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Ch4.ppt - 1 CHAPTER 4 Types of Chemical Reactions and...

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Unformatted text preview: 1 CHAPTER 4 Types of Chemical Reactions and Solution Stoichiometry 4.1. 4.1. Aqueous Aqueous Solutions Solutions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. 2 Solution Solution Components Components A SOLUTION CONTAINS: • SOLVENT - a larger component • SOLUTE - a smaller component Aqueous solutions are solutions in which the solvent is water (H2O). 3 3 4.3. Concentrations in Solution The relative amount of solute in a solution is given by its concentration. concentration • A common way to represent concentration is molarity. • • Molarity is defined as moles of solute per liter of solution. n M= V Molarity # of moles Volume of solution The volume of solution must be in Liters! 4 Molarity n M= V • Units of molarity: mol/L “M” after a number is an abbreviation for the units of molarity: 3 M = 3 mol/L 5 Sample Problem 1 PROBLEM: PLAN: Calculating the Molarity of a Solution Glycine (H2NCH2COOH) is an amino acid. What is the molarity of an aqueous solution that contains 0.715 mol of glycine in 495 mL? Molarity is the number of moles of solute per liter of solution. mL of solution 103mL = 1L L of solution divide moles by volume molarity(mol/L) glycine SOLUTION: 1L 495 mL soln x 1000mL 0.715 mol glycine 0.495 L soln = 0.495 L soln = 1.44 M glycine 6 Preparing Preparing Solutions Solutions To prepare a solution of known molarity, weigh out a solid solute and dissolve in a given quantity of solvent. 7 8 PROBLEM PROBLEM 1: 1: 5.00 5.00 gg of of NiCl NiCl22•6 •6 H H22O O were were dissolved dissolved in in enough enough water water to to make make 250 250 mL mL of of solution. solution. Calculate Calculate molarity. molarity. Step 1: Calculate moles of NiCl2•6H2O 1 mol 5.00 g • = 0.0210 mol 237.7 g Step 2: Calculate molarity 0.0210 mol = 0.0841 M 0.250 L [NiCl2•6 H2O ] = 0.0841 M 9 PROBLEM PROBLEM 2: 2: What What mass massof ofoxalic oxalicacid, acid, HH22CC22O O44,,is is required requiredto tomake make250. 250.mL mLof of aa0.0500 0.0500M Msolution? solution? Because Conc (M) = moles/volume = mol/V this means that moles = M•V Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g 10 SAMPLE PROBLEM 2 PROBLEM: Calculating Molarity What is the molarity of a solution prepared by dissolving 32.0 g of CaCl2 in 250 mL of solution? PLAN: We have to convert the grams of CaCl2 to moles and the milliliters of solution to liters. Then we substitute into the equation for molarity. SOLUTION: 32.0 g CaCl2 1 mole CaCl2 x 250 ml 110.98 g CaCl2 x 1L 103 mL = 0.288 mol CaCl2 = 0.250 L 0.288 mole CaCl2 molarity = 0.250 L = 1.15 M CaCl2 11 Sample PROBLEM 3 PROBLEM: Calculating volume from Molarity What volume of a 0.750 M solution will contain 1.25 mol of solute? PLAN: Solve the molarity formula to give the volume. SOLUTION: M= n V M = 0.750 mol/L n = 1.25 mol n= MxV V= n M 1.25 mol x 1L 0.75 mol = 1.67 L 12 Summary of mass-mole-number-volume relationships in solution. MASS (g) of compound in solution M.M. (g/mol) AMOUNT (mol) of compound in solution Avogadro’s number (molecules/mol) MOLECULES of compound in solution Molarity (mol/L) VOLUME (L) of solution 13 Preparing Preparing a a Solution Solution by by Dilution Dilution Diluting a concentrated solution produces a solution that is dilute (less concentrated). 