© 2012 Pearson Education, Inc.

A. H2O(l) B. K+(aq) and H2O(l) C. CN-(aq) D. K+(aq) and CN-(aq)
Answer: D

A. H2O(l) B. Na+(aq) and H2O(l) C. ClO4-(aq) D. Na+(aq) and ClO4-(aq)
Answer: D

Electrolytes and Nonelectrolytes
Soluble ionic compounds tend to be electrolytes. Molecular compounds tend to be non-electrolytes, except for acids and bases. © 2012 Pearson Education, Inc.

Answer: A A. NaCl(aq) B. CH3OH(aq)

D. Cannot determine from Figure 4.2
A. CH3OH(aq) B. NaOH(aq) C. CH3COOH(aq) D. Cannot determine from Figure 4.2 Answer: B Editor: This question requires the availability of Figure It should be provided with this question.

Strong Electrolytes Are…
Strong acids Strong bases Soluble ionic salts © 2012 Pearson Education, Inc.

Sample Exercise 4.1 Relating Relative Numbers of Anions and Cations to Chemical Formulas
The accompanying diagram represents an aqueous solution of either MgCl2, KCl, or K2SO4. Which solution does the drawing best represent? The diagram shows twice as many cations as anions, consistent with the formulation K2SO4. Practice Exercise If you were to draw diagrams representing aqueous solutions of (a) NiSO4, (b) Ca(NO3)2, (c) Na3PO4, (d) Al2(SO4)3, how many anions would you show if each diagram contained six cations? Answer: (a) 6, (b) 12, (c) 2, (d) 9

Precipitation Reactions
When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed. © 2012 Pearson Education, Inc.

Answer: C K+ and I– C. K+ and NO3– Pb2+ and I– D. Pb2+ and NO3–

Sample Exercise 4.2 Using Solubility Rules
Classify these ionic compounds as soluble or insoluble in water: (a) sodium carbonate, Na2CO3, (b) lead sulfate, PbSO4. According to Table 4.1, most carbonates are insoluble. But carbonates of the alkali metal cations (such as sodium ion) are an exception to this rule and are soluble. Thus, Na2CO3 is soluble in water. Table 4.1 indicates that although most sulfates are water soluble, the sulfate of is an exception. Thus, PbSO4 is insoluble in water. Practice Exercise Classify the following compounds as soluble or insoluble in water: (a) cobalt II hydroxide, (b) barium nitrate, (c) ammonium phosphate. Answers: (a) insoluble, (b) soluble, (c) soluble 16

Metathesis (Exchange) Reactions
Metathesis comes from a Greek word that means “to transpose.” It appears as though the ions in the reactant compounds exchange, or transpose, ions: AgNO3(aq) + KCl(aq)  AgCl(s) + KNO3(aq) © 2012 Pearson Education, Inc.

BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2 KCl(aq)
Sample Exercise 4.3 Predicting a Metathesis Reaction (a) Predict the identity of the precipitate that forms when aqueous solutions of BaCl2 and K2SO4 are mixed. (b) Write the balanced chemical equation for the reaction. (a) The reactants contain Ba2+, Cl–, K+, and SO42– ions. Exchanging the anions gives us BaSO4 and KCl. According to Table 4.1, most compounds of SO42– are soluble but those of Ba2+ are not. Thus, BaSO4 is insoluble and will precipitate from solution. KCl is soluble. (b) From part (a) we know the chemical formulas of the products, BaSO4 and KCl. The balanced equation is BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2 KCl(aq) Practice Exercise (a) What compound precipitates when aqueous solutions of Fe2(SO4)3 and LiOH are mixed? (b) Write a balanced equation for the reaction. (c) Will a precipitate form when solutions of Ba(NO3)2 and KOH are mixed? Answer: (a) Fe(OH)3, (b) Fe2(SO4)3(aq) + 6 LiOH(aq)  2 Fe(OH)3(s) + 3 Li2SO4(aq), (c) no (both possible products, Ba(OH)2 and KNO3, are water soluble) 18

Solution Chemistry It is helpful to pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution). If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction. © 2012 Pearson Education, Inc.

