Chapter 4 Chemical Quantities and Aqueous Reactions

Chapter 4 Chemical Quantities and Aqueous Reactions

Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 4 Chemical Quantities and Aqueous Reactions Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Copyright 2011 Pearson Education, Inc. 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? 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? Tro: Chemistry: A Molecular Approach, 2/e 3 Copyright 2011 Pearson Education, Inc.

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 4 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. 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. 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 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 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 Tro: Chemistry: A Molecular Approach, 2/e 8 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 mol H2O C6H12O6 + 6 O2 6 CO2 + 6 H2O 1 mol C6H12O6 : 6 mol H2O Solution: Check:

0.6 mol H2O = 0.60 mol H2O 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 gasolne Assuming that gasoline is octane, C8H18, the equation for the reaction is 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) 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

Tro: Chemistry: A Molecular Approach, 2/e 10 Copyright 2011 Pearson Education, Inc. Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasoline Given: 3.4 x 1015 g C8H18 Find: g CO2 Conceptual Plan: Relationships: 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2 Solution: Check: because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e

11 Copyright 2011 Pearson Education, Inc. Which Produces More CO2; Volcanoes or Fossil Fuel Combustion? Our calculation just showed that the world 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 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 13 kg 1.110

Tro: Chemistry: A Molecular Approach, 2/e 100% 1.8% yr 12 Copyright 2011 Pearson Education, Inc. Example 4.1: How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis? Given: Find: Conceptual Plan: Relationships: 37.8 g CO2, 6 CO2 + 6 H2O C6H12O6+ 6 O2 g C6H12O6 1mol 44.01 g

1mol C6H12O6 6 mol CO2 180.2 g 1 mol 1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6 : 6 mol CO2 Solution: 1 mol C6H12O 6 180.2 g C6H12O 6 1 mol CO2 37.8 g CO2 44.01 g CO2 6 mol CO 2 1 mol C6H12O 6 25.8 g C6H12O 6 Check: because 6x moles of CO2 as C6H12O6, but the molar mass

of C6H12O6 is 4x CO2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 13 Copyright 2011 Pearson Education, Inc. 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) Tro: Chemistry: A Molecular Approach, 2/e 14 Copyright 2011 Pearson Education, Inc. 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)

Given: Find: 100.0 g PbO2, 2 PbO2 2 PbO + O2 g O2 Conceptual Plan: Relationships: 1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 : 2 mol PbO2 Solution: Check: because moles of O2 as PbO2, and the molar mass of PbO2 is 7x O2, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 15 Copyright 2011 Pearson Education, Inc.

Stoichiometry Road Map Pure Substance Solution aAbB % A(aq) ppm A(aq) M A(aq) M = moles L Moles A MM 22.4 L

mass A density equation Moles B MM 22.4 L Volume A(g) equation volume A (l) Tro: Chemistry: A Molecular Approach, 2/e M B(aq) % B(aq)

ppm B(aq) Volume B(g) mass B density volume B(l) 16 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. 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. 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! Tro: Chemistry: A Molecular Approach, 2/e 19 Copyright 2011 Pearson Education, Inc. The 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 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) 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. 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

of O2, which is the limiting reactant? Tro: Chemistry: A Molecular Approach, 2/e 22 since less CO2 can be made from the O2 than the CH4, so the O2 is the limiting reactant 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

23 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: 1.20 mol Si, 1.00 mol N2 mol Si3N4 Conceptual Plan: Relationships: 2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N4 Solution: Limiting reactant

Theoretical yield Tro: Chemistry: A Molecular Approach, 2/e 24 Copyright 2011 Pearson Education, Inc. More Making Pizzas Lets 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. 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 4.4: Finding limiting reactant, theoretical yield, and percent yield Tro: Chemistry: A Molecular Approach, 2/e 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 28 Copyright 2011 Pearson Education, Inc. 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)

Write down the given quantity and its units Given: 28.6 kg C 88.2 kg TiO2 42.8 kg Ti produced Tro: Chemistry: A Molecular Approach, 2/e 29 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) 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. 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. Write a conceptual plan } kg C kg TiO2 smallest amount is from limiting reactant smallest mol Ti

Tro: Chemistry: A Molecular Approach, 2/e 31 Copyright 2011 Pearson Education, Inc. 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 Collect needed relationships

