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CHM 1025 THIRD PROBLEM SET INSTRUCTOR: DON SWIETER OFFICE: K–223 PHONE: 395–5303 QUESTIONS ABOUT LEWIS STRUCTURES (Read handout on Drawing Lewis Structures before doing problems below) 1. Write Lewis electron structures for the molecules most likely to be formed from the simple combination of nonmetals (Note: More than one atom of each element may be required): a) H, Cl b) H, S c) Cl, O d) N, Br e) Br, Br f) C, S g) P, I h) C, F 2. The following molecules contain one or more multiple bond(s). The sequence of bonding is indicated. Write the correct Lewis structure for each. a) H, N, N, H b) S, C, S c) C, O d) S, C, O e) Cl, C, C, Cl f) H, C, N 3. Write ALL acceptable Lewis structures for each of the following molecules or polyatomic ions. Include formal charges where required. + + & a) NO b) NO 2 c) CO 2 d) C6H 14 e) NF 2 & f) COCl 2 g) NH 2 h) PH 3 i) C3H 4 j) *H 2O 3 k) PF + l) *HNO m) *HOCl n) CH O o) C H 4 2 2 2 4 p) C H F q) C H F r) C H F s) C H F t) C H O 2 3 2 2 2 2 5 2 4 2 2 6 * H’s are attached to O QUESTIONS ABOUT GEOMETRY AND VSEPR See handout at the end of this problem set forVSEPR summary. 4. For each of the species listed in problem 3, predict the geometrybased on VSEPR theory and state whether the species is expected to be polar, non-polar or whether the idea doesn’t apply? 5. Each of the following molecules only contain one covalent bond. Therefore, the individual bond polarityis indicative of net molecular polarity. List the following molecules in order of increasing molecular polarity: I-Cl, Br-Cl, Cl-Cl, Br-F, Cl-F QUESTIONS INVOLVING SOLUTIONS 6. In which physical state (solid, liquid, or gas) is it easiest to carry out chemical reactions? Explain. 7. What is the definition of the word “solution?” 8. What are the components of a solution? How are they distinguished? 9. Why are most chemical reactions carried out in liquid solution? HW3_CHM1025.wpd 03/13/14 2 10. Generally, what is the most “convenient” concentration unit for chemists to use? Why? 11. What is the definition of “Molarity?” 12. Whatisthemolarityofsugar,C H O ,i123.22o11ugararedissolvedtogive746mLofsolution? 13. What is the molarity of KCl if 1.45 g of KCl are dissolved to give 50.0 mL of solution? 14. How many grams of NaNO are th3re in 75.0 mL of 1.00 M NaNO solution? 3 15. Discuss an experimental method to determine whether or not a compound is molecular or ionic. 16. Consider the formulas: HOH, NaOH, HCl, NaCl. Which are predicted to be molecular materials? Which are expected to be ionic materials? Explain. 17. What is unusual about the composition of salts such as NH Cl o4 (NH ) SO ?4 2 4 18. For solutions, what are electrolytes? What are nonelectrolytes? 19. What is necessary for a substance to conduct electricity? Explain. 20. Why does solid NaCl not conduct electricity but an aqueous solution of NaCl does? 21. Give all possible ways one can get ions into solution. Explain! 22. Ifasoluteisknowntobeanonelectrolytein solution, what,ifanything,canoneinferaboutthetype of bonding in the pure solute? 23. If a solute is known to be an electrolyte in solution, what, if anything, can one infer about the type of bonding in the pure solute? 24. The processing of determining the identity and concentration of solutes present in appreciable amounts in a solution is called inventorying. Inventory the following solutions. a) 0.25 M NaCl b) 0.50 M Ca(NO ) 3 2 c) 0.15 M (NH ) S4 2 4 QUESTIONS ABOUT ACIDS AND BASES 25. WhatisthedefinitionofaBrønsted-Lowryacid? What is thedefinitionofaBrønsted-Lowrybase? 26. WhatchemicalfeatureMUSTbepresentIFaspeciesistofunctionasaBrønsted-Lowryacid? This is not to saythat anychemical that possesses this feature WILLfunction as a Brønsted-Lowryacid. 27. WhatchemicalfeatureMUSTbepresentIFaspeciesistofunctionasaBrønsted-Lowrybase? This is not to saythatanychemical that possesses this feature WILLfunction as a Brønsted-Lowrybase. 28. What are the products of any acid/base reaction? 3 29. What are the conjugate acids of the following species? a) H 2 b) OH S c) NH 3 d) X S e) HX 30. What are the conjugate bases of the following species? a) H 2 b) OH S c) NH 3 d) HX e) H 3 + 31. In terms of their reaction with water, how are strong acids defined? How are weak acids defined? 32. Write the equation for the reaction of a strong acid, HX, with water. Write the equation for the reactionofaweakacid,HX,withwater. Whatisthedifferencebetweenthetwoequations? Explain. 33. What are the six common strong acids? Memorize them! How then, are all other acids classified? 34. Remember: An inventory is a list of the formulas and concentrations of the solute ions and molecules present in appreciable amounts in a solution. What is the inventory of 1M HCl? What is the inventory of a 1 M solution of any strong acid, HX? A 1 M solution of any weak acid, HY? 35. Explain why liquid HCl does not conduct electricity but a 1M HCl solution does conduct. 36. How would the conductivity of a 1M HCl solution differ from the conductivity of a 1M HC H O 2 3 2 solution? Explain. 37. If the conductivity of an aqueous solution of acid HZ were such that the 150 watt bulb lit brightly while the aqueous solution of acid HY lit only the 7.5 watt bulb dimly, what conclusion can one safely draw from these observations about the relative acid strengths of HZ and HY? Explain. 