Ionic Equation Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq) 
In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. This more accurately reflects the species that are found in the reaction mixture: Ag+(aq) + Cl−(aq)  AgCl(s) To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. The only things left in the equation are those things that change (i.e., react) during the course of the reaction. Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions: Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq)  AgCl(s) + K+(aq) + NO3−(aq) © 2012 Pearson Education, Inc.

Writing Net Ionic Equations
Write a balanced molecular equation. Dissociate all strong electrolytes. Cross out anything that remains unchanged from the left side to the right side of the equation. Write the net ionic equation with the species that remain. © 2012 Pearson Education, Inc.

D. No spectator ions are involved
A. Ag+(aq) and Cl-(aq) B. NO3-(aq) and Cl-(aq) C. Na+(aq) and NO3-(aq) D. No spectator ions are involved Answer: C

Sample Exercise 4.4 Writing a Net Ionic Equation
Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of calcium chloride and sodium carbonate are mixed. CaCl2(aq) + Na2CO3(aq)  CaCO3(s) + 2 NaCl(aq) Ca2+(aq) + 2 Cl–(aq) + 2 Na+(aq) + CO32–(aq)  CaCO3(s) + 2 Na+(aq) + 2 Cl–(aq) Ca2+(aq) + CO32–(aq)  CaCO3(s) Practice Exercise Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of silver nitrate and potassium phosphate are mixed. Answer: 3 Ag+(aq) + PO43–(aq)  Ag3PO4(s) 24

Acids The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H+ when dissolved in water. Both the Danish chemist J. N. Brønsted and the British chemist T. M. Lowry defined them as proton donors. There are only seven strong acids: Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO3) Sulfuric (H2SO4) Chloric (HClO3) Perchloric (HClO4) © 2012 Pearson Education, Inc.

Answer: D A B C D E. 4

Bases Arrhenius defined bases as substances that increase the concentration of OH− when dissolved in water. Brønsted and Lowry defined them as proton acceptors. The strong bases are the soluble metal salts of hydroxide ion: Alkali metals Calcium Strontium Barium © 2012 Pearson Education, Inc.

A. Al(OH)3 is not basic in water. B. Al(OH)3 is insoluble in water.
C. Al(OH)3 is a strong acid in water, not basic. D. Al(OH)3 is a weak acid in water, not basic. Answer: B

Sample Exercise 4.5 Comparing Acid Strengths
The following diagrams represent aqueous solutions of acids HX, HY, and HZ, with water molecules omitted for clarity. Rank the acids from strongest to weakest. The order is HY > HZ > HX. HY is a strong acid because it is totally ionized (no HY molecules in solution), whereas both HX and HZ are weak acids, whose solutions consist of a mixture of molecules and ions. Because HZ contains more ions and fewer molecules than HX, it is a stronger acid. Practice Exercise Imagine a diagram showing 10 Na+ ions and 10 OH– ions. If this solution were mixed with the one pictured above for HY, what species would be present in a diagram that represents the combined solutions after any possible reaction? Answer: The diagram would show 10 Na+ ions, 2 OH– ions, 8 Y– ions, and 8 H2O molecules. 30