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 32 Copyright 2011 Pearson Education, Inc. 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 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 limiting reactant Tro: Chemistry: A Molecular Approach, 2/e smallest moles of Ti 33 Copyright 2011 Pearson Education, Inc. 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 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 Tro: Chemistry: A Molecular Approach, 2/e 34

Copyright 2011 Pearson Education, Inc. 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 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 35 Copyright 2011 Pearson Education, Inc. 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 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 36 Copyright 2011 Pearson Education, Inc. 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? Tro: Chemistry: A Molecular Approach, 2/e

37 Copyright 2011 Pearson Education, Inc. 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 Plan: Choose smallest 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 38 Copyright 2011 Pearson Education, Inc. 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? Solution: Theoretical yield 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. 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 Tro: Chemistry: A Molecular Approach, 2/e

40 Copyright 2011 Pearson Education, Inc. Describing Solutions Because solutions are mixtures, the composition can vary from one sample to another pure substances have constant composition saltwater 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 are identical Tro: Chemistry: A Molecular Approach, 2/e 41

Copyright 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 42 Copyright 2011 Pearson Education, Inc.

ConcentrationsQuantitative 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 occasionally amount of solvent Tro: Chemistry: A Molecular Approach, 2/e 43 Copyright 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 44 Copyright 2011 Pearson Education, Inc. Preparing 1 L of a 1.00 M NaCl Solution Tro: Chemistry: A Molecular Approach, 2/e 45 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: 25.5 g KBr, 1.75 L solution molarity, M Conceptual Plan: Relationships: 1 mol KBr = 119.00 g, M = moles/L Solution: Check: 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? Tro: Chemistry: A Molecular Approach, 2/e 47 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 mol NH3 Conceptual g NH3 Plan: mL soln L soln 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 48 Copyright 2011 Pearson Education, Inc.

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 Tro: Chemistry: A Molecular Approach, 2/e 49 Copyright 2011 Pearson Education, Inc. Example 4.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH?

Given: 0.125 M NaOH, 0.255 mol NaOH Find: liters, L Conceptual Plan: Relationships: 0.125 mol NaOH = 1 L solution Solution: Check: 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 50 Copyright 2011 Pearson Education, Inc. Practice Determine the mass of CaCl2

(MM = 110.98) in 1.75 L of 1.50 M solution Tro: Chemistry: A Molecular Approach, 2/e 51 Copyright 2011 Pearson Education, Inc. 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: Relationships: 1.50 mol CaCl2 = 1 L solution; 110.98 g CaCl2 = 1 mol Solution: Check: 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. 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 Conceptual Plan: Relationships: 1.00 L soln = 1.00 mol; 1 mL = 0.001 L; 1 mol = 249.69 g Solution: Dissolve 62.4 g of CuSO45H2O in enough water to total 250.0 mL

Check: the unit is correct, the magnitude seems reasonable as the volume is of a liter Tro: Chemistry: A Molecular Approach, 2/e 53 Copyright 2011 Pearson Education, Inc. Practice How would you prepare 250.0 mL of 0.150 M CaCl2 (MM = 110.98)? Tro: Chemistry: A Molecular Approach, 2/e 54 Copyright 2011 Pearson Education, Inc. Practice How would you prepare 250.0 mL of 0.150 M CaCl2?

Given: 250.0 mL solution Find: mass CaCl2, g Conceptual Plan: Relationships: 1.00 L soln = 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 of a liter Tro: Chemistry: A Molecular Approach, 2/e 55 Copyright 2011 Pearson Education, Inc. 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 doesnt 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 M1V1 = M2V2 Tro: Chemistry: A Molecular Approach, 2/e 56 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: 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?

Tro: Chemistry: A Molecular Approach, 2/e 58 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: 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. Practice How would you prepare 200.0 mL of 0.25 M NaCl solution from a 2.0 M solution? Tro: Chemistry: A Molecular Approach, 2/e 60 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?