38. In terms of their reaction with water, how are strong bases defined? How are weak bases defined? 39. Write the equation for the reaction of a strong base, B, with water. Write the equation for the reactionofaweak base,B,withwater. Whatisthedifferencebetweenthetwoequations? Explain. 40. What arethe ten common strongbases? Memorize them! How then, areall other bases classified? 41. Whatistheinventoryofa1M Na Osolut2on? What is theinventoryofa1M solutionofanystrong base? A 1 M solution of any weak base? 42. Explain whyliquid NH doe3not conductelectricitybuta1MNH solutiondo3sconduct(slightly). 43. Explain why conductivity can be used to determine the relative strengths amongst a group of molecular acidsor amongst agroupofmolecular bases but cannot be used to determine the relative strengths amongst a group of ionic acids or amongst a group of ionic bases. 44. IftheconductivityofanaqueoussolutionofbaseZwassuchthatthe7.5wattbulblit brightlywhile theaqueoussolutionofbaseYlitonlythe7.5wattbulbdimly,whatconclusioncanonesafelydraw from these observations about the relative base strengths of Z and Y? Explain. 4 45. You aregiven the hypothetical acids HR, HG, and HS which are coSorless, in the pure Sorm. Their respective bases however, are highlycolored; R being red, G being green, and Y being yellow in solution. Consider the following experiments. i. When HR is mixed with water, the resulting solution is only a faint pink. ii. When a solution of HR (eg. 1M HR) is mixed with a solution containing G (eg. 1M NaG) the resulting solution is a deep red color. iii. When HY is mixed with water, the resulting solution is a deep yellow. A) Write the equations which correspond to the changes in each of the above reactions and indicate the relative extent to which each occurred. B) For each of the equations in A, decide which acid is the stronger, the acid on the left side or the acid on the right side. C) For each of the equations in A, decide which base is the stronger, thebase on the left side or the base on the right side. D) Now prepare an acid/base table similar to the one presented in class which lists your FOUR acidsandtheirFOUR conjugate bases. Consult yournotesto seehowto properlylist them! E) If a solution of a fourth acid, HB (eg. 1M HB) (which is colorless), is mixed with a solution of R (eg. 1M NaR) and the resultingsolution is a deep red, where does the acid HB belong inyourtable? Canitbeplacedunambiguously? Ifnotwhatreactionorreactionswouldyou perform in order to correctly place acid HB in your chart? 5 Review: Writing Correct Chemical Equations The word "correct," in the above statement may appear to be an unnecessary adjective. This is because one maylegitimatelyask the question, "How could an incorrect chemical equation ever be considered achemical equationin the firstplace?" Nevertheless,itisanunfortunaterealitythatthe chemical literature does contain incorrect chemical equations. How canthis be the case? After all, if we establish an overall balance of total atoms and total charge between product and reactant substance,dowenotthenhaveacorrectchemicalequation? Tofindout,let'sreexaminethecriteria for an equation. Clearly, if we write anysuch balanced statements, we obtain an algebraically-acceptable equation. But, does this mean necessarily that we have a chemically-acceptable equation? To this, we must recognize thattheansweris "No!" Anequationis hardlyofvalueunlesstheequationdealswiththe chemicalproblemathand. Recall,thatanequationmustbeinaccordwiththechemicalfactsand must contain only thechemically pertinentspecies. Hence,thereal problemin writingchemical equations stems from the necessity to correctlyidentifythe reactant and product substances, a task which frequently is difficult because the chemical nature of any material can be known only from reliable experimental results. To be sure, then, the exact makeup of anychemical substance is known onlywithin the boundaries of experimental accuracy. Thus, when we write formulas in equations, these formulas should constitute the simplest and best chemical representation that we know for the reactants and products. A beginner in chemistry, therefore, must have learned from either some prior experience what the products and reactants are, or he must look up this information if it is available, or he must be told! For without first-hand knowledge as to what are the products and reactants, one CAN NOT write a correct, meaningful, chemical equation. Let us now consider some examples. A. Supposethatyouareinalaboratoryandaredirectedtothereagentshelf,whereuponyouobserve twolargebottles,eachofwhichcontainsaclear,colorless,homogenousliquid. Onebottlebears the label 0.5 M Na C2 and3the other, 0.5 M CaCl . Now,2you are told (in case you might not know) that both Na CO2and3CaCl are w2ite, crystalline, ionic solids under standard laboratory conditions. Your next direction is to mix two mL of the 0.5 M Na CO 2ith 3wo mL of the 0.5 M CaCl . As 2 soon as you make the mix, you discover that the mixture becomes opaque white and that on standing a white powdery solid separates in rather copious quantity leaving a clear colorless liquid. Clearly then, a reaction occurred upon mixing the solutions. What is the “correct equation” for this reaction? Now, if you're careful, you'll back off, don your thinking cap, and recognize that from this very brief laboratory experience you're already in a position to assemble quite a bit of chemical information. 