Sample Exercise 4.6 Identifying Strong, Weak, and Nonelectrolytes
Classify these dissolved substances as strong, weak, or nonelectrolyte: CaCl2, HNO3, C2H5OH (ethanol), HCOOH (formic acid), KOH. Ionic compounds: CaCl2 and KOH all ionic compounds are strong electrolytes. Molecular compounds: HNO3, HCOOH, and C2H5OH. Nitric acid, HNO3, is a common strong acid, and therefore is a strong electrolyte. Formic acid HCOOH is a weak acid, and therefor a weak electrolyte. Ethanol, C2H5OH, is neither an acid nor a base, so it is a nonelectrolyte. Practice Exercise Consider solutions in which 0.1 mol of each of the following compounds is dissolved in 1 L of water: Ca(NO3)2 (calcium nitrate), C6H12O6 (glucose), NaCH3COO (sodium acetate), and CH3COOH (acetic acid). Rank the solutions in order of increasing electrical conductivity, based on the fact that the greater the number of ions in solution, the greater the conductivity. Answer: C6H12O6 (nonelectrolyte) < CH3COOH (weak electrolyte, existing mainly in the form of molecules with few ions) < NaCH3COO (strong electrolyte that provides two ions, Na+ and CH3COO–) < Ca(NO3)2 (strong electrolyte that provides three ions, Ca2+ and 2 NO3–)

Neutralization Reactions
Generally, when solutions of an acid and a base are combined, the products are a salt and water: CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) © 2012 Pearson Education, Inc.

Neutralization Reactions
When a strong acid reacts with a strong base, the net ionic equation is HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) H+(aq) + Cl−(aq) + Na+(aq) + OH−(aq)  Na+(aq) + Cl−(aq) + H2O(l) H+(aq) + OH−(aq)  H2O(l) © 2012 Pearson Education, Inc.

C. Magnesium hydroxide reacts only slowly with hydrochloric acid.
Answers: B A. Insufficient chloride ions are added to react with all of the solid magnesium hydroxide. B. Insufficient hydrogen ions are added to react with all of the solid magnesium hydroxide. C. Magnesium hydroxide reacts only slowly with hydrochloric acid. D. A neutralization reaction does not occur.

Sample Exercise 4.7 Writing Chemical Equations for a Neutralization Reaction
For the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OH)2, write (a) the balanced molecular equation, (b) the complete ionic equation, (c) the net ionic equation. The salt contains the cation of the base (Ba2+) and the anion of the acid (CH3COO–). Thus, the salt formula is Ba(CH3COO)2. According to Table 4.1, this compound is soluble in water. The unbalanced molecular equation for the neutralization reaction is CH3COOH (aq) + Ba(OH)2 (aq)  H2O (l) + Ba(CH3COO)2 (aq) 2 CH3COOH (aq) + Ba(OH)2 (aq)  2 H2O (l) + Ba(CH3COO)2 (aq) 2 CH3COOH (aq) + Ba2+(aq) + 2 OH- (aq)  2 H2O (l) + Ba CH3COO- (aq) 2 CH3COOH (aq) + 2 OH- (aq)  2 H2O (l) + 2 CH3COO- (aq) CH3COOH (aq) + OH- (aq)  H2O (l) + CH3COO- (aq) 35

Sample Exercise 4.7 Writing Chemical Equations for a Neutralization Reaction
Continued Practice Exercise For the reaction of phosphorous acid (H3PO3) and potassium hydroxide (KOH), write (a) the balanced molecular equation and (b) the net ionic equation. Answer: (a) H3PO3(aq) + 3 KOH(aq)  3 H2O(l) + K3PO3(aq), (b) H3PO3(aq) + 3 OH–(aq)  3 H2O(l) + PO33–(aq). (H3PO3 is a weak acid and therefore a weak electrolyte, whereas KOH, a strong base, and K3PO3, an ionic compound, are strong electrolytes.) 36

Gas-Forming Reactions
Some metathesis reactions do not give the product expected. In this reaction, the expected product (H2CO3) decomposes to give a gaseous product (CO2): When a carbonate or bicarbonate reacts with an acid, the products are a salt, carbon dioxide, and water: CaCO3(s) + HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) NaHCO3(aq) + HBr(aq) NaBr(aq) + CO2(g) + H2O(l) © 2012 Pearson Education, Inc.