Given: M1 = 2.0 M, M2 = 0.25 M, V2 = 200.0 mL Find: V1, L Conceptual Plan: Relationships: 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. 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 Tro: Chemistry: A Molecular Approach, 2/e 62 Copyright 2011 Pearson Education, Inc. Example 4.8: What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)? Given: Find: 0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2 L KCl Conceptual

Plan: 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 63 Copyright 2011 Pearson Education, Inc. Practice Solution stoichiometry 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 + Ba(OH)2 BaCl2 + 2 H2O Tro: Chemistry: A Molecular Approach, 2/e 64 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) Given: 0.0438 43.8 mL L of 0.107 M HCl, 0.0376 37.6 mLLBa(OH) Ba(OH)22 Find: M MBa(OH) Ba(OH)22 Conceptual

Plan: 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: Check: the units are correct, the number makes sense because there are fewer moles than liters Tro: Chemistry: A Molecular Approach, 2/e 65 Copyright 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 66 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 67 Copyright 2011 Pearson Education, Inc. 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 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 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 ionic compounds dissociate into ions when they dissolve Tro: Chemistry: A Molecular Approach, 2/e molecular compounds do not dissociate when

they dissolve 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) Acids that only ionize a small percentage are called weak acids HF(aq) H+(aq) + F(aq)

Tro: Chemistry: A Molecular Approach, 2/e 71 Copyright 2011 Pearson Education, Inc. 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 72 Copyright 2011 Pearson Education, Inc. Classes of Dissolved Materials Tro: Chemistry: A Molecular Approach, 2/e 73 Copyright 2011 Pearson Education, Inc. Dissociation and Ionization 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) 2 H+(aq) + SO42(aq) Tro: Chemistry: A Molecular Approach, 2/e 74 Copyright 2011 Pearson Education, Inc. Practice Write the equation for the process that

occurs when the following strong electrolytes dissolve in water 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.

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 76 Copyright 2011 Pearson Education, Inc.

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 Tro: Chemistry: A Molecular Approach, 2/e 77 Copyright 2011 Pearson Education, Inc. Solubility Rules Compounds that Are Generally Soluble in Water

Tro: Chemistry: A Molecular Approach, 2/e 78 Copyright 2011 Pearson Education, Inc. Solubility Rules Compounds that Are Generally Insoluble in Water Tro: Chemistry: A Molecular Approach, 2/e 79 Copyright 2011 Pearson Education, Inc. 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 80 Copyright 2011 Pearson Education, Inc. Precipitation Reactions Precipitation reactions are 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 81

Copyright 2011 Pearson Education, Inc. 2 KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2 KNO3(aq) Tro: Chemistry: A Molecular Approach, 2/e 82 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 83 Copyright 2011 Pearson Education, Inc.

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 Tro: Chemistry: A Molecular Approach, 2/e 84

Copyright 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 85 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 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 balance charges K2CO3(aq) + NiCl2(aq) KCl + NiCO3 Tro: Chemistry: A Molecular Approach, 2/e

86 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 3. Determine the solubility of each product KCl is soluble NiCO3 is insoluble 4. If both products are soluble, write no reaction does not apply because NiCO3 is insoluble Tro: Chemistry: A Molecular Approach, 2/e 87 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) KCl(aq) + NiCO3(s) Tro: Chemistry: A Molecular Approach, 2/e 88 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 89 Copyright 2011 Pearson Education, Inc. 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 (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 90 Copyright 2011 Pearson Education, Inc. 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 materials structure when dissolved are called complete ionic equations aqueous strong electrolytes are written as ions 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) K+(aq) + 2NO3(aq) + Mg(OH)2(s) Tro: Chemistry: A Molecular Approach, 2/e 91 Copyright 2011 Pearson Education, Inc. Ionic Equations Ions that are both reactants and products are called spectator ions 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) Tro: Chemistry: A Molecular Approach, 2/e 92 Copyright 2011 Pearson Education, Inc. Practice Write the ionic and net ionic equation for each SO44(aq) (aq) ++ 22 AgNO AgNO33(aq) (aq) 22 KNO KNO33(aq) (aq) ++ Ag Ag22SO

SO44(s) (s) KK22SO 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) 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) 2 + CO (aq) + 2 H (aq) COCopyright

H2Pearson O(l)Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 2011 93 3 2(g) + Acid-Base Reactions Also called neutralization reactions because the acid and base neutralize each others properties 2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l) 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 Tro: Chemistry: A Molecular Approach, 2/e

94 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+ 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 95 Copyright 2011 Pearson Education, Inc. Common Acids Tro: Chemistry: A Molecular Approach, 2/e 96 Copyright 2011 Pearson Education, Inc. Common Bases Tro: Chemistry: A Molecular Approach, 2/e 97 Copyright 2011 Pearson Education, Inc.