6 Let's note the pertinent stuff: 1. Both salts, Na CO2and 3aCl , are v2ry soluble in water. The simple fact that they can be dissolved to an extent of 0.5 M tells you this. 2. Therefore, you mayquantitativelyinventoryeachoriginalsolution. So whatisaninventory of a solution? It is a list of all species which arepresent in appreciable concentration (other than H 2; since H O 2s a solvent - it's known to be present in overwhelming abundance, so we don't list it even though we certainly know that it is present). The inventoryis: [Na ] = 1 M & [CO ] = 0.5 M [Ca ] = 0.5 M & [Cl ] = 1 M 3 For 0.5 M Na C2 , 3 For 0.5 M CaCl , 2 (Note: is calledsquarebracketnotation, and itmeans concentration in molesperliter,M.) + S2 3. You know to write Na , CO , etc.,3as you have been told that these salts are ionic solids. That is, since they are ionic, you know that when they dissolve in water, they simply dissociateintotheirconstituentions(unlesstheconstituentsreactwithH Otoalargee2tent- a process we have not yet discussed). Now,withthisformationandyourpracticalexperiencethatNaCl(commontablesalt)isvery solubleinwater,youcanwritethecorrectequation. Thatis,yourealizethatyouhavemixed a solution with an abundance of Na and CO 3S2with a solution with an abundance of Ca +2 S and Cl , which resulted in the immediate formation of a white solid. So what is this solid? Bysimply permuting the added ingredients, you see that the only possibilities are Na CO , 2 3 CaCl 2NaCl,andCaCO . Andfr3mthislistyoumayimmediatelydischargeNa CO ,CaCl , 2 3 2 and NaCl, since you know each of these salts to be very soluble in H O. Or, 2s a chemist + S S2 +2 S might say, Na is compatible with both Cl and CO 3 and Ca is compatible with Cl in an aqueous environment. Clearly, then, you mayrecognize logically, without doing anymore experiments, that the white solid formed (precipitated) must be CaCO (chalk or 3 oyster shells). Hence the equation is: +2 S2 Ca (aq) + CO 3 (aq) xxv CaCO (s3 Recognize, therefore, that it would have been absolutely incorrect to have written . . . (i) Na CO + CaCl xxv CaCO + 2 NaCl 2 2 2 3 or to have written . . . (ii) Na + CO 3S2+ Ca + Cl S xxv CaCO + 3aCl, since neither of these statements reflect the actual change which took place when you made the mix. That is, (i) says to a chemist that two solids (or, at the minimum, two pure substances) were mixed, resulting in the production of two different solids. And, (ii) says that upon thesimultaneouscombinationofthesefourions (trueso far!)thattwosolids were formed (not true, no solid NaCl was formed at all!). So realize that the simple combination of two (or more) chemical systems in no way guarantees that a reaction will take place. For instance, ifyouwereto mix 0.1 M NaCl with 0.1 M KNO (another colorless solution), you would observe no change at all other than an 3 increase in overall solution volume. 7 Consequently, to write something like . . . NaCl + KNO 3 xxv NaNO + 3Cl or + S + S Na + Cl + K + NO 3 xxv NaNO + K3l would be utterly ridiculous as no such chemical changes (reactions) would take place upon mixing thesesolutions. Therefore, thelogicalconsequenceofall this is to realize that equations areto bewrittenonlyforsituations whereinthereis evidencethatis chemical change has indeed taken place. B. Recall the laboratory experiment, "Solution Stoichiometry." In this experiment, you were formally introduced to "concentration." Specifically, you dealt with the "moles per liter" (or "molar,"abbreviatedasM)concentrationofasolutionofsodiumhydroxide(NaOH). Molar(M) meanstheamountofsoluteinmoleswhicharecontainedineveryliterofthatparticularsolution. Thesolution whichwasusedwas0.50MNaOH(totwosignificantfigures). So,ifyoucarefully measuredout1.0Lofthissolution,andboiledgentlyuntilallthewaterwasremoved,youwould obtain0.50moles(20g)ofNaOH: awhite,crystallinesolid,whichisratherhygroscopic(which means the solid absorbs water from the air). In this experiment you determined the percent of HC H O in2a 3in2gar solution byreacting the HC H2O 3it2 a NaOH solution of known concentration. Now, if you didn't get this message when you did the experiment, recognize why we stated that "at the equivalence point of the titration (i.e., when the solution in the flask developed a faint, but perceptible and lasting pink color) the moles of NaOH added must equal the moles of HC H O in the 25.00 mL of the S 2 3 2 vinegar sample." This is because OH and HC H O combin2 c3em2cally on a mole for mole basis. The correct equation shows this is OH (aq) + HC H O (aq) xxv H O ( R) + C H O (aq) 2 3 2 2 2 3 2 Why not? NaOH + HC H O 2 3 2 xxv NaC H2O 3 H2O, o2 Na + C H O + H2O?3 2S 2 Again, what is NaOH? It is a white, crystalline, hygroscopic solid under standard laboratory conditions. Did you add such a material to the vinegar solution? No! You added a solution + S which contained Na and OH . And when you added this solution to the vinegar, did anywhite solids form? Noagain! Therefore, sinceNaC H O isawhitecrystalline solid undernormallab 2 3 2 conditions, we do not write NaC H