ger 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges Leo Tro: Chemistry: A Molecular Approach, 2/e 109 Copyright © 2011 Pearson Education, Inc. do not involve O2, 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 Tro: Chemistry: A Molecular Approach, 2/e Rules for Assigning Oxidation States 110 Copyright © 2011 Pearson Education, Inc. Rules for Assigning Oxidation States • Rules are in order of priority 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion 1. free elements have an oxidation state = 0 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1 2. monatomic ions have an oxidation state equal to their charge 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 and Cl = −1 in NaCl Na = +1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 4. (b) Group II metals have an oxidation state of +2 in all their compounds Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0 Mg = +2 in MgCl2 Tro: Chemistry: A Molecular Approach, 2/e 111 Copyright © 2011 Pearson Education, Inc. Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below Tro: Chemistry: A Molecular Approach, 2/e 112 Copyright © 2011 Pearson Education, Inc. Example: Determine the oxidation states of all the atoms in a propanoate ion, C3H5O2– • There are no free elements or free ions in nonmetals higher on the table take priority • propanoate, so the first rule that applies is Rule 3b (C3) + (H5) + (O2) = −1 Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order: H = +1 O = −2 (C3) + 5(+1) + 2(−2)Note: = −1unlike charges, (C3) = −2 oxidation states can be fractions! C = −⅔ Tro: Chemistry: A Molecular Approach, 2/e 113 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 114 Copyright © 2011 Pearson Education, Inc. 19 10/7/2014 Practice – Assign an oxidation state to each element in the following Oxidation and Reduction Another Definition • Br2 Br = 0, (Rule 1) • Oxidation occurs when an atom’s oxidation • K+ K = +1, (Rule 2) • Reduction occurs when an atom’s oxidation • LiF Li = +1, (Rule 4a) & F = −1, (Rule 5) • CO2 O = −2, (Rule 5) & C = +4, (Rule 3a) • SO4 O = −2, (Rule 5) & S = +6, (Rule 3b) state increases during a reaction 2− state decreases during a reaction CH4 + 2 O2 → CO2 + 2 H2O −4 +1 • Na2O2 Na = +1, (Rule 4a) & O = −1 , (Rule 3a) Tro: Chemistry: A Molecular Approach, 2/e 115 Copyright © 2011 Pearson Education, Inc. 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 +4 –2 0 +1 −2 oxidation reduction Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. 116 Example: Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions: Reducing agent Oxidizing agent Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O 0 +7 −2 +1 +3 +4 −2 +1 −2 the oxidizing agent contains the element that is reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent Tro: Chemistry: A Molecular Approach, 2/e 117 Copyright © 2011 Pearson Education, Inc. Practice – Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions: Sn4+ + Ca → Sn2+ + Ca2+ +4 0 +2 +2 Ca is oxidized, Sn4+ is reduced Ca is the reducing agent, Sn4+ is the oxidizing agent Reduction Oxidation Tro: Chemistry: A Molecular Approach, 2/e 118 Copyright © 2011 Pearson Education, Inc. Solution Stoichiometry • Because 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 F2 + S → SF4 0 0 +4−1 S is oxidized, F is reduced S is the reducing agent, F2 is the oxidizing agent Tro: Chemistry: A Molecular Approach, 2/e 119 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 120 Copyright © 2011 Pearson Education, Inc. 20 10/7/2014 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(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)? Given: Find: Conceptual Plan: Practice — Solution stoichiometry • 43.8 mL of 0.107 M HCl is needed to 0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2 neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base? L KCl L Pb(NO3)2 mol Pb(NO3)2 mol KCl L KCl 2 HCl + Ba(OH)2 BaCl2 + 2 H2O Relationships: 1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl Solution: Check: because you need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x the volume of Pb(NO3)2 Tro: Chemistry: A Molecular Approach, 2/e 121 Copyright © 2011 Pearson Education, Inc. Practice – 43.8 mL of 0.107 M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base? 2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(aq) Tro: Chemistry: A Molecular Approach, 2/e Titration MBa(OH) Ba(OH)22 Find: M L HCl mol HCl mol Ba(OH)2 M Ba(OH)2 L Ba(OH)2 Relationships: 1 mL= 0.001 L, 1 L HCl = 0.107 mol, 1 mol Ba(OH) 2 : 2 mol HCl Solution: Copyright © 2011 Pearson Education, Inc. • Often in the lab, a solution’s concentration Given: 0.0438 43.8 mL L of 0.107 M HCl, 0.0376 37.6 mLLBa(OH) Ba(OH)22 Conceptual Plan: 122 • 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 Check: the units are correct, the number makes sense because there are fewer moles than liters Tro: Chemistry: A Molecular Approach, 2/e 123 Copyright © 2011 Pearson Education, Inc. a long glass tube with precise volume markings that allows small additions of solution Tro: Chemistry: A Molecular Approach, 2/e Acid-Base Titrations 124 Copyright © 2011 Pearson Education, Inc. Titration • 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 also known as the equivalence point Tro: Chemistry: A Molecular Approach, 2/e 125 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 126 Copyright © 2011 Pearson Education, Inc. 21 10/7/2014 Titration The titrant is the base solution in the burette. As the titrant is added to the flask, 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. Tro: Chemistry: A Molecular Approach, 2/e Copyright © 2011 Pearson Education, Inc. 127 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? Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH • Write down the quantity to find, and/or its units Find: 129 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? L NaOH mL HCl L HCl Tro: Chemistry: A Molecular Approach, 2/e 12.54 mL of 0.100 M NaOH Tro: Chemistry: A Molecular Approach, 2/e 128 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? Copyright © 2011 Pearson Education, Inc. Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 1 mole HCl = 1 mole NaOH 0.100 M NaOH 0.100 mol NaOH 1 L sol’n Copyright © 2011 Pearson Education, Inc. Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L • Write a conceptual plan mL NaOH • Write down the given quantity and its units Given: 10.00 mL HCl • Collect needed equations and conversion factors concentration HCl, M Tro: Chemistry: A Molecular Approach, 2/e 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? mol NaOH mol HCl Tro: Chemistry: A Molecular Approach, 2/e 130 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? Copyright © 2011 Pearson Education, Inc. Information Given: 10.00 mL HCl 12.54 mL of 0.100 M NaOH Find: M HCl Rel: 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 • Apply the conceptual plan = 1.25 x 103 mol HCl 131 Copyright © 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 132 Copyright © 2011 Pearson Education, Inc. 22 10/7/2014 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? Information Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl Rel: 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 • Apply the conceptual plan 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? Information Given: 10.00 mL HCl 12.54 mL NaOH Find: M HCl Rel: 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 • Check the solution HCl solution = 0.125 M The units of the answer, M, are correct. The magnitude of the answer makes sense because the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated. Tro: Chemistry: A Molecular Approach, 2/e 133 Copyright © 2011 Pearson Education, Inc. Practice − What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4? H2SO4 + 2 NaOH Na2SO4 + 2 H2O 50.0 mL 27.5 mL 0.1015 M ? M Tro: Chemistry: A Molecular Approach, 2/e 135 Copyright © 2011 Pearson Education, Inc. Copyright © 2011 Pearson Education, Inc. Practice — What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of 0.1015 M H 2SO4? 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(aq) Given: Find: 50.0 mLLof 0.0500 of0.1015 0.1015 M MHH22SO SO44,,27.5 0.0275 mLLNaOH NaOH M NaOH Conceptual L H SO 2 4 Plan: mol H2SO4 mol NaOH M NaOH L NaOH Relationships: 1 mL= 0.001L, 1 LH2SO4 = 0.1015mol, 2mol NaOH : 1mol H2SO4 Solution: Check: Tro: Chemistry: A Molecular Approach, 2/e 134 the units are correct, the number makes because the volume of NaOH is about ½ the H2SO4, but the stoichiometry says you need twice the moles of NaOH as H2SO4 Tro: Chemistry: A Molecular Approach, 2/e 136 Copyright © 2011 Pearson Education, Inc. 23