SOLUTIONS - CHAPTER 9 Problems

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1 SOLUTIONS - CHAPTER 9 Problems NOTE: Exam 3 will be on Tuesday, November 22 nd, from 8:30pm to 10:30pm. Room assignments for the exam are the same as for the first two exams. The exam will cover the following chapters: Chapter 6, sections 6.4 to 6.6; Chapter 7, all; Chapter 8, all; Chapter 9, all). 1) (Burdge, 9.2) What is the difference between a nonelectrolyte and an electrolyte? Between a weak electrolyte and a strong electrolyte? An electrolyte is a substance that produces ions when dissolved in water. A nonelectrolyte is a substance that does not produce ions when dissolved in water (such as sugar). Note that water itself is considered a nonelectrolyte, although it does produce a very small (~ 10-7 mol/l) concentration of ions at room temperature due to selfionization. A strong electrolyte completely ionizes when dissolved in water to form two or more ions per formula unit of compound. A weak electrolyte ionizes to only a small extent, producing much less than two ions per formula unit of dissolved compound. Strong acids and ionic compounds are strong electrolytes, while weak acids and weak bases are weak electrolytes. Note that insoluble ionic compounds are strong electrolytes, because for whatever amount of the compound dissolves in water two or more ions are formed. Figure 9.2 on page 322 gives a useful flow chart for classifying substances as strong electrolytes, weak electrolytes, or nonelectrolytes. 2) While an aqueous solution of sodium chloride (NaCl) conducts electricity, solid sodium chloride does not. Explain this observation. Solid sodium chloride consists of cations and anions arranged in a crystal structure. Since the ions cannot move, they cannot transfer electricity, and so solid sodium chloride is an insulator. When sodium chloride is dissolved in water it forms free cations and anions (Na + (aq) and Cl - (aq)) that can freely move throughout the solution. The ions can therefore be used to transport electricity. A solution of NaCl in water is therefore an electrical conductor. 3) (Burdge, 9.12) Identify each of the following substances as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. a) H2O nonelectrolyte b) KCl strong electrolyte (ionic compound) c) HNO3 strong electrolyte (strong acid) d) HC 2 H 3 O 2 weak electrolyte (weak acid) e) C12H22O11 nonelectrolyte (a disaccharide, a form of sugar) 1

2 4) Identify each of the following compounds as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. a) C6H12O6 nonelectrolyte (a monosaccharide, a molecular compound) b) HNO2 weak electrolyte (a weak acid) c) KOH strong electrolyte (a strong soluble base) d) Fe(OH)3 strong electrolyte (an insoluble strong base. Recall that all ionic compounds are considered strong electrolytes, because even if only a small amount of the compound dissolves in water, the amount that does dissolve breaks apart to form Fe +3 and OH -.) e) CH3CH2OH nonelectrolyte (an alcohol, and so a molecular compound) 5) (Burdge, 9.14) Describe hydration. What properties of water enable its molecules to interact with ions in solution? Hydration is the process where particles (often ions) are surrounded by water molecules. Because water molecules have a partial positive charge on the hydrogen atoms and a partial negative charge on the oxygen atom, the molecules can strongly interact with anions and cations. Water molecules can pull ions out of a crystal structure, and by surrounding the ions, minimize interactions between cations and anions. 6) (Burdge, 9.20) Characterize the following compounds as soluble or insoluble in water: a) CaCO3 Insoluble b) ZnSO4 Soluble (most sulfate compounds are soluble) c) Hg(NO3)2 Soluble (nitrate compounds are soluble) d) HgSO4 Soluble (most sulfate compounds are soluble. Note that Hg2SO4 (mercury (I) sulfate) is insoluble) e) NH4ClO4 Soluble (both ammonium and perchlorate compounds are soluble) 7) Classify the following compounds as soluble or insoluble in water. a) AgCl Insoluble (most silver compounds are insoluble in water) b) Ba(C2H3O2)2 Soluble (compounds containing acetate ion are soluble) c) Cu(OH)2 Insoluble (most hydroxide compounds are insoluble) d) CaF2 Insoluble (most compounds containing F - ion are insoluble) e) ZnCl2 Soluble (most chloride compounds are soluble) f) Ni(ClO3)2 Soluble (most chlorate compounds are soluble) 8) (Burdge, 9.22) Write total ionic and net ionic equations for the following reactions: a) Na2S(aq) + ZnCl2(aq) 2 NaCl(aq) + ZnS(s) TOTAL NET 2 Na + (aq) + S 2- (aq) + Zn 2+ (aq) + 2 Cl - (aq) 2 Na + (aq) + 2 Cl - (aq) + ZnS(s) Zn 2+ (aq) + S 2- (aq) ZnS(s) 2