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 98 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 99 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 100 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) H2O(l) H+(aq) + OH(aq) H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 101 Copyright 2011 Pearson Education, Inc. Practice Predict the products and balance the equation HCl(aq) + Ba(OH)2(aq) (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

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 102 Copyright 2011 Pearson Education, Inc. Titration Often in the lab, a solutions 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 Tro: Chemistry: A Molecular Approach, 2/e 103 Copyright 2011 Pearson Education, Inc. 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 also known as the equivalence point Tro: Chemistry: A Molecular Approach, 2/e 104 Copyright 2011 Pearson Education, Inc. Titration Tro: Chemistry: A Molecular Approach, 2/e 105 Copyright 2011 Pearson Education, Inc.

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 titrations 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 106 Copyright 2011 Pearson Education, Inc.

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 Tro: Chemistry: A Molecular Approach, 2/e 107 Copyright 2011 Pearson Education, Inc. 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: concentration HCl, M Tro: Chemistry: A Molecular Approach, 2/e 108 Copyright 2011 Pearson Education, Inc.

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 Find: M HCl Collect needed equations and conversion factors HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 1 mole HCl = 1 mole NaOH 0.100 M NaOH 0.100 mol NaOH 1 L soln Tro: Chemistry: A Molecular Approach, 2/e

109 Copyright 2011 Pearson Education, Inc. 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 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 L NaOH mL HCl L HCl Tro: Chemistry: A Molecular Approach, 2/e mol NaOH 110 mol

HCl Copyright 2011 Pearson Education, Inc. 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 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 Tro: Chemistry: A Molecular Approach, 2/e 111 Copyright 2011 Pearson Education, Inc. 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 Tro: Chemistry: A Molecular Approach, 2/e 112 Copyright 2011 Pearson Education, Inc.

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 113 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

114 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? 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(aq) Given: Find: 50.0 mLLof 0.0500 of0.1015 0.1015M 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: the units are correct, the number makes because the volume of NaOH is about the H2SO4, but the stoichiometry says you need twice the moles of NaOH as H2SO4

Tro: Chemistry: A Molecular Approach, 2/e 115 Copyright 2011 Pearson Education, Inc. Gas-Evolving Reactions Some reactions form a gas directly from the ion 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) Tro: Chemistry: A Molecular Approach, 2/e 116 Copyright 2011 Pearson Education, Inc.

Electron Bookkeeping For reactions that are not metal + nonmetal, or 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 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 Tro: Chemistry: A Molecular Approach, 2/e 117 Copyright 2011 Pearson Education, Inc. Rules for Assigning Oxidation States

Rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(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 Tro: Chemistry: A Molecular Approach, 2/e 118 Copyright 2011 Pearson Education, Inc. 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 NO3, (+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 MgCl2 Tro: Chemistry: A Molecular Approach, 2/e 119 Copyright 2011 Pearson Education, Inc. 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

Tro: Chemistry: A Molecular Approach, 2/e 120 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 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) = 1 (C3) = 2 C = Tro: Chemistry: A Molecular Approach, 2/e

121 Note: unlike charges, oxidation states can be fractions! Copyright 2011 Pearson Education, Inc. Practice Assign an oxidation state to each element in the following Br2 Br = 0, (Rule 1) K+ K = +1, (Rule 2) LiF Li = +1, (Rule 4a) & F = 1, (Rule 5)

CO2 O = 2, (Rule 5) & C = +4, (Rule 3a) SO42 O = 2, (Rule 5) & S = +6, (Rule 3b) Na2O2 Na = +1, (Rule 4a) & O = 1 , (Rule 3a) Tro: Chemistry: A Molecular Approach, 2/e 122 Copyright 2011 Pearson Education, Inc. Oxidation and Reduction Another Definition

Oxidation occurs when an atoms oxidation state increases during a reaction Reduction occurs when an atoms oxidation state decreases during a reaction CH4 + 2 O2 CO2 + 2 H2O 4 +1 0 +4 2 +1 2 oxidation reduction Tro: Chemistry: A Molecular Approach, 2/e 123

Copyright 2011 Pearson Education, Inc. OxidationReduction 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) + 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 124

Copyright 2011 Pearson Education, Inc. 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 Reduction Oxidation Tro: Chemistry: A Molecular Approach, 2/e 125 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 F2 + S SF4 0 0 +41 S is oxidized, F is reduced S is the reducing agent, F2 is the oxidizing agent Tro: Chemistry: A Molecular Approach, 2/e 126 Copyright 2011 Pearson Education, Inc.

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