O 2s 3 p2oduct of the titration reaction, because it isn't! Perhaps you questionedhowit wasthat weknowthatin thevinegarsolution we had a substance (other than H O)2whose formula was best represented as HC H O . That i2, 3f 2onic solids dissolveintosimpleringredients(theconstituentions),mightnotHC H O dissolveinwaterinto 2 3 2 simpler ingredients? Yes,this is possible! Andin fact this doesoccurto asmall but measurable extent, when HC H O is dissolved in H O, the transformation is: 2 3 2 2 + S HC H2O 3 H2O 2 xxv H 3 + C H O2 3 2 (this goes about 0.1% for 1 M HC H O 2 3 2 8 and we shall recognize that materials which are capable of giving up H (proton) in a chemical process are known as acids. Consequently, we see that HC H O is an acid (H O is too!). So 2 3 2 2 when we wrote the titration equation, why didn't we write S + S + OH + H O ?3or OH + HC H O + H O ?2 3 2 3 Theanswertothisissimplythatbecausethereissuchaverysmall[H O ]inthevinegarsolution, + 3 there is no meaningful point in including H O in the t3tration equation. That is, the primary S reaction which occurs when we add NaOH solution to vinegar is between the base OH and the acid HC H O 2 3 2 So let us sophisticate our appreciation and understanding of equation writing by noting that although more than one reaction may occur when we mix chemical systems, we only write equations for the principal (significant) changes which take place. What experiment could you (or might you) do to prove that the substance principallypresent in vinegar, other than H O, was HC H O (i.e., NOT H O and C H O )? See Problem 36. S 2 2 3 2 3 2 3 2 QUESTIONS ON EQUATION WRITING FOR ACID BASE REACTIONS 46. Write the equation for the principal acid/base reaction which occurs when equal volumes of the following solutions are mixed. Also, comment on the extent of the reaction as written. (Note: Prior to writing anyequations, always begin with a quantitative inventoryof each of the solutions which are to be mixed.) a. 0.1 M HCl with 0.1 M KOH b. 0.1 M HNO with 031 M NH 3 c. 0.1 M H SO 2ith 4.1 M NaC H O 2 3 2 d. 0.1 NaHSO with 4.1 M NaC H O 2 3 2 e. 0.1 M NH Cl wi4h 0.1 M NaOH f. 0.1 M NaHCO with 0.1 M LiOH 3 g. 0.1 M HC H O w2th30.2 M NaC H O 2 3 2 h. 0.1 M HC H O with 0.1 M MgSO (NOTE: Mg(H O) +2 is a weaker acid than HC H O ) 2 3 2 4 2 6 2 3 2 i. 0.1 M HC H O w2th3H 2 (i.e., v2negar is diluted) j. Pure liquid HNO with p3re solid NaOH 9 QUESTIONS ON OXIDATION AND REDUCTION 47. What is the definition of oxidation? What is the definition of reduction? 48. What is the definition of an oxidizer? What is the definition of a reducer? 49. What are the products of any redox reaction? 50. Whatistherelationship,ifany,betweentheconceptsofoxidationstateandformalcharge? Explain. 51. What is the theoretical maximum and minimum oxidation state exhibited byeach of the following elements? a. N b. Xe c. Cl d. Na e. Se f. Ca g. C h. B 52. Assign oxidation numbers to each of the elements in the following compounds. a. CsClO b. SO c. UF 4 3 6 d. XeF 4 e. KNO 3 f. Al 2SO )4 3 g. CsH h. Na Cr O i. OsO 2 2 7 4 j. (NH )4 2 3 k. BaSe l. S 8 m. Mg UO n. NO + o. SrO 3 6 2 2 p. XeO 6S4 q. S 2 7S2 r. AsH 3 53. Identify the following changes as either oxidation or reduction: S S +3 a. MnO to2MnO 4 b. BiO 3 to Bi c. SO 2o SO 3 d. OCl to ClO 3 e. N 2 t4 N O 2 f. H 2 t2 H O 2 54. What is the oxidizing agent (oxidant) and the reducing agent (reductant) in each of the following equations? a. Cl2+ 2 Br S xxv 2 Cl + Br 2 b. 2 NO + 72H 2 xxv 2 NH 3 4 H O 2 c. 5 SO 3S2+ 2 MnO + 4 H O 3 + xxv 5 SO 4S2+ 2 Mn + 9 H O 2 d. 2 Cr(H 2) (3H) + 3 ClO + 4 OH S xxv 2 CrO 4S2+ 3 Cl + 11 H O2 e. Cl2+ 2 H O2 xxv HClO + H O 3 Cl S 55. Consider the electrolysis of a 1M CuCl sol2tion. As it proceeds,youwouldmakethese observations: (1) The original blue-green color of the solution would gradually fade. (2) A shiny light-brown metallic material would deposit on the surface of the cathode. (3) A strong-smelling (odor of bleach!) greenish-yellow gas would be formed and discharged at the surface of the anode. Write the equations for the two half-reactions and the equation for the overall reaction. (Hint: aqueous CuCl is 2est inventoried as Cu(H O) 2 6+2and Cl .) 10 DETERMINING THE PRODUCTS OF REDOX REACTIONS Suppose we add a bit of K to a few hundred mL of distilled H O 2hich results in the production of a gas and yields a solution which was strongly conducting and which turns red litmus to blue. Also, recall that this reaction is an example of redox (i.e., electrons are interchanged). To "write the equation," what questions might a chemist ask? Or, better yet, what might you as a neophyte chemist ask in conjunction with information with which you are already familiar? a. What might the gas be? H ? 2 ? B2th? Are there any other reasonable possibilities? b. The resulting solution conducts, . . . therefore . .? Ions must have been produced! S c. What aretheion+? Recall thatKisanotoriouslygoode "giver-upper”;i.e.,Kis adandyreducer. This makes K a very good bet! And, since positive (K ) ions were produced, it follows that negative ions were also produced. S Hence, H O2(nothSng else preSent initially) must have consumed the e 's from the oxidation of K. How about OH ? Will OH turn red litmus blue? d. Let's return to the gas to see if we can decide exactly what it is. Examine H2O with respect to + S2 theoxidationstatesoftheconstituentatoms,viz. H andO . InH andO ,th2oxidati2nstates forHandOarezero,respectively. Henceit followsthatthegaseousproductmust beH ! That 2 is, if H O consumes electrons (as it does/has in this case), one (or more) of its constituent 2 elements must have been reduced. So . . . S 0 H 2 + xe xxv H 2+ other stuff) is reasonable, whereas H O + ye S xxv O (+ other stuff) is not reasonable. 