14 Converting a concentrated solution to a dilute solution. PROBLEM: PROBLEM: You You have have 50.0 50.0 mL mL of of 3.0 3.0 M M NaOH NaOH and and you you want want 0.50 0.50 M M NaOH. NaOH. What What do do you you do? do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution! H2O 3.0 M NaOH Concentrated 0.50 M NaOH Dilute 15 16 PROBLEM: PROBLEM: You You have have 50.0 50.0 mL mL of of 3.0 3.0 M M NaOH NaOH and and you you want want 0.50 0.50 M M NaOH. NaOH. What What do do you you do do?? But how much water should we add? The important point is : mass of NaOH in ORIGINAL solution = mass of NaOH in FINAL solution therefore moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution PROBLEM: PROBLEM: You You have have 50.0 50.0 mL mL of of 3.0 3.0 M M NaOH NaOH and and you you want want 0.50 0.50 M M NaOH. NaOH. What What do do you you do? do? Amount of NaOH in original solution = M•V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 mL 17 PROBLEM: PROBLEM: You You have have 50.0 50.0 mL mL of of 3.0 3.0 M M NaOH NaOH and and you you want want 0.50 0.50 M M NaOH. NaOH. What What do do you you do? do? Conclusion: H2O add 250 mL of water to 3.0 M NaOH Concentrated 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M 0.50 M NaOH NaOH. Dilute 18 19 Dilution Dilution A shortcut Minitial • Vinitial = Mfinal • Vfinal 20 Sample PROBLEM 4 PROBLEM: Using dilution equation What volume of a 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? PLAN: Use the dilution equation. SOLUTION: M1V1 = M2V2 M1 = 16 mol/L M2 = 0.10 mol/L V1 = V1 = M2V2 V2 = 1.5 L M1 0.10 x 1.5 16 = 0.0094 L = 9.4 mL Aqueous Aqueous Solutions Solutions Aqueous solutions are solutions in which the solvent is water (H2O). KMnO4 in water K+(aq) + MnO4-(aq) 21 How How do do Ionic Ionic Compounds Compounds Dissolve? Dissolve? • Hydration - the process that occurs when salts dissolve in water. • Water attracts the charged ions b/c water is a polar molecule. 22 23 Ionic compound in a water solution 4.2. Electrolytes • Does pure water conduct electric current? No. • Do aqueous solutions conduct electric current? It depends on the solute. • An electrolyte is a substance whose aqueous solution conducts electricity. 24 The electrical conductivity of ionic solutions. Solid ionic compound Distilled water Ionic compound dissolved in water 25 Electrolytes Solutes Strong Electrolytes Weak Electrolytes Their aqueous solutions conduct electricity strongly Their aqueous solutions conduct electricity weakly Non-electrolytes Their aqueous solutions don’t conduct electricity 26 Strong Electrolytes NaCl, HCl, and NaOH are strong electrolytes. They dissociate completely (or nearly so) into ions. 27 Weak Electrolytes Acetic acid dissociates (ionizes) only to a small extent, so it is a weak electrolyte. CH3COOH(aq) ---> CH3COO-(aq) + H+(aq) 28 Nonelectrolyt es Nonelectrolytes dissolve in water, but do not dissociate into ions, and do not conduct electricity. most covalent compounds (except for acids) are nonelectrolytes Examples include: sugar, ethanol, pure water 29 30 Dissociation • • Dissociation - breaking down of a compound