3 b) 2 K3PO4(aq) + 3 Sr(NO3)2 6 KNO3(aq) + Sr3(PO4)2(s) TOTAL NET 6 K + (aq) + 2 PO4 3- (aq) + 3 Sr 2+ (aq) + 6 NO3 - (aq) 6 K + (aq) + 6 NO3 - (aq) + Sr3(PO4)2(s) 3 Sr 2+ (aq) + 2 PO4 3- (aq) Sr3(PO4)2(s) c) Mg(NO3)2(aq) + 2 NaOH(aq) 2 NaNO3(aq) + Mg(OH)2(s) TOTAL NET Mg 2+ (aq) + 2 NO3 - (aq) + 2 Na + (aq) + 2 OH - (aq) 2 Na + (aq) + 2 NO3 - (aq) + Mg(OH)2(s) Mg 2+ (aq) + 2 OH - (aq) Mg(OH)2 9) (Burdge, 9.27) Give the Arrhenius and the Bronsted definition of an acid and a base. Why are the Bronsted definitions more useful in describing acid-base properties? A Bronsted acid acts as a proton donor (H + donor) in a chemical reaction, while a Bronsted base acts as a proton acceptor. For example, in the equation HCl(aq) + NH3(aq) NH4 + (aq) + Cl - (aq) HCl is donating a proton, and so is a Bronsted acid, and NH3 is accepting a proton, and so is a Bronsted base. The Bronsted definition of acids and bases is usually (although not always) more useful than the Arrhenius definition because it is less restrictive. It allows classification of reactions as acid-base reactions (in the Bronsted sense) even for processes not taking place in water. 10) (Burdge, 9.30) What factors qualify a compound as a salt? Specify which of the following compounds are salts: A salt is an ionic compound that can be formed by an acid-base reaction. In practical terms, all common ionic compounds except hydroxides and oxides are salts. In the answers below, I have given an acid-base reaction that will form the compounds classified as salts. a) CH4 not a salt (a molecular compound) b) NaF a salt ( NaOH + HF NaF + H2O) c) NaOH not a salt (although an ionic compound) d) CaO not a salt (although an ionic compound) e) BaSO4 a salt (Ba(OH)2 + H2SO4 BaSO4 + 2 H2O) f) HNO3 not a salt (a strong acid) g) NH3 not a salt (a weak base) h) KBr a salt (KOH + HBr KBr + H2O) 3

4 11) (Burdge, 9.31) Identify each of the following as a weak or strong acid or base: a) NH3 weak base b) H3PO4 weak (polyprotic) acid c) LiOH strong soluble base d) HCOOH (formic acid) weak acid e) H2SO4 strong acid f) HF weak acid g) Ba(OH)2 strong soluble base 12) (Burdge, 9.32) Identify each of the following species as a Bronsted acid, a Bronsted base, or both. a) HI Bronsted acid. b) C 2 H 3 O 2 - Bronsted base. To see this, consider the reaction of this ion with water. C2H3O2 - (aq) + H2O(l) HC2H3O2(aq) + OH-(aq) c) H2PO4 - Both (and so amphoteric). Consider the following reactions H2PO4 - (aq) + H2O(l) HPO4 2- (aq) + H3O + (aq) H2PO4 - (aq) + HCl(aq) H3PO4(aq) + Cl - (aq) In the first reaction the hydrogen phosphate ion donates a proton and so acts as a Bronsted acid. In the second reaction the ion accepts a proton and so acts as a base. d) HSO4 - Both (and so amphoteric). 13) (Burdge, 9.34) Balance the following unbalanced equations, and write the molecular equations, total ionic equations, and net ionic equations (if appropriate): a) HBr(aq) + NH3(aq) MOLECULAR HBr(aq) + NH3(aq) NH4Br(aq) TOTAL IONIC H + (aq) + Br - (aq) + NH3(aq) NH4 + (aq) + Br - (aq) NET IONIC H + (aq) + NH3(aq) NH4 + (aq) b) Ba(OH)2(aq) + H3PO4(aq) MOLECULAR 3 Ba(OH)2(aq) + 2 H3PO4(aq) Ba3(PO4)2(s) + 6 H2O(l) TOTAL IONIC 3 Ba 2+ (aq) + 6 OH - (aq) + 2 H3PO4(aq) Ba3(PO4)2(s) + 6 H2O(l) NET IONIC the same as the total ionic (no spectator ions) 4

5 c) HClO4(aq) + Mg(OH)2(s) MOLECULAR 2 HClO4(aq) + Mg(OH)2(s) Mg(ClO4)2(aq) + 2 H2O(l) TOTAL IONIC 2 H + (aq) + 2 ClO4 - (aq) + Mg(OH)2(s) Mg 2+ (aq) + 2 ClO4 - (aq) + 2 H2O(l) NET IONIC 2 H + (aq) + Mg(OH)2 Mg 2+ (aq) + 2 H2O(l) 14) Balance the following unbalanced equations, and write the molecular equations, total ionic equations, and net ionic equations (if appropriate): a) AgNO3(aq) + CuCl2(aq) MOLECULAR TOTAL IONIC NET IONIC 2 AgNO3(aq) + CuCl2(aq) Cu(NO3)2(aq) + 2 AgCl(s) 2 Ag + (aq) + 2 NO3 - (aq) + Cu 2+ (aq) + 2 Cl - (aq) Cu 2+ (aq) + 2 NO3 - (aq) + 2 AgCl(s) Ag + (aq) + Cl - (aq) AgCl(s) b) CuCl2(aq) + NaClO4(aq) MOLECULAR TOTAL IONIC NET IONIC CuCl2(aq) + 2 NaClO4(aq) Cu(ClO4)2(aq) + 2 NaCl(aq) Cu 2+ (aq) + 2 Cl - (aq) + 2 Na + (aq) + 2 ClO4 - (aq) Cu 2+ (aq) + 2 Cl - (aq) + 2 Na + (aq) + 2 ClO4 - (aq) no reaction All of the ions are spectator ions, and so no reaction occurs. 15) (Burdge, 9.42) For the complete redox reactions given here, write the half-reactions and identify the oxidizing and reducing agents: a) 4 Fe + 3 O2 2 Fe2O3 oxidation 4 Fe 4 Fe e - reduction 3 O e - 6 O 2- oxidizing agent is O2; reducing agent is Fe b) Cl2 + 2 NaBr 2 NaCl + Br2 oxidation 2 Br - Br e - reduction Cl2 + 2 e - 2 Cl - oxidizing agent is Cl2; reducing agent is Br - c) Si + 2 F2 SiF4 oxidation Si Si e - reduction 2 F2 + 4 e - 4 F - oxidizing agent is F2; reducing agent is Si 5

6 d) H2 + Cl2 2 HCl oxidation H2 2 H e - reduction Cl2 + 2 e - 2 Cl - oxidizing agent is Cl2; reducing agent is H2 16) (Burdge, 9.42) Phosphorus forms many oxoacids (ternary acids containing H, O, and P). Indicate the oxidation number of phosphorus in each of the following acids: Note that in all of the compounds below hydrogen has an oxidation number of +1 and oxygen has an oxidation number of -2. a) HPO3 +5 b) H3PO2 +1 c) H3PO3 +3 d) H3PO4 +5 e) H4P2O7 +5 f) H5P3O ) (Burdge, 9.48) Give the oxidation number of the underlined atoms in each of the following molecules and ions: a) Mg3N2 Mg is +2, and so N is -3 b) CsO2 Cs is +1, and so O is - 1 /2 (weird, but fractional oxidation numbers occasionally occur) c) CaC2 Ca is +2, and so C is -1 d) CO3 2- O is -2, and so C is +4 e) C2O4 2- O is -2, and so C is +3 f) ZnO2 2- Zn is +2, and so O is -2 g) NaBH4 Na is +1, H is -1, and so B is +3 (I got this answer using a more sophisticated method for finding oxidation numbers, and so would not expect you to be able to do this one) h) WO4 2- O is -2, and so W is +6 18) Give the oxidation numbers for all atoms in the following molecules or ions: a) C2H5OH H is +1, O is -2, and so C is -2 b) CO2 O is -2, and so C is +4 c) ClO4 - O is -2, and so Cl is +7 d) PbCl2 Cl is -1, and so Pb is +2 e) XeF6 F is -1, and so Xe is +6 f) H3PO4 H is +1, O is -2, and so P is +5 g) IO3 - O is -2, and so I is +5 h) H2O2 H is +1, and so O is -1 (this occurs in peroxide compounds) 6

7 19) (Burdge, 9.52) Predict the outcomes of the reactions represented by the following equations by using the activity series, and balance the equations: a) Cu(s) + HCl(aq) no reaction (copper is below hydrogen in the activity series) b) Au(s) + NaBr(aq) no reaction (gold is below sodium in the activity series) c) Mg(s) + CuSO4(aq) MgSO4(aq) + Cu(s) d) Zn(s) + KBr(aq) no reaction 20) (Burdge, 9.54) Classify the following redox reactions as combination, decomposition, or displacement: a) P Cl2 4 PCl5 combination b) 2 NO N2 + O2 decomposition c) Cl2 + KI 2 KCl + I2 displacement (Cl displaces I) d) 3 HNO2 HNO3 + H2O + 2 NO decomposition 21) (Burdge, 9.66) Calculate the molarity of each of the following solutions: a) 6.57 g of methanol (CH3OH) in 1.50 x 10 2 ml of solution. M(CH3OH) = g/mol moles CH3OH = 6.57 g 1 mol = mol g [CH3OH] = mol = 1.37 mol/l L b) 10.4 g of calcium chloride (CaCl2) in 2.20 x 10 2 ml of solution M(CaCl2) = g/mol moles CaCl2 = 10.4 g 1 mol = mol g [CaCl2] = mol = mol/l L c) 7.82 g of naphthalene (C10H8) in 85.2 ml of benzene solution M(C10H8) = g/mol moles C10H8 = 7.82 g 1 mol g = mol [C10H8] = mol = mol/l L 7

8 22) (Burdge, 9.70) Water is added to 25.0 ml of a M KNO3 solution until the volume of the solution is exactly 500. ml. What is the concentration of the final solution? McLc = MdLd Md = (Lc/Ld) Mc = (25.0 ml/500. ml) (0.866M) = M 23) (Burdge, 9.98) A sample of g of an unknown compound containing barium ions (Ba 2+ ) is dissolved in water and treated with excess Na2SO4. If the mass of the BaSO4 precipitate formed is g, what is the percent by mass of Ba in the original unknown compound? The reaction taking place is Ba 2+ (aq) + SO4 2- (aq) BaSO4(s) M(BaSO 4 ) = g/mol moles precipitate = g M(Ba 2+ ) = g/mol 1 mol =1.759 x 10-3 mol g grams Ba 2+ ion = x 10-3 mol BaSO4 1 mol Ba g = g Ba 2+ 1 mol BaSO4 1 mol % Ba by mass = g 100. % = 35.7 % g 24) For each of the following aqueous solutions find the ph of the solution, and classify the solution as acidic, basic, or neutral. a) [H + ] = 3.7 x 10-3 M ph = - log10(3.7 x 10-3 ) = 2.43 acidic b) [H + ] = 9.2 x M ph = - log10(9.2 x ) = basic c) [OH - ] = 5.9 x 10-5 [H + ] = (1.00 x )/(5.9 x 10-5 ) = 1.7 x M ph = - log10(1.7 x ) = 9.77 basic 8

9 25) (Burdge, 9.102) Calculate the concentration (in molarity) of an NaOH solution if 25.0 ml of the solution is needed to neutralize 17.4 ml of a M HCl solution. The neutralization reaction is HCl + NaOH NaCl + H2O moles NaOH = ml HCl soln mol HCl 1 mol NaOH = x 10-3 mol L soln 1 mol HCl [NaOH] = 5.429x 10-3 mol = mol/l L 26) (Burdge, 9.114) Sodium carbonate (Na2CO3) is available in very pure form and can be used to standardize acid solutions. What is the molarity of an HCl solution if 28.3 ml of the solution is required to react with g of Na2CO3? The reaction taking place is 2 HCl + Na2CO3 2 NaCl + H2CO3 M(Na2CO3) = g/mol moles HCl = g Na2CO3 1 mol Na2CO3 2 mol HCl g 1 mol Na2CO3 = x 10-3 mol [HCl] = x 10-3 mol = mol/l L 27) (Burdge, 9.146) A g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO 3, to yield g of AgCl. Calculate the percent by mass of each compound in the mixture. This problem requires a bit of thought. The reaction of interest involves chloride ion Ag + (aq) + Cl - (aq) AgCl(s) Based on the mass of precipitate formed we can determine the grams of Cl -. Since the only sources of chloride ion are the NaCl and KCl in the mixture, it follows that total mass Cl - = mass from NaCl + mass from KCl mass Cl - from NaCl = (mass of NaCl) (fraction Cl in NaCl) mass Cl - from KCl = (mass KCl) (fraction Cl in KCl) 9

10 M(AgCl) = g/mol M(NaCl) = g/mol M(KCl) = g/mol fraction Cl in NaCl) = g Cl/mol NaCl = g NaCl/mol NaCl fraction Cl in KCl) = g Cl/mol KCl = g NaCl/mol KCl total mass Cl = g AgCl 1 mol AgCl 1 mol Cl g Cl = g Cl g 1 mol AgCl 1 mol Cl Now, let x = mass of NaCl. Then ( x) = mass of KCl. Using the equation for the total mass of Cl, we get g = x (0.6066) + ( x) (0.4755) = g x x = ( g g) = g The percent by mass NaCl in the sample is % NaCl(by mass) = g NaCl 100. % = 43.9 % NaCl by mass g sample 28) (Burdge, 9.164) The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of a stock solution. Instead, it is done by a series of dilutions. A sample of g of KMnO4 was dissolved in water and made up to a volume of ml in a volumetric flask. A ml sample of this solution was transferred to a ml volumetric flask and diluted to the mark with water. Next, ml of the diluted solution was transferred to a ml flask and diluted to the mark with water. a) Calculate the concentration (in molarity) of the final solution. We may use McLc = MdLd, which gives Md = (Lc/Ld) Mc, in all of the dilution steps. For the initial concentration M(KMnO4) = g/mol moles KMnO4 = g 1 mol = x 10-3 mol g Initial concentration = x 10-3 mol = x 10-2 mol/l L 10

11 1 st dilution Md = (2.000 ml/ ml)( x 10-2 M) = x 10-5 mol/l 2 nd dilution Md = (10.00 ml/250.0 ml) (2.079 x 10-5 M) = 8.32 x 10-7 mol/l This is the molarity of the final solution b) Calculate the mass of KMnO4 needed to prepare the final solution if it was prepared directly. The mass of KMnO4 needed to prepare ml of a 8.32 x 10-7 M solution directly is mass KMnO4 = L soln 8.32 x 10-7 mol g KMnO4 = 3.29 x 10-5 g L soln mol It would be difficult to precisely measure out such a small mass of this compound directly, which is why the dilute solution is prepared by serial dilution. 29) (Burdge, 9.76) The maximum level of fluoride ion that the EPA allows in U.S. drinking water is 4.0 mg/l. Convert this concentration into molarity. [F - ] = 4.0 x 10-3 g 1 mol = 2.11 x 10-4 M 1. L g 30) A ml sample of a stock solution of nitric acid (HNO3) is titrated with a M solution of potassium hydroxide (KOH). After the addition of ml of the KOH solution the neutralization reaction is complete. What is the concentration of HNO3 in the stock solution? The balanced neutralization reaction is HNO3(aq) + KOH(aq) KNO3(aq) + H2O(l) The moles of KOH present when the neutralization reaction is complete is moles KOH = x 10-3 L KOH mol KOH = x 10-3 mol KOH L soln moles HNO3 = x 10-3 mol KOH 1 mol HNO3 = x 10-3 mol HNO3 1 mol KOH molarity of stock HNO3 soln = x 10-3 mol HNO3 = M HNO3 soln x 10-3 L 11

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