Gas-Forming Reactions
Similarly, when a sulfite reacts with an acid, the products are a salt, sulfur dioxide, and water: This reaction gives the predicted product, but you had better carry it out in the hood, or you will be very unpopular! Just as in the previous examples, a gas is formed as a product of this reaction: SrSO3(s) + 2HI(aq) SrI2(aq) + SO2(g) + H2O(l) Na2S(aq) + H2SO4(aq)  Na2SO4(aq) + H2S(g) © 2012 Pearson Education, Inc.

A. SO2(g) B. H2(g) C. CO2(g) D. H2S(g) Answer: A

Oxidation-Reduction Reactions
An oxidation occurs when an atom or ion loses electrons. A reduction occurs when an atom or ion gains electrons. One cannot occur without the other. © 2012 Pearson Education, Inc.

Oxidation Numbers To determine if an oxidation–reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity. Elements in their elemental form have an oxidation number of 0. The oxidation number of a monatomic ion is the same as its charge. © 2012 Pearson Education, Inc.

Oxidation Numbers Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. Oxygen has an oxidation number of −2, except in the peroxide ion, in which it has an oxidation number of −1. Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal. Fluorine always has an oxidation number of −1. The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. © 2012 Pearson Education, Inc.

Sample Exercise 4.8 Determining Oxidation Numbers
Determine the oxidation number of sulfur in (a) H2S, (b) S8, (c) SCl2, (d) Na2SO3, (e) SO42–. (a) hydrogen is+1 (rule 3b). H2S is neutral. (rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of –2. (b) Because this is an elemental form of sulfur, the oxidation number of S is 0 (rule 1). (c) Chlorine is –1 (rule 3c). SCl2 is neutral (rule 4). Letting x equal the oxidation number of S, we have x + 2(–1) = 0. Consequently, the oxidation number of S must be +2. (d) Sodium is +1 (rule 2). Oxygen is –2 (rule 3a). Letting x equal the oxidation number of S, we have 2(+1) + x + 3(–2) = 0. Therefore, the oxidation number of S in this compound is +4. (eO is –2 (rule 3a) the net charge of the SO42– ion is -2 (rule 4). Thus, we have x + 4(–2) = –2. From this relation we conclude that the oxidation number of S in this ion is +6. Practice Exercise What is the oxidation state of the boldfaced element in (a) P2O5, (b) NaH, (c) Cr2O72–, (d) SnBr4, (e) BaO2? Answer: (a) +5, (b) –1, (c) +6, (d) +4, (e) –1 44

A. +1 B. -1 C. -2 D. -3 Answer: D

A. +6 B. +5 C. +4 D. -1 Answer: B

Displacement Reactions
In displacement reactions, ions oxidize an element. The ions, then, are reduced. © 2012 Pearson Education, Inc.

Sample Exercise 4.9 Writing Equations for Oxidation-Reduction Reactions
Write the balanced molecular and net ionic equations for the reaction of aluminum with hydrobromic acid. The reactants are Al and HBr. The cation formed by Al is Al3+, and the anion from hydrobromic acid is Br–. Thus, the salt formed in the reaction is AlBr3.Writing the reactants and products and then balancing the equation gives the molecular equation: 2 Al(s) + 6 HBr(aq)  2 AlBr3(aq) + 3 H2(g) Both HBr and AlBr3 are soluble strong electrolytes. Thus, the complete ionic equation is 2 Al(s) + 6 H+(aq) + 6 Br–(aq)  2 Al3+(aq) + 6 Br–(aq) + 3 H2(g) Because Br– is a spectator ion, the net ionic equation is 2 Al(s) + 6 H+(aq)  2 Al3+(aq) + 3 H2(g) Practice Exercise (a) Write the balanced molecular and net ionic equations for the reaction between magnesium and cobalt(II) sulfate. (b) What is oxidized and what is reduced in the reaction? Answer: (a) Mg(s) + CoSO4(aq)  MgSO4(aq) + Co(s) Mg(s) + Co2+(aq)  Mg2+(aq) + Co(s), (b) Mg is oxidized and Co2+ is reduced. 48