2 2 (Oxygen in H O has an oxidation state of S2 and therefore, must LOSE electrons to get to O ) 2 2 Therefore, the appropriate half-reactions are: K xxv K + e (oxidation: the reducing half-reaction) S S 2 H 2 + 2 e xxv H 2 2 OH (reduction: the oxidizing half-reaction) . . . and the net reaction is . . . + S 2 K + 2 H 2 xxv 2 K + H +22 OH . + S Doyou seewhyK+2 H O 2 xxv K +H +22H cannot becorrect? (Nochargebalance!) Ofcourse,ifKOHwasaninsolublematerial,wewouldhaveobservedtheformationofawhite crystalline solidduringthereaction,andwewouldthenhavewrittenKOHasaproduct,not K + S and OH . 11 56. Consider the observations in the following experiments. i. K reacts explosively with H O (2ee previous discussion on determining products). ii. Ca reacts vigorously (but not violently like K) with H O to pr2duce the gas and a colorless solution. iii. Cu reacts, for all intents and purposes, not at all with H O2under normal laboratory conditions (What would be the expected products? Cu(H O) , OH , 2 ).6+2 S 2 iv. When a piece of Cu is placed in a 1M AgNO solution, the original colorless solution 3 turns a blue-green color as the piece of Cu disappears. A silvery-colored metallic solid is also formed. With these experimental findings in mind: A) Write the equations which correspond to the changes in each of the above reactions and indicate the relative extent to which each occurred. B) ForeachoftheequationsinA,decidewhichreduceristhestronger; thereduceronthe left side or the reducer on the right side. C) For each of the equations in A, decide which oxidizer is the stronger; the oxidizer on the left side or the oxidizer on the right side. D) Now prepare a redox table similar to the one presented in class which lists your FIVE reducersandtheirFIVEconjugateoxidizers. Consultyournotestoseehowtoproperlylist them! 57. Complete and balance the following reactions assuming that the oxidizer ends up at its minimum oxidation state and the reducer ends up with its highest oxidation state. Also, label the products as either ionic or molecular. a. Na + I 2 b. Cs + S 8 c. Mg + P 4 d. Al + O 2 e. Be + O 2 f. Li + O 2 g. C + O h. Zn + Br i. Sc + O 2(xs) 2 2 j. K + H 2 k. Ba + H 2 l. P 4 F 2 m. H +2Br 2 n. H 2 S 8 o. H 2 O 2 13 Relative Acid/Base Strengths -- in Conjugate Pairs Acid Base & HClO 4 ClO 4 & ˜° H 2O 4 HSO 4 / & / Strong HI I too weak / (to act as bases / Acids HBr Br & in aqueous sol’n) / / HCl Cl& / / HNO NO & A / / 3 3 + / H 3 H 2 / & / H 2O 3 HSO 3 / & &2 / HSO 4 SO 4 Weak / / H 3O 4 H 2O 4& Bases / / HNO 2 NO 2& / / HF F& / & Increasing HCOOH HCOO Increasing ACID & BASE Strength CH C3OH CH C3O Strength +3 +2 , Al(H O2 6 Al(H O2 (5H) , & / H 2O 3 HCO 3 / / H 2 HS & / / H PO & HPO &2 / 2 4 4 / Weak NH + NH / 4 3 & &2 / Acids HCO 3 CO 3 / + / CH N3 3 CH N3 2 / &2 &3 / HPO 4 PO 4 / / H 2 OH & / / / / HS & S&2 / too weak Strong ¨. / (to act as acids NH NH & Bases °‹ in aqueous sol’n) 3 2 & &2 / 9 OH O CHM 1025 THIRD ANSWER KEY INSTRUCTOR: DON SWIETER OFFICE: K-223 PHONE: 395-5303 QUESTIONS ABOUT LEWIS STRUCTURES 1. a) b) c) d) e) f) g) h) 2. a) b) c) d) e) f) 3. Note: Lewis structures are not intended to depict geometry - only electron distributions. A * by the structure indicate that resonance is present. a) b) c) d1) d2) d3) d4) d5) e) f) g) h) HW3_ANS_CHM1025.wpd 03/13/14 2 i1) i2) i3) j) k) l) m) n) o) p) q1) q2) q3) r) s1) s2) t1) t2) QUESTIONS ABOUT GEOMETRY AND VSEPR 4. a) linear on both N and O: DNA. b) linear on N: DNA. c) linearonC:Nonpolar. d) tetrahedralonall C’s in every isomer: Polar. e) angular on N: DNA. f) trigonal planar on C: Polar. g) angular on N: DNA. h) trigonal pyramidal on P: Polar. i1) trigonal planar/linear/trigonal planar: Nonpolar. i2) linear/linear/tetrahedral: Nonpolar i3) trigonal planar on 2 C’s and tetrahedral on 1 C: Polar. j) trigonal planar on C and angular on O: Polar. k) tetrahedral on P: DNA. l) angular on both N and O: Polar. m) angular: Polar. n) trigonal planar: Polar. o) trigonal planar on both C: Nonpolar. p) trigonal planar on both C’s: Polar. q1) trigonal planar on both C’s: Polar q2) trigonal planar on both C’s: Nonpolar q3) trigonal planar on both C’s: Polar r) tetrahedral on both C’s: Polar. s1 & s2) tetrahedral on both C’s: Polar. t1 & t2) tetrahedral on both C’s and angular on O: Polar. 3 5. Since these molecules are diatomic, the bond polarity is equal to the overall molecular polarity. BOND I)))Cl Br)))Cl Cl)))Cl Br)))F Cl))F EN of Atoms 2.5 , 3.0 2.8 , 3.0 3.0 , 3.0 2.8 , 4.0 3.0 , 4.0 Ä(EN) 0.5 0.2 0.0 1.2 1.0 Polarity of Bonds or Cl)))Cl Br)))Cl I)))Cl Cl)))F Br))F Molecules < < < < QUESTIONS INVOLVING SOLUTIONS 6. The liquid state is usually the easiest state in which to carry out reactions. Physical State Advantages Disadvantages Gas Particles in rapid motion. Therefore, Gases occupy large volumes. particle mix and react rapidly. Special container needed to contain gases. Solid Solidsoccupysmallvolumes. Particle movement nearlyabsent. No special container needed. Therefore, reaction very slow. Liquid Particles in motion. Therefore, ???????? particles mix and react reasonably fast. Volumes occupied by liquids are almost as small as solids. No special equipment needed. 