into ions in a solution. anions and cations are separated from each other When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion. Examples of Dissociation 1) potassium iodide KI(aq) K+(aq) + I-(aq) K 2) I copper(II) sulfate CuSO4(aq) Cu SO4 3) I- K+ Cu2+(aq) + SO42-(aq) SO42- Cu2+ potassium sulfate K2SO4(aq) K SO4 K 2 K+(aq) + K+ K+ SO42-(aq) SO42- 31 Electrolytes Three major types of electrolytes: 1) soluble ionic compounds Ex.: NaCl 2) acids – release H+ in aqueous solution Ex.: HCl 3) bases – release OH- in aqueous solution Ex.: NaOH Note: acids can dissociate into ions, even though they are not ionic! HCl(aq) ---> H+(aq) + Cl-(aq) 32 The Nature of a CuCl2 Solution: Ion Concentrations 33 CuCl2(aq) --> Cu2+(aq) + 2 Cl-(aq) If [CuCl2] = 0.30 M, then [Cu2+] = ? [Cl-] = ?0.30 M 2 x 0.30 = 0.60 M Sample Problem 5 PROBLEM: 34 Determining Moles of Ions in Aqueous Ionic Solutions How many moles of each ion are in the following solutions? (a) 5.0 mol of ammonium sulfate dissolved in water (b) 35 mL of 0.84 M zinc chloride PLAN: Use the stoichiometry of a process when the substance dissolves in water. SOLUTION: (a) (NH4)2SO4(s) 5.0 mol (NH4)2SO4 x 2NH4+(aq) + SO42-(aq) 2 mol NH4+ = 10. mol NH4+ 1 mol (NH4)2SO4 (b) ZnCl2(aq) 35 mL ZnCl2 x 5.0 mol SO42- Zn2+(aq) + 2Cl-(aq) 1L 0.84 mol ZnCl2 -2 x = 2.9x10 mol ZnCl2 3 L 10 mL = 2.9x10-2 mol Zn2+ = 5.8x10-2 mol Cl- 4.4. Types of Chemical Reactions In double replacement, the anions exchange places between cations. AX + BY AY + BX Pb(NO3)2(aq) + 2 KI(aq) ----> PbI2(s) + 2 KNO3 (aq) 35 Types of Chemical Reactions Precipitation Reactions Gas-Forming Reactions REACTIONS REDOX REDOX REACTIONS REACTIONS Acid-Base Reactions 36 Types of reactions 37 • A precipitation reaction involves formation of an insoluble compound — a precipitate. Pb(NO3)2(aq) + 2 KI(aq) 2 KNO3(aq) + PbI2(s) Solid (not soluble) Net ionic equation Pb2+(aq) + 2 I-(aq) PbI2(s) Water Solubility of Ionic Compounds Soluble Ionic Compounds 1. All compounds of Group 1A ions (Li+, Na+, K+, etc.) and ammonium ion (NH4+) are soluble. 2. All nitrates (NO3-) and acetates (C2H3O2-) are soluble. 3. Most of chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble, except those of Ag+, Pb2+, and Hg22+. 4. Most of sulfates are soluble except those of Ca2+, Sr2+, Ba2+, Pb2+. Insoluble Ionic Compounds 1. Most of metal hydroxides are insoluble, except those of Group 1A ions (Na+, K+) and the larger members of Group 2A (Sr2+, Ba2+) . 2. Most of carbonates (CO32-), phosphates (PO43-), and sulfides (S2-), are insoluble, except those of Group 1A ions and NH4+. see also Table 4.1 38 Are They Soluble in Water? • KOH Soluble, because the cation is K+ • CaCl2 Soluble, most chlorides are soluble • AgBr Insoluble, even though most bromides are soluble, this is an exception • Pb(NO3)2 • PbSO4 Soluble, because the anion is NO3- Insoluble, even though most sulfates are soluble, this is an exception 39 40 Solubility of Solids vs Temperatu Solubility of solids increases at a higher T ACIDS and BASES Two sets of definitions for acids and bases: 1) Arrhenius definiton; 2) Bronsted-Lowry definiton. Arrhenius definition of an acid: Acid Acid is is aa substance substance that that produces produces H H++ ion ion in in an an aqueous aqueous