Al + H+  Al3+ + H2 When the oxidation-reduction reaction above is correctly balanced, the coefficients are 1, 2  1, 1. 1, 3  1, 2. 2, 3  2, 3. d. 2, 6  2, 3. Answer: d

Displacement Reactions
In this reaction, silver ions oxidize copper metal: Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) The reverse reaction however, does not occur: Cu2+(aq) + 2Ag(s)  Cu(s) + 2Ag+(aq) x © 2012 Pearson Education, Inc.

Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
Sample Exercise 4.10 Determining When an Oxidation-Reduction Reaction Can Occur Will an aqueous solution of iron(II) chloride oxidize magnesium metal? If so, write the balanced molecular and net ionic equations for the reaction. Because Mg is above Fe in the table, the reaction occurs. Mg(s) + FeCl2(aq)  MgCl2(aq) + Fe(s) The net ionic equation is Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s) Practice Exercise Which of the following metals will be oxidized by Pb(NO3)2: Zn, Cu, Fe? Answer: Zn and Fe 52

? Yes No Answer: Yes Editor: Table 4.5 is required to answer this question. It should be provided as a separate ppt for instructor to use.

Yes No Answer: No

Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution: moles of solute volume of solution in liters Molarity (M) = © 2012 Pearson Education, Inc.

Mixing a Solution To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute. The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask. © 2012 Pearson Education, Inc.

21.0 g of NaF dissolved in water to make 500 mL of solution
Answer: B Molarity = number of moles solute/volume of solution in liters.

Sample Exercise 4.11 Calculating Molarity
Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution. The number of moles of Na2SO4 is obtained by using its molar mass: Converting the volume of the solution to liters: Thus, the molarity is Practice Exercise Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL of solution. Answer: M 58

Sample Exercise 4.12 Calculating Molar Concentrations of Ions
What is the molar concentration of each ion present in a M aqueous solution of calcium nitrate? Calcium nitrate is composed of calcium ions (Ca2+) and nitrate ions (NO3–), so its chemical formula is Ca(NO3)2. Because there are two NO3– ions for each Ca2+ ion, each mole of Ca(NO3)2 that dissolves dissociates into 1 mol of Ca2+ and 2 mol of NO3–. Thus, a solution that is M in Ca(NO3)2 is M in Ca2+ and 2  M = M in NO3–: Practice Exercise What is the molar concentration of K+ ions in a M solution of potassium carbonate? Answer: M 59

Sample Exercise 4.13 Using Molarity to Calculate Grams of Solute
How many grams of Na2SO4 are required to make L of M Na2SO4? Calculating the moles of Na2SO4 using the molarity and volume of solution gives Because each mole of Na2SO4 has a mass of 142 g, the required number of grams of Na2SO4 is Practice Exercise (a) How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4? (b) How many milliliters of 0.50 M Na2SO4 solution are needed to provide mol of this salt? Answer: (a) 1.1 g, (b) 76 mL 60

Dilution One can also dilute a more concentrated solution by Using a pipet to deliver a volume of the solution to a new volumetric flask, and Adding solvent to the line on the neck of the new flask. The molarity of the new solution can be determined from the equation Mc  Vc = Md  Vd, where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions. © 2012 Pearson Education, Inc.