7. Asolutionisdefinedasahomogeneousmixture. Thisdefinitiondoesnotspecifythephysical state. Therefore, a solution may be a solid, liquid or a gas mixture (as long as it is homogeneous). 8. The two components of a solution are the solvent and the solute. The distinction between them is arbitrary. The distinction is often based on the relative amounts of the two components with the solvent being present in a larger amount. Sometimes, the difference is based on the final physical state ofthemixture. Forexample, ifoneofthecomponents is a solid and the other a liquid then the distinction is made based on the state of the mixture. If the mixture is a liquid then the solvent is considered to be the liquid. If the mix is a solid then the solid is the solvent. 9. Reactions are generallycarried out in the liquid state for the advantages listed in problem 6. If one or more of the reactants are gases or solids, theyare converted into a liquid solution bydissolving them into a liquid solvent. 10. Since chemicals combine by number and not by mass it follows that one would wish to determine the moles of the various reactants present in the reaction. Furthermore, since the most convenient propertyto measure for a liquid is volume (see reasons for using liquid solutions above), it follows thatonewouldneedaconversionfactortoconvertbetweenthevolumeofliquidusedandthemoles ofsolutedissolvedinthesolution. Therefore,molarity(M)isthemostconvenientconcentrationunit for chemists. 4 11. 12. 13. 14. 15. The classification of a sample as ionic or molecular is not always easy. It is often done through a series of observations. The first step is to determine the physical state of the pure substance. If the puresubstanceisagasorliquidundernormallaboratoryconditions,thenthisisconsideredasproof that it is a molecular compound. However, if it is a solid, the sample can still be either ionic or molecular. At this point the chemist usuallyresorts to conductivitymeasurements. If the sample is soluble in water and the resulting solution does not conduct, this is evidence that it is molecular. If the sample does conduct, it may still be either ionic or molecular (see problem 21 below). At this point the chemist might melt the sample (if possible) and determine if the melt is conductive. If it does, it is classified as ionic and if it is non-conductive it is considered molecular. 16. H OandHClareexpectedtobemolecularcompoundsastheyarecomposedofallnonmetalatoms. 2 NaOH and NaCl are expected to be ionic since they are composed of metal and nonmetal atoms. 17. The fact that they are ionic is remarkable since they are composed of all nonmetal atoms. Each compound contains the ammonium ion NH , a ca4ion made up of nonmetal elements only and not the usual metal atoms that typically form cations. 18. Electrolytes are substances that conduct electricity when dissolved in solution. Nonelectrolytes are substances that do not conduct electricity when dissolved in solution. 19. Sincecurrentelectricityissimplytheflowofchargeasamplemustcontainchargesthatcanflow. 20. A sample of solid NaCl does not conduct electricity because even though the sample does contain charges (the ions) those charges (the ions) cannot move (or flow). In an aqueous solution of NaCl, the charge or ions are free to flow in the liquid state (the water solution) and therefore the solution is an excellent conductor of electricity. 21. There are two basic ways ions can get into solution, by dissociation and by ionization. Dissociation referstheprocess in whichtheions dis-associatefromeachotherastheydissolve into a liquid solution. Ionization refers to the process of reaction between chemical entities to create ions. 5 22. Since the solute has actually dissolved into the solution then the solute particles must be separated and mobile. If the solution does not conduct electricity then the solution does not contain ions. Thus only molecules exist in the solution and therefore the solute must be molecular in nature. 23. Since the solute has actually dissolved into the solution then the solute particles must be separated and mobile. If the solution does conduct electricity then the solution does contain mobile ions. However, this does not mean that the solute must be ionic. It is true that IF an ionic solute does dissolve then mobile ions will be produced and the resulting solution will conduct electricity. However, it is possible for a reaction to occur between a molecular solute and a molecular solvent to create ions (seeIonization in problem 21) and since this solution would also contain mobile ions then this solution would conduct electricity too. Thus if all we know is that a solution conducts electricity we cannot tell if the solute is ionic or molecular in nature. 24. a) [Na ] = 0.25 M, [Cl ] = 0.25 M b) [Ca ] = 0.50 M, [NO ] = 1.0 M 3 + &2 c) [NH ]4= 0.30 M, [SO ] = 4.15 M QUESTIONS ABOUT ACIDS AND BASES + 25. A Brønsted-Lowry acid is defined as a proton (H ) donor. A Brønsted-Lowry base is defined as a proton (H ) acceptor. 