solution. solution. HCl(aq) ---> H+(aq) + Cl-(aq) 41 Hydronium Ion How many protons, neutrons, and electrons does the H+ ion have? H+ ion is just 1 proton! So we will call it “proton”. Whenever H+ ion is released in an aqueous solution, it links with a water molecule to produce hydronium ion H3O+ : H+(aq) + H2O(l) ---> H3O+(aq) 42 43 ACIDS ACIDS ++ An acid -------> H O An acid -------> H33O in in water water HCl H 2O Cl- hydronium ion H 3O+ We can divide acids and bases into STRONG or WEAK ones. STRONG and WEAK ACIDS a strong electrolyte • • 100% ionized in water a weak acid is a strong acid is a weak electrolyte only a small percentage of the molecules ionize 44 Reaction of zinc with a strong and a weak acid 1M HCl(aq) 1M CH3COOH(aq) 45 Strong Strong and and Weak Weak ACIDS ACIDS Strong acid Some strong acids are HCl hydrochloric H2SO4 sulfuric HNO3 nitric HClO4 perchloric Weak acid 46 47 BASES BASES Arrhenius definition of a base: Base Base is is aa substance substance that that produces produces hydroxide hydroxide ion ion (OH (OH--)) in in an an aqueous aqueous solution. solution. -Base ---> OH Base ---> OH in in water water NaOH(aq) ---> Na+(aq) + OH-(aq) BASES BASES -Base ---> OH Base ---> OH in in water water NaOH is example of a strong base All soluble metal hydroxides are strong bases (KOH, Ba(OH)2, etc) 48 Ammonia, NH3 is a Base Example of a weak base is ammonia, NH3 49 50 Strong and Weak Acids and Bases Know the strong acids & bases! & many more weak acids! 51 4.6. Describing Reactions In Solution For reactions in water solutions, we can write 3 types of equations: Molecular equation shows all of the reactants and products as whole compounds. This is a regular chemical equation, like those discussed earlier. Complete ionic equation shows all strong electrolytes (soluble ionic compounds, strong acids, and strong bases) as dissociated into ions. Net ionic equation eliminates the spectator ions from the complete ionic equation, and shows the net chemical change. Complete Complete Ionic Ionic Equations Equations 52 Molecular equation: Mg(s) + 2 HCl(aq) H2(g) + MgCl2(aq) Which of the compounds are strong electrolytes? HCl (acid) and MgCl2 (soluble ionic compound) Lets break them down into ions! Mg(s) + 2 H+(aq) + 2 Cl-(aq) H2(g) + Mg2+(aq) + 2 Cl-(aq) This is a complete ionic equation. Net Net Ionic Ionic Equations Equations 53 Mg(s) + 2 H+(aq) + 2 Cl-(aq) H2(g) + Mg2+(aq) + 2 Cl-(aq) The two Cl­ ions are SPECTATOR IONS — they are the same on both sides of equation. They did not really participate. As a contrast, Mg and H are not the same on both sides: Mg(s) Mg2+(aq); 2 H+(aq) H2(g) We can cancel the spectator ions out: Mg(s) + 2 H+(aq) H2(g) + Mg2+(aq) This is the NET IONIC EQUATION Net Net Ionic Ionic Equations Equations 1. When predicting products - make sure all the formulas are consistent with ion charges, before you balance the equation. 2. Use the solubility rules to identify insoluble compounds. Do NOT dissociate insoluble solids or gases into ions. 3. Dissociate all soluble ionic compounds, strong acids, and strong soluble bases into ions. 54 Driving Forces of Chemical Reactions 55 Precipitation Reactions (formation of solid) Gas-Forming Reactions REACTIONS Acid-Base Neutralization Reactions (formation of water) Formation of a Weak Acid or Base Ionic Equations for Precipitation Reactions Molecular equation (unbalanced): NaI + Pb(NO3)2 PbI2 + NaNO3 Molecular equation (balanced): 2NaI + Pb(NO3)2 PbI2 + 2NaNO3 Which product is insoluble? 2NaI(aq) + Pb(NO3)2(aq) PbI2(s) + 2NaNO3(aq) Complete ionic equation: 2Na+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3- (aq) PbI2(s) + 2Na+(aq) + 2NO3- (aq) Net ionic equation: 2I- (aq) + Pb2+(aq) PbI2(s) 56 Sample Problem 6 PROBLEM: Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations In each of the following cases, write balanced molecular, complete ionic, and net ionic equations, and predict whether a reaction occurs. 57 (a) sodium sulfate(aq) + barium nitrate(aq) (b) ammonium nitrate(aq) + sodium bromide(aq) PLAN: SOLUTION: (a) exchange anions & cations Na2SO4 + Ba(NO3)2 write formulas of products Na2SO4 + balance equation check for insolubility Table 4.1 eliminate spectator ions for net ionic equation NaNO3 + BaSO4 Ba(NO3)2 2NaNO3 + BaSO4 Na2SO4(aq) + Ba(NO3)2 (aq) 2NaNO3(aq) + BaSO4(s) 2Na+(aq) +SO42-(aq)+ Ba2+(aq)+2NO3-(aq) 2Na+(aq) +2NO3-(aq)+ BaSO4(s) SO42-(aq)+ Ba2+(aq) BaSO4(s) Sample Problem 6 PROBLEM: Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations In each of the following cases, write balanced molecular, complete ionic, and net ionic equations, and predict whether a reaction occurs. 58 (a) sodium sulfate(aq) + barium nitrate(aq) (b) ammonium nitrate(aq) + sodium bromide(aq) PLAN: exchange anions & cations write formulas of products balance equation check for insolubility Table 4.1 eliminate spectator ions for net ionic equation SOLUTION: (b) NH4NO3 + NaBr NH4NO3(aq) + NaBr (aq) NH4Br + NaNO3 NH4Br (aq) + NaNO3(aq) All reactants and products are soluble, so no precipitate forms. All ions are spectator ions, therefore there’s no net ionic equation. No observed reaction. Predicting if Reaction is a Precipitation Reaction 1. Determine what ions each aqueous reactant has 2. Exchange Ions (+) ion from one reactant with (-) ion from other 3. Determine Solubility of Each Product in Water using the Solubility Rules – – if product is insoluble, it will precipitate if neither product will precipitate, not a precipitation reaction 59 Precipitation Reactions The “driving force” is the formation of precipitate. 60 4.8. Acid-Base Reactions Acid-base reaction - reaction between an acid and a base. • Acid-base reactions produce: 1) water; 2) an ionic compound called “salt”. Example: HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) • how can we show it’s double replacement? Re-write H2O as HOH: HCl(aq) + NaOH(aq) HOH(l) + NaCl(aq) Net ionic equation: H+(aq) + OH-(aq) H2O(l) 61 Acid-Base Reactions 62 HCl(aq) + NaOH(aq) ---> NaCl(aq) + H22O(l) Acid-Base reactions are also called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end. • • The key event is the formation of water: H+(aq) + OH-(aq) ---> H2O(l) • This net ionic equation applies to ALL reactions of STRONG acids and bases. • The reaction is exothermic (it warms up the solution). Sample Problem 7 PROBLEM: 63 Writing Ionic Equations Write balanced molecular, complete ionic, and net ionic equations for the reaction between sodium hydroxide(aq) and sulfuric acid(aq). PLAN: SOLUTION: reactants are strong acids a) Write molecular equation and balance it. and bases and therefore 2NaOH(aq) + H2SO4(aq) 2H2O(l) + Na2SO4(aq) completely ionized in b) Write complete ionic equation. Break down water acid (H2SO4), base (NaOH), and soluble ionic salt (Na2SO4) into ions. One of the products is water 2Na+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq) The other product contains spectator ions 2H2O(l) +2Na+(aq)+SO42-(aq) c) Write net ionic equation. Identify and cancel spectator ions to find the net chemical change. 