A. Concentration (molarity) remains the same.
B. The new concentration is 0.25 M. C. The new concentration is 1.00 M. D. The new concentration is 2.50 M. Answer: B

Sample Exercise 4.14 Preparing a Solution by Dilution
How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4? Calculating the moles of H2SO4 in the dilute solution: Calculating the volume of the concentrated solution that contains mol H2SO4: Converting liters to milliliters gives 15 mL. We get the same result using M1V1 = M2V2 Practice Exercise (a) What volume of 2.50 M lead(II) nitrate solution contains mol of Pb2+? (b) How many milliliters of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution? (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting stock solution? Answer: (a) L = 20.0 mL, (b) 5.0 mL, (c) 0.40 M 63

Sample Exercise 4.15 Using Mass Relations in a Neutralization Reaction
How many grams of Ca(OH)2 are needed to neutralize 25.0 mL of M HNO3? The product of the molar concentration of a solution and its volume in liters gives the number of moles of solute: Because this is a neutralization reaction, HNO3 and Ca(OH)2 react to form H2O and the salt containing Ca2+ and NO3–: Thus, 2 mol HNO3 ≈ mol Ca(OH)2. Therefore, Practice Exercise (a) How many grams of NaOH are needed to neutralize 20.0 mL of M H2SO4 solution? (b) How many liters of M HCl(aq) are needed to react completely with mol of Pb(NO3)2(aq), forming a precipitate of PbCl2(s)? Answer: (a) g, (b) L 65

A. Increase by one-half the volume used for titration with NaOH.
Answer: A A. Increase by one-half the volume used for titration with NaOH. B. Increase by two the volume used for titration with NaOH. C. Decrease by two the volume used for titration with NaOH. D. Decrease by one-half the volume used for titration with NaOH.

Sample Exercise 4.16 Determining Solution Concentration by an Acid–Base Titration
One commercial method used to peel potatoes is to soak them in a NaOH solution for a short time, then remove them and spray off the peel. The NaOH concentration is normally 3 to 6 M, and the solution must be analyzed periodically. In one such analysis, 45.7 mL of M H2SO4 is required to neutralize 20.0 mL of NaOH solution. What is the concentration of the NaOH solution? The number of moles of H2SO4 is the product of the volume and molarity of this solution: Acids react with metal hydroxides to form water and a salt. Thus, the balanced equation for the neutralization reaction is According to the balanced equation, 1 mol H2SO4 ≈ 2 mol NaOH. Therefore, Knowing the number of moles of NaOH in 20.0 mL of solution allows us to calculate the molarity of this solution: 68

Sample Exercise 4.16 Determining Solution Concentration by an Acid–Base Titration
Continued Practice Exercise What is the molarity of an NaOH solution if 48.0 mL neutralizes 35.0 mL of M H2SO4? Answer: M 69

Ag+(aq) + Cl–(aq)  AgCl(s)
Sample Exercise 4.17 Determining the Quantity of Solute by Titration The quantity of Cl– in a municipal water supply is determined by titrating the sample with Ag+. The precipitation reaction taking place during the titration is Ag+(aq) + Cl–(aq)  AgCl(s) The end point in this type of titration is marked by a change in color of a special type of indicator. (a) How many grams of chloride ion are in a sample of the water if 20.2 mL of M Ag+ is needed to react with all the chloride in the sample? (b) If the sample has a mass of 10.0 g, what percent Cl– does it contain? (a) From the balanced equation we see that 1 mol Ag+ ≈ 1 mol Cl-. Using this information and the molar mass of Cl, we have (b) 70

MnO4–(aq) + 5 Fe2+(aq) + 8 H+(aq) Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
Sample Exercise 4.17 Determining the Quantity of Solute by Titration Continued Practice Exercise A sample of an iron ore is dissolved in acid, and the iron is converted to Fe2+. The sample is then titrated with mL of M MnO4– solution. The oxidation-reduction reaction that occurs during titration is MnO4–(aq) + 5 Fe2+(aq) + 8 H+(aq) Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l) (a) How many moles of MnO4– were added to the solution? (b) How many moles of Fe2+ were in the sample? (c) How many grams of iron were in the sample? (d) If the sample had a mass of g, what is the percentage of iron in the sample? Answer: (a)  10–3 mol MnO4–, (b)  10–3 mol Fe2+, (c) g, (d) 33.21% 71

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