26. In order to function as a proton (H ) donor the species must have at least one H in the formula. 27. In order to function as a proton (H ) acceptor the species must have at least one unshared pair of + electrons to bond to the incoming proton (H ). 28. The products are the conjugate acid (of the reactant base) and the conjugate base (of the reactant acid). The conjugate acid is defined as the acid created when a reactant base accepts one H . The + conjugate base is defined as the base created when a reactant acid donates one H . + + + 29. a. H O3 b. H 2 c. NH 4 d. HX e. H 2 30. a. OH & b. O &2 c. NH & d. X & e. H O 2 2 31. In terms of their reaction with water, a strong acid is defined as any acid that reacts with water to form H O and the conjugate base to a large extent. A weak acid is then defined as any acid that 3 + reacts with water to form H O3and the conjugate base to a small extent. + & 32. Reaction of a strong acid and water: HX + H O 2 xxv H 3 + X + & Reaction of a weak acid and water: HX + H O 2 xxv H 3 + X There is no differencebetween the equation for a strongacid reactingwith waterand the equation for a weak acid reacting with water. The only difference between a strong and a weak acid is the extent of reaction (see problem 30). 6 33. The six common strong acids are: HCl, HBr, HI, HNO , H SO , and3HClO2. T4e other acids4are classified as weak acids. 34. The inventory of 1 M HCl is [H O ] = 1 M and [Cl ] = 1 M. & 3 + & The inventory of a 1 M solution of any strong acid, HX, is [H O ] = 1 3 and [X ] = 1 M. The inventory of a 1 M solution of any weak acid, HY, is [HY] = 1 M. 35. Liquid HCl does not conduct because HCl is a molecule and therefore without being involved in some kind of ionization process (like an acid/base reaction) the liquid HCl contains no ions to support conductivity. When the HCl was added to the water to form the 1 M HCl solution a chemical reaction occurred to a large extent to form H O and3Cl . Thus, the solution contains many ions to support conductivity. 36. Although HCl and HC H O bot2 r3ac2 as acids with water and the initial amount of HCl and HC H2O 3er2thesame,theresultingsolutions areverydifferent. Asmentionedabove,HClreacts with watertoalargeextent to formH O andCl ions andthus theresultingsolution containsmany 3 + ions to support conductivity. On the other hand, HC H O react2 w3th2water to form H O and 3 & C 2 O3io2s to only a small extent and therefore the resulting solution contains only a relatively smallnumberofionstosupportconductivity. Therefore,1MHClisagoodconductorofelectricity and 1 M HC H O2is3a 2oor conductor of electricity. 37. As mobile ions are necessary for a solution to conduct electricity, it follows that more ions will enable a solution to conduct electricity better. Thus, if the solution of HZ conducts better than the solution of HY, it follows that the HZsolution contains more ions than the HY solution. Therefore the acid HZmust have reacted more with the waterto produceions (H O and Z ) t3an the acid HY + & reacted to produce ions (H O a3d Y ). Thus, acid HZ must be a stronger acid than HY. 38. In terms of their reaction with water, a strong base is defined as any base that reacts with water to & form OH and the conjugate acid to a large extent. A weak base is then defined as any base that reacts with water to form OH and the conjugate acid to a small extent. + & 39. Reaction of a strong base and water: B + H O 2 xxv HB + OH Reaction of a weak acid and water: B + H O 2 xxv HB + OH & Thereis nodifferencebetween the equation for a strongbasereactingwith waterandtheequation for a weak base reacting with water. The only difference between a strong and a weak base is the extent of reaction (see problem 37). &3 &3 &3 &2 &2 &2 &2 & & &2 40. The ten common strong bases are: N , P , As , O , S , Se , Te , H , NH , and NH . 2 The other bases are classified as weak bases. 7 41. The inventory of 1 M Na O is [Na ] = 2 M and [OH ] = 2 M. 2 Recall that the O ion is a strong base and therefore reacts with water to form its conjugate acid & & (OH ) and the conjugate base of water (OH )! + & The inventory of a 1 M solution of any strong base, B, is [HB ] = 1M and [OH ] = 1 M. The inventory of a 1 M solution of any weak base, B, is [B] = 1 M. 42. Liquid NH d3es not conduct because NH is a 3olecule and therefore without being involved in some kind of ionization process (like an acid/base reaction) the liquid NH con3ains no ions to support conductivity. When the NH wa3 added to the water to form the 1 M NH soluti3n a chemical reaction occurred + & to a small extent to form NH a4d OH . Thus, the solution contains a few ions to support some conductivity. 43. The conductivity of solutions of molecular acids and molecular bases is due to these materials reacting with the water solvent to create ions. The stronger the acid (or base), the more conductive the resulting solution will be. Therefore the degree of conductivitygives a measure of the strength of the acid (or base). However, for ionic acids (or bases) the solutions will always be highly conductive since the solutions will always contain many ions regardless of strength (after all, the solutesareionic). Therefore,conductivityisuselessinassessingtheacid(orbase)strengthsofionic acids (or bases). 