2OH-(aq)+ 2H+(aq) 2H2O(l) 64 Gas-Forming Reactions Acids react with metal carbonates to produce a gas. Antacid tablet has citric acid + NaHCO3 CaCO3(s) + 2 HCl (aq) ---> CaCl2(aq) + H2CO3(aq) Carbonic acid is unstable and forms CO2 & water: H2CO3(aq) ---> CO2 (g) + H2O(l) Overall equation: CaCO3(s) + 2 HCl (aq) ---> CaCl2(aq) + CO2 (g) + H2O(l) 65 Gas-Forming Reactions Gas-Forming Reactions • 66 Example: NH4Cl + NaOH ? NH4OH + NaCl NH3(g) + H2O + NaCl Total ionic: NH4+ + Cl- + Na+ + OH- NH3 + H2O + Na + + ClNet ionic: NH4+ + OH- NH3 + H2O 67 A reaction of weak acid that forms a gaseous product Molecular equation NaHCO3(aq) + CH3COOH(aq) CH3COONa(aq) + CO2(g) + H2O(l) Complete ionic equation Na+(aq)+ HCO3-(aq) + CH3COOH(aq) CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l) Net ionic equation HCO3-(aq) + CH3COOH(aq) CH3COO-(aq) + CO2(g) + H2O(l) Ammonia in exchange reactions 68 • In water solutions, ammonia is in equilibrium with ammonium hydroxide: • NH3(aq) + H2O(l) <---> NH4OH (aq) Whenever ammonium hydroxide is produced, it decomposes to NH3 and water: NH4Cl(aq) + NaOH (aq) NH3(g) + H2O(l) + NaCl(aq) • When ammonia acts as a reactant in double replacement reactions, think of it as NH4OH : 2H2O(l) + 2 NH3(aq) + MgCl2(aq) 2 NH4Cl(aq) + Mg(OH)2(s) NH4OH Ammonia in exchange reactions 2NH4OH (aq) + MgCl2(aq) 2NH4Cl(aq) + Mg(OH)2(s) NH4OH (aq) <--> NH3(aq) + H2O(l) <--> NH4+ + OH preferred form Complete ionic: 2 NH3 + 2 H2O + Mg2+ + 2Cl- 2NH4+ + 2Cl- + Mg(OH)2 Net ionic: 2 NH3 + 2 H2O + Mg2+ 2NH4+ + Mg(OH)2 69 70 4.7. Solution Stoichiometry PROBLEM: SOLUTION: When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed. Step 1: Write the chemical equation, with correct formulas of each compound: Na2SO4(aq) + Pb(NO3)2 (aq) --> PbSO4(s) + 2NaNO3 (aq) Step 2: Balance the chemical equation. Step 3: Calculate moles of each reactant: moles Pb(NO3)2 = 1.25 L x 0.0500 mol/L = 0.0625 mol moles Na2SO4 = 2.00 L x 0.0250 mol/L = 0.0500 mol Solution Stoichiometry Na2SO4(aq) + Pb(NO3)2 (aq) --> PbSO4(s) + 2 NaNO3 (aq) 0.0500 mol Na2SO4 vs 0.0625 mol Pb(NO3)2 Step 4: Determine the limiting reactant: Na2SO4 0.0500 mol moles Na2SO4 how much PbSO4? Step 5: Calculate the number of moles of PbSO4, based on the limiting reactant: moles of PbSO4 = moles Na2SO4 = 0.0500 mol Step 6: Convert moles of PbSO4 to grams: (0.0500 mol )(303.3 g/mol) = 15.2 g 71 SOLUTION STOICHIOMETRY • Zinc reacts with acids to produce H2 gas. • Have 10.0 g of Zn • What volume of 2.50 M HCl is needed to convert the Zn completely? 72 GENERAL GENERAL PLAN PLAN FOR FOR SOLUTION SOLUTION STOICHIOMETRY STOICHIOMETRY CALCULATIONS CALCULATIONS Mass HCl Mass zinc Mole ratio Moles zinc Moles HCl Volume HCl 73 74 Zinc Zincreacts reactswith with aci...
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