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Ch 8: Aqueous Solutions: Chemistry of the Hydrosphere H2S + Cu2+ → CuS(s) + 2H+ (Fe, Ni, Mn also) HS- + 2 O2 HSO4- + energy (supports life) 1 2 Figure taken from “Principles of Biochemistry”, 2nd Ed. By Lehninger, Nelson, and Cox Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange 3 Concentration of Solutions - Molarity The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Mass-to-mass ratios: mg/kg solvent Mass-to-volume ratios: mg/L solvent 1 g = mg = mg 103 g L 106 g Parts per million: dH2O = 1.00 g/mL @ 25 oC Molarity (M) = moles of solute liters of solution Sample Exercise 8.1: Converting Mass-perVolume Concentrations into Molarity Vinyl chloride (C2H3Cl, MW = 62.49) is one of the most widely used industrial chemicals. It is also one of the most toxic, and it’s a known carcinogen. The maximum concentration of vinyl chloride allowed in drinking water in the U.S. is 0.002 mg/L. What is that concentration in moles per liter? Plan: mg/L g/L mol/L = Molarity 1 mol 0.002 mg x 1 g x 62.49 g = 3.2 x 10-8 1000 mg L mol/L 4 Sample Exercise 8.2: Converting Mass-per-Mass Concentrations into Molarity A water sample from the Great Salt Lake in Utah contains 83.6 mg Na+ per gram of lake water (soln). What is the molar concentration of Na+ ions if the density of the lake water is 1.160 g/mL? AW Na = 22.99 mg Na+ g mol Plan: mol = M L 1 g soln mL L dsoln = 1.160 g/mL 1 mol Na+ 1g 83.6 mg Na+ x 1000 mg x 22.99 g = 3.636 x 10-3 mol Na+ 1 g soln mol L M = 1 mL soln x 1.160 g = 1L x 1000 mL = 8.6221 x 10-4 L 3.636 x 10-3 mol Na+ = 4.22 M 8.6221 x 10-4 L Sample Exercise 8.3: Calculating the Quantity of Solute Needed to Prepare a Solution An aqueous solution called phosphate-buffered saline (PBS) is used in biology research to wash and store living cells. It contains ionic solutes including 10.0 mM Na2HPO42H2O (MW = 178.0). How many grams of this solute would you need to prepare 10.0 L of PBS? MM Plan: mM M mol g use the volume Useful equation: M x V = moles 178.0 g 1 mol 10.0 mmol x x 10.0 L x mol 1000 mmol L = 17.8 g M x V = moles 5 Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) = Moles of solute after dilution (f) MiVi = MfVf 6 Dilution Equation: Mi Vi = Mf Vf Sample Exercise 8.4: Diluting Solutions The solution used in hospitals for intravenous infusion—called physiological or normal saline - is 0.155 M NaCl. It may be prepared by diluting a commercially available standard solution that is 1.76 M NaCl. What volume of standard solution (Vi) is required to prepare 10.0 L of physiological saline? Collect data: Mi = 1.76 M Vi = ? Mf = 0.155 M Vf = 10.0 L add H2O Add 881 mL of 1.76 M NaCl, and then “dilute to the mark” (10.0 L) Mi Vi = Mf Vf Vi = Mf Vf Mi Vi = (0.155 M )(10.0 L) (1.76 M) = 0.881 L = 881 mL 7 Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange 8 Strong Electrolytes: • Nearly 100% dissociated into ions • Conduct current efficiently Examples: ionic compounds such as NaCl, strong acids like HCl • NaCl(s) → Na+(aq) + Cl−(aq) • HCl(aq) + H2O(l) H3O+ + Cl-(aq) Figure from John J. Fortman, J. Chem. Ed. Vol. 71, No. 1, 1994, p. 27-28 9 Weak Electrolytes: • Only partially dissociate into ions • Slightly conductive Example: weak acids such as acetic acid Hydronium Ion: H3O+ The higher the concentration of H3O+ in solution, the stronger the acid. + 10 Non-electrolytes Substances in which no ionization occurs. There is no flow of electrical current through the solution Examples: molecular or covalent compounds such as ethanol 11 Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange 12 Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. Cause color changes in plant dyes (litmus paper). React with certain metals to produce hydrogen gas. 2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g) React with carbonates and bicarbonates to produce carbon dioxide gas 2HCl (aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g) + H2O (l) Aqueous acid solutions conduct electricity. Limestone dissolving in acid = cave Most metals dissolve in acids to produce H2 13 Strong Acids and Bases Strong Acids/Bases: • Completely ionized in aqueous solution (i.e., strong electrolytes) • Memorize the 6 strong acids • All other acids are weak KNOW! 14 Bases Have a bitter taste. Feel slippery. Many soaps contain bases. Cause color changes in plant dyes (litmus paper). Aqueous base solutions conduct electricity. A Brønsted acid is a proton donor A Brønsted base is a proton acceptor base acid conjugate acid conjugate base For every acid there is a conjugate base, and for every base there is a conjugate acid (conj. acid-base pairs). 15 Water: Acid or Base? Water as Base: HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) Water as Acid: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq) Amphiprotic: Acts as an Acid or Base. Preface: Double-Displacement Reactions Many of the reactions in this chapter follow this pattern: (+) (-) (+) (-) (+) (-) (+) (-) AB + CD → AD + CB 16 Acid-Base (Neutralization) Reactions acid + base HCl(aq) + NaOH(aq) H2SO4(aq) + 2 KOH(aq) salt + water HOH(aq) + NaCl(aq) 2 HOH(aq) + K2SO4(aq) Net Ionic Equations 1. Write the balanced molecular equation. 2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 3. Cancel the spectator ions on both sides of the ionic equation Tips: • Do not break apart solids, liquids (such as H2O), and gases (e.g. CO2) • Do not break apart weak electrolytes (such as weak acids, e.g. HC2H3O2) • Carbonic acid, H2CO3, decomposes into CO2 and H2O • An acid such as H2SO4 breaks apart into 2H+(aq) + SO42-, not H22+ + SO42- 17 1. Write the balanced molecular equation. H2SO4(aq) + 2 KOH(aq) strong 2 H2O(aq) + K2SO4(aq) strong Not a strong electrolyte!! strong 2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 2 H+ SO42- 2 K+ 2OH- 2 H2O 2 K+ SO42- 3. Cancel the spectator ions on both sides of the ionic equation Net ionic equation: 2 H+ + 2 OH- 2 H2O H+ + OH- H2O Sample Exercise 8.5: Writing the Net Ionic Equation for a Neutralization Reaction Write the net ionic equation describing the reaction that takes place when acid rain containing sulfuric acid reacts with a marble statue (calcium carbonate) 1. Write the balanced molecular equation. H2SO4(aq) + CaCO3(s) strong solid H2O(l) + CO2(g) + CaSO4(aq) Not strong electrolytes!! strong* 2. Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions. 2H+(aq) + SO42-(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca2+(aq) + SO42-(aq) *generally speaking, CaSO4 is insoluble in water, but not in strong acid. More on this later when we study solubility rules. 18 Sample Exercise 8.5: Writing the Net Ionic Equation for a Neutralization Reaction 3. Cancel the spectator ions on both sides of the ionic equation 2H+(aq) + SO42-(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca2+(aq) + SO42-(aq) Net ionic equation: 2H+(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca2+(aq) Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange 19 Double-Displacement Reactions Again Precipitation reactions in this section follow the same pattern as acid-base reactions: (+) (-) (+) (-) (+) (-) (+) (-) AB + CD → AD + CB Types of Solutions Saturated solution: • Contains the maximum amount of solute that can dissolve in a given volume at a given temperature. • Solubility = g solute/100 mL solvent. Supersaturated solution: • Contains more dissolved solute than predicted at a given temperature. 20 A Saturated Solution Example Supersaturated Solution Sodium acetate precipitates from a supersaturated solution. 21 22 Precipitate: • Solid product formed from reactants in solution. • AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s) Can predict formation of precipitates based on solubility “rules.” We learned in Ch 4, p. 134, that the electrostatic energy (strength) of an ionic bond depended on the product of the charges, divided by the internuclear distance. The solubility of ionic compounds generally follows the trend that the greater the charge, and the shorter the internuclear distance, the more insoluble the compound is. α (alpha) means “is proportional to” Q+ = charge on the cation Q- = charge on the anion d = internuclear distance - + 𝑄+ 𝑥 𝑄− 𝐸 α 𝑑 d cations +1 compounds alkali metals NH4+ +2 +3 Al3+ anions compounds solubility -1 NO3- Always soluble CH3COO-2 SO42- Soluble with exceptions -2 S2-, O2- Insoluble with exceptions -3 PO43- Insoluble with exceptions 23 Always soluble Soluble w/exceptions and Ag+ Add to table Insoluble w/exceptions Table 8.3 - Soluble Compounds Soluble Cations: • Group I ions (alkali metals) and NH4+ Soluble Anions: • NO3− and CH3COO− (acetate) • Halides (Group 17) » Exceptions: Cu+, Hg22+, Ag+, Pb2+ • Sulfates (SO42−) » Exceptions: Hg22+, Ag+, Pb2+, Ca2+, Ba2+, Sr2+ Add to table 24 Table 8.3 - Insoluble Compounds All hydroxides (OH−) except: • Group IA = Li, Na etc and NH4+ • Slightly soluble: Group IIA = Ca(OH)2, Sr(OH)2, and Ba(OH)2 All sulfides (S2−) All carbonates (CO32−) except IA, NH4+ All phosphates (PO43−) All chromates (CrO42-) Mnemonic device to help remember the exceptions to the soluble compounds - Cute HAPpy halides HAPpy sulfates "2" Cl-, Br-, I- Cu+ Sr2+ Hg22+ Ag+ Ca2+ Pb2+ Hg22+ Ag+ Pb2+ Ba2+ from Group 2 (slightly) 25 Mnemonic device to help remember the insoluble compounds - Shop for a chrome car u l f i d e s S2- y d r o x i d e s OH- h o s p h a t e s h r o m a t e s a r b o n a t e s CrO42- CO32- PO43- Applying the Solubility Rules – Will a Precipitate Form? Does a precipitate form when sodium chloride is mixed with silver nitrate? If so, write the net ionic equation for the formation of the precipitate: Cute HAPpy halides Cl-, Br-, I- Cu+ Hg22+ Ag+ Pb2+ NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq) Na+(aq) + Ag+(aq) + Cl-(aq) + NO3-(aq) AgCl(s) + Na+(aq) + NO3-(aq) Ag+(aq) + Cl-(aq) AgCl(s) 26 Formation of an Insoluble Product NaI(aq) + Pb(NO3)2(aq) → The soluble compounds break up into ions and recombine - K+(aq) I-(aq) Pb2+(aq) NO3-(aq) Sample Exercise 8.6 A precipitate forms when aqueous solutions of ammonium sulfate and barium chloride are mixed. Write the net ionic equation for the reaction. (NH4)2SO4(aq) + BaCl2(aq) 2 NH4Cl(aq) + BaSO4(s) 2NH4+(aq) + SO42-(aq) + Ba2+(aq) + 2Cl-(aq) BaSO4(s) + 2NH4+(aq) + 2Cl-(aq) Ba2+(aq) + SO42-(aq) BaSO4(s) 27 Sample Exercise 8.7 – Prediction the Mass of a Precipitate Barium sulfate is used to enhance Xray imaging of the upper and lower GI tract. In upper GI imaging, patients drink a suspension of solid barium sulfate in H 2O. The compound is not toxic because of its limited solubility. To make pure barium sulfate, a precipitation reaction is employed: aqueous solutions of soluble barium nitrate and sodium sulfate are mixed together, and solid barium sulfate is separated by filtration. How many grams of barium sulfate (MW = 233.40) will be produced if exactly 1.00 L of 1.55 M barium nitrate is reacted with excess sodium sulfate? Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaNO3(aq) V = 1.00 L 1.55 M Plan: g BaSO4 = Excess no LR ?g MW = 233.40 ratio M x V = mol Ba(NO3)2 mol BaSO4 g BaSO4 233.40 g BaSO4 1.55 mol 1 mol BaSO4 x 1.00 L x x = 362 g BaSO4 1 mol BaSO4 1 mol Ba(NO ) L 3 2 Sample Exercise 8.8 – Calculating Solute Concentration from Mass of Precipitate To determine the concentration of chloride ion in a 100.0 mL sample of groundwater, a chemist adds a large enough volume of a solution of AgNO 3 to precipitate all the Cl- ions as silver chloride. The mass of the resulting precipitate is 71.7 mg. What is the Cl- concentration in the sample in mg/L? AgNO3(aq) + Cl-(aq) excess no LR AgCl(s) + NO3-(aq) V = 100.0 mL 71.7 mg = 0.100 L = 0.0717 g ? mg/L ratio Plan: mg AgCl g AgCl mol AgCl mol Cl- g Cl- Then mg/L Cl- = g Cl- mg Cl- mg/L Divide by liters of sample 28 Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange 29 (electron transfer reactions) Section 8.6 - Oxidation-Reduction Reactions Oxidation-Reduction Reactions (Redox): Characterized by gain or loss of electrons by atoms involved in the reaction. Oxidation: Historical definition = gain of oxygen (sometimes with a simultaneous loss of hydrogen) e.g. C2H6 + O2 CO2 + H2O Modern definition = loss of electrons Reduction: Historical definition = loss of oxygen (sometimes with a simultaneous gain in hydrogen) e.g. Fe2O3 + 3 CO 2 Fe + 3 CO2 “oil rig” Modern definition = gain of electrons Table 8.4 - Oxidation Number Rules The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred, i.e. applies to both ionic and covalent compounds. 1. The O.N. before a reaction for pure elements = 0 (including diatomics). Mg HH O O 2. O.N = the charge on monovalent ions Cl Cl + e Mg Mg2+ + 2 e 3. O.N. of fluorine = -1 for all of its compounds 30 4. O.N. of oxygen = -2 in nearly all of its compounds O +2e 2- O O Exceptions are peroxides (O.N. = -1) and superoxides (O.N. = -1/2), e.g. H2O2, KO2 5. O.N. of hydrogen = +1 in nearly all of its compounds HH 2 H+ + 2 e Exception = hydrides, e.g. CaH2 6. O.N. values of the atoms in a neutral molecule sum up to zero CaCl2 +2 + 2(Cl) = 0 so Cl = -1 7. O.N. values of the atoms in a polyatomic ion sum up to the charge on the ion 3- PO4 P + 4(-2) = -3 so P = +5 Sample Exercise 8.9 What are the oxidation states for sulfur in (a) S8, (b) SO2, (c) Na2S, and (d) CaSO4 ? 31 Electron Transfer in Redox Reactions What follows is a method that keeps track of the flow of electrons within the atoms and molecules taking part in a chemical reaction. O.N. = final O.N. - initial O.N. oxidation lose n electrons O.N. = +n reduction gain n electrons O.N. = -n Pb0 Pb2+ + 2 e O.N. = (2 – 0) = +2 Cu2+ + 2 e Cu0 O.N. = (0 – 2) = -2 32 What is the net ionic equation for the reaction of zinc metal plus cupric sulfate? Oxidizing Agent = oxidizes something else while being reduced (Cu 2+ Cu0) Reducing Agent = reduces something else while being oxidized (Zn Zn2+) 33 Sample Exercise 8.10: Identifying Oxidizing and Reducing Agents and Determining Number of Electrons Transferred Energy released by the reaction of hydrazine and dinitrogen tetroxide is used to orient and maneuver spacecraft and to propel rockets into space. Identify the elements that are oxidized and reduced, the oxidizing agent, and the reducing agent, and determine the number of electrons transferred in the balanced chemical equation. (+4) (-2) (-2) (+1) 2N + 4(+1) = 0 So N = -2 2N + 4(-2) = 0 So N = +4 (0) Element in its natural state (+1) (-2) O.N. rules H = +1 O = -2 O.N. = 0 – (-2) = +2 lost electrons oxidation (-2) (+4) (0) O.N. = (0 – 4) = -4 gained electrons reduction 34 Net number of electrons transferred during the reaction: = (#molecules) x (#atoms in molecule) x O.N. O.N. from previous slide 2 x 2 x +2 = +8 (oxidation) 1 x 2 x -4 = -8 (reduction) O.N. from previous slide So the net number of electrons flowing is 8 Balancing Redox Reactions in Acid and Base (We’re going to use a simpler method than the book called the Method of Half Reactions) Examples of Half Reactions: copper wire immersed in a silver nitrate solution: (+1) (+5) (-2) (+2) (+5) (-2) (0) (0) Cu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2 Ag(s) O.N. Cu = +2 – (0) = +2 oxidation O.N. Ag = 0 – (1) = -1 reduction The half-reactions are: ox: Cu(s) Cu2+(aq) + 2 e red: Ag+(aq) + e Ag(s) Cu(s) Ag(s) AgNO3(aq) (clear soln) Cu(NO3)2(aq) (blue soln) 35 Examples of Half Reactions (cont’d): copper wire immersed in a silver nitrate solution: The total reaction is obtained by adding the half reactions together, after balancing electrons ox: Cu(s) Cu2+(aq) + 2 e red: Ag+(aq) + e Ag(s) ox: x2 Cu(s) Cu2+(aq) + 2 e red: 2 Ag+(aq) + 2 e 2 Ag(s) Net: Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) The next topic is how to balance a redox reaction that occurs in acid or base. Procedure for balancing a redox reaction in acid or base. 1. 2. 3. 4. 5. Determine the oxidation numbers for the reactants and compare to the products. Write the oxidation and reduction half-reactions without electrons (yet) Balance everything but oxygen and hydrogen Balance oxygen by adding water Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions, continue as if in acidic solution, but at the end each H + ion will be neutralized by adding OH- ions 6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons will be on the right, for the reduction half-reaction, the electrons will appear on the left 7. Make sure the number of electrons in each half-reaction are the same. Then add the half reactions together 8. Make sure that the equation is balanced for mass and for charge NOTE: sometimes you have to cancel H2O’s that are on each side, as well as H+ or OH- 36 Sample Exercise 8.11 Balancing Redox Equations I: Acidic Solutions Microorganisms called denitrifying bacteria that grow in waterlogged soil convert NO3- ions into N2O gas as they feed on dead plant tissue (empirical formula = CH2O), converting it into CO2 and H2O. Write a balanced net ionic equation describing this conversion of dissolved nitrates to N 2O gas. Assume the reaction occurs in slightly acidic water. 1. Determine the oxidation numbers for the reactants and compare to the products. (+5) (-2) (0) (+1) (-2) (+1) (-2) (+4) (-2) (+1) (-2) NO3-(aq) + CH2O(s) N2O(g) + CO2(g) + H2O(l) Given: O.N. C = +4 – (0) = +4 oxidation O.N. N = 1 – (5) = -4 reduction 2. Write the oxidation and reduction half-reactions without electrons (yet) ox: CH2O(s) CO2(g) red: NO3-(aq) N2O(g) 3. Balance everything but oxygen and hydrogen ox: CH2O(s) CO2(g) red: 2 NO3-(aq) N2O(g) 4. Balance oxygen by adding water ox: H2O + CH2O(s) CO2(g) red: 2 NO3-(aq) N2O(g) + 5 H2O 37 5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions, continue as if in acidic solution, but at the end each H + ion will be neutralized by adding OH- ions ox: H2O + CH2O(s) CO2(g) + 4 H+(aq) red: 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O 6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons will be on the right, for the reduction half-reaction, the electrons will appear on the left ox: H2O + CH2O(s) CO2(g) + 4 H+(aq) + 4 e red: 8 e + 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O 7. Make sure the number of electrons in each half-reaction are the same. Then add the half reactions together ox: H2O + CH2O(s) CO2(g) + 4 H+(aq) + 4 e x2 red: 8 e + 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O Net: 2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 5H2O NOTE: sometimes you have to cancel H2O’s that are on each side, as well as H+ or OH2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 5H2O 2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 3H2O 2CH2O(s) + 2H+(aq) + 2NO3-(aq) 2CO2(g) + N2O(g) + 3H2O 38 8. Make sure that the equation is balanced for mass and for charge 2CH2O(s) + 2H+(aq) + 2NO3-(aq) 2CO2(g) + N2O(g) + 3H2O C=2 C=2 H=6 O=8 H=6 O=8 N=2 N=2 Mass: Charge: 2(+1) + 2(-1) = 0 =0 Homework Problem 98(c) Balancing Redox Equations II: Basic Solutions Permanganate ion is used in water purification to remove oxidizable substances. Complete and balance the reaction for the removal of sulfite ion. 1. Determine the oxidation numbers for the reactants and compare to the products. (+7) (-2) (+4) (-2) (+4) (-2) (+6) (-2) MnO4-(aq) + SO32-(aq) MnO2(s) + SO42-(aq) O.N. S = +6 – (+4) = +2 oxidation O.N. Mn = +4 – (+7) = -3 reduction 2. Write the oxidation and reduction half-reactions without electrons (yet) ox: SO32-(aq) red: MnO4-(aq) SO42-(aq) MnO2(s) 39 3. Balance everything but oxygen and hydrogen ox: SO32-(aq) red: MnO4-(aq) SO42-(aq) MnO2(s) S and Mn are already balanced 4. Balance oxygen by adding water ox: H2O + SO32-(aq) SO42-(aq) red: MnO4-(aq) MnO2(s) + 2H2O 5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions, continue as if in acidic solution, but at the end each H+ ion will be neutralized by adding OH- ions ox: H2O + SO32-(aq) SO42-(aq) + 2H+(aq) red: 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O 6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons will be on the right, for the reduction half-reaction, the electrons will appear on the left ox: H2O + SO32-(aq) SO42-(aq) + 2H+(aq) + 2 e red: 3 e + 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O 7. Make sure the number of electrons in each half-reaction are the same. Then add the half reactions together ox: H2O + SO32-(aq) SO42-(aq) + 2H+(aq) + 2 e x3 red: 3 e + 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O x 2 Net: 3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4-(aq) 3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O 40 NOTE: sometimes you have to cancel H2O’s that are on each side, as well as H+ or OH3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4-(aq) 3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O 3SO32-(aq) + 8H+(aq) + 2MnO4-(aq) 3SO42-(aq) + 6H+(aq) + 2MnO2(s) + H2O 3SO32-(aq) + 2H+(aq) + 2MnO4-(aq) 3SO42-(aq) + 2MnO2(s) + H2O + 2OH+ 2OH2H2O 3SO32-(aq) + H2O(l) + 2MnO4-(aq) 3SO42-(aq) + 2MnO2(s) + 2OH-(aq) 8. Make sure that the equation is balanced for mass and for charge 3SO32-(aq) + H2O(l) + 2MnO4-(aq) 3SO42-(aq) + 2MnO2(s) + 2OH-(aq) Mass: Charge: S=3 S=3 O = 18 H=2 O = 18 H=2 Mn = 2 Mn = 2 3(-2) + 2(-1) = -8 3(-2) +2(-1) = -8 41 Chapter Outline 8.1 Solutions and Their Concentrations 8.2 Dilutions 8.3 Electrolytes and Nonelectrolytes 8.4 Acids, Bases, and Neutralization Reactions 8.5 Precipitation Reactions 8.6 Oxidation-Reduction Reactions 8.7 Titrations 8.8 Ion Exchange Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 42 Titration Example, p. 344-345 H2SO4 + 2NaOH → Na2SO4 + 2H2O (titrant) End Point H2SO4 (unknown) Titration Calculations, p. 344 Suppose it takes 22.40 mL of 0.00100 M sodium hydroxide to react completely with the sulfuric acid in a 100.0 mL sample of acidic mine drainage. What was the molarity of the sulfuric acid? 2NaOH(aq) + H2SO4(aq) 2H2O + Na2SO4 0.00100 M 22.40 mL = 0.02240 L 100.0 mL = 0.1000 L M=? 1 mol H2SO4 2 mol NaOH Plan: M x V = mol NaOH mol H2SO4 V M mol H2SO4 = 0.00100 mol NaOH x 0.02240 L NaOH x 1 mol H2SO4 L 2 mol NaOH = 1.120 x 10-5 mol M = 1.120 x 10-5 mol 0.1000 L = 1.120 x 10-4 M 43 Sample Exercise 8.13: Apple cider vinegar is an aqueous solution of acetic acid (MW = 60.05) made from fermented apple juice. Commercial vinegar must contain at least 4 grams of acetic acid per 100 mL of vinegar. Suppose the titration of a 25.00 mL sample of vinegar requires 12.15 mL of 1.885 M sodium hydroxide. What is the molarity of acetic acid in the vinegar sample? Does it meet the 4 g/100 mL standard? HC2H3O2(aq) + NaOH(aq) H2O + NaC2H3O2 25.00 mL M=? mg/L = ? 12.15 mL = 0.01215 L M = 1.885 1 mol HC2H3O2 1 mol NaOH Plan: M x V = mol NaOH M HC2H3O2 = 1.885 mol NaOH L mol HC2H3O2 MW V M g/mL x 0.01215 L NaOH x 1 mol HC2H3O2 x 1 1 mol NaOH 0.02500 L = 0.9161 M g L = 0.9161 mol x 60.05 g 1mol L x 1L 1000 mL = 0.05501 g/mL = 5.501 g/100 mL 44