44. As discussed in problem 42 above, mobile ions are necessary for a solution to conduct electricity. More ions will enable a solution to conduct electricity better. Thus, if the solution of Z conducts better than the solution of Y, it follows that the Z solution contains more ions than the Y solution. + & ThereforethebaseZmusthavereactedmorewiththewatertoproduceions(HZ andOH )thanthe base Y reacted to produce ions (HY and OH ). Thus, base Z must be a stronger base than Y. + & 45. A. Reaction (i): HR + H O 2 xxv H 3 + R Small Extent Reaction (ii): HR + G & xxv HG + R & Large Extent Reaction (iii): HY + H O xxv H O + Y & Large Extent 2 3 B. Reaction (i): H O > HR 3 Reaction (ii): HR > HG + Reaction (iii): HY > H O3 & C. Reaction (i): R > H O2 Reaction (ii): G > R & Reaction (iii): H 2 > Y & 8 D. Acids Bases & Stronger HY Y Weaker + .H 3 H 2 . . .HR R& . Weaker .HG G & . Stronger E. The reaction involved is: Reaction (iv): HB + R & xxv HR + B & Small Extent & Since the solution is deep red then there must be a lot of R present and therefore the reaction only went to a small extent. Thus, B > R as bases and HR > HB as acids. This arrangement places HB below HR on the table. It is unclear whether it is stronger or weaker than HG. To decideon the relative strength of HB and HG as acids (also B and G as bases) ONE of the two following reactions must be performed and the extent of reaction determined. & & Reaction (va): HB + G xxv HG + B Extent ? OR & & Reaction (vb): HG + B xxv HB + G Extent ? QUESTIONS ON EQUATION WRITING + & 46. a. Inventory of 0.1 M HCl: [H O ] 3 0.1 M , [Cl ] = 0.1 M. Inventory of 0.1 M KOH: [K ] = 0.1 M , [OH ] = 0.1 M. Reaction: H O (aq) + OH (aq) & xxv 2 H O ( R) Large Extent 3 2 b. Inventory of 0.1 M HNO : 3 [H 3 ] = 0.1 M , [NO ] 3 0.1 M. Inventory of 0.1 M NH : 3NH ] = 3.1 M. + + Reaction: H O3(aq) + NH (aq) 3 xxv H 2 ( R) + NH (4q) Large Extent c. Inventory of 0.1 M H SO : [H O ] = 0.1 M , [HSO ] = 0.1 M. 2 4 3 + 4 & Inventory of 0.1 M NaC H 2 3 [2a ] = 0.1 M , [C H O 2 =30.2 M. Reaction: H O3(aq) + C H O 2a3) 2& xxv H 2 ( R) + HC H2O 3aq2 Large Extent + & d. Inventory of 0.1 M NaHSO : [N4 ] = 0.1 M , [HSO ] = 0.14M. Inventory of 0.1 M NaC H 2 3 [2a ] = 0.1 M , [C H O 2 =30.2 M. Reaction: HSO (aq) + C H O (aq) & xxv SO &2 (aq) + HC H O (aq) Large Extent 4 2 3 2 4 2 3 2 + & e. Inventory of 0.1 M NH Cl:4[NH ] = 4.1 M , [Cl ] = 0.1 M. Inventory of 0.1 M NaOH: [Na ] = 0.1 M , [OH ] = 0.1 M. + & Reaction: NH (a4) + OH (aq) xxv H 2 ( R) + NH (3q) Large Extent f. Inventory of 0.1 M NaHCO : [Na ] = 0.1 M , [HCO ] = 0.1 M. 3 + & 3 Inventory of 0.1 M KOH: [Li ] = 0.1 M , [OH ] = 0.1 M. & & &2 Reaction: HCO (a3) + OH (aq) xxv H2O ( R) + CO 3 (aq) Large Extent 9 g. Inventory of 0.1 M HC H O 2 [3C 2 O ] = 2.13M.2 + & Inventory of 0.1 M NaC H O 2 [3a 2 = 0.1 M , [C H O ] = 021 3. 2 Reaction: HC H O2(a3) 2 C H O (aq)2 3 2 xxv C 2 3 (2q) + HC H O (a2) 3 2 NOTE: This reaction is between the strongest acid and the strongest base present in appreciable amount but as the reactants and products are the same, the reaction goes nowhere! h. Inventory of 0.1 M HC H O 2 [3C 2 O ] = 2.13M.2 Inventory of 0.1 M MgSO : [Mg(H O) ] = 0.1 M , [SO ] = 0.1 M. &2 4 2 6 4 &2 & & Reaction: HC H O2(a3) 2 SO 4 (aq) xxv C2H 3 (2q) + HSO (aq) Sm4ll Extent i. Inventory of 0.1 M HC H O 2 [3C 2 O ] = 2.13M.2 & + Reaction: HC H O2(a3) 2 H O ( 2 R) xxv C 2 O3(2q) + H O (aq)3 Small Extent j. Inventory of pure HNO is H3O 3. Inventory of pure NaOH is NaOH. Reaction: HNO ( R) + NaOH (s) xxv H O ( R) + Na (aq) + NO (aq) Large Extent 3 2 3 QUESTIONS ON OXIDATION AND REDUCTION 47. Oxidation is the process in which electrons are formally lost. Reduction is the process in which electrons are formally gained. 48. An oxidizer is the chemical that causes some other chemical to undergo oxidation (the loss of electrons). The oxidizer does this by taking away (or gaining) electrons from the other chemical. A reducer is the chemical that causes some other chemical to undergo reduction (the gain of electrons). The reducer does this by giving away (or losing) electrons to the other chemical. 49. The products of any redox reaction are another oxidizer and another reducer (called the conjugate oxidizer and conjugate reducer). Note the similarity to the products of an acid/base reaction. 50. Other than the fact that both oxidation states and formal charges are calculated numbers and therefore do not really exist, there is no connection between the concepts of oxidation states and formal charges. Formal charges are used to determine the plausibility of a given Lewis structure while oxidation states are used to follow electrons in redox reactions. 51. ForaMAINGROUPELEMENT(sandpblockelement)thatreacts,thetheoreticalmaximumand minimumoxidationstatesaredeterminedfromtheelementspositionrelativetothenoblegasfamily. The maximum oxidation state is given by: Max Ox State = +(Group Number) The minimum oxidation state is given by: Min Ox State = (Group Number & 8) a. N (group 5) Max Ox State = Group# = +5 Min Ox State = (Group # & 8) = (5 & 8) = &3 10 b. Xe (group 8) Max Ox State = Group# = +8 Min Ox State = (Group # & 8) = (8 & 8) = 0 c. Cl (group 7) Max Ox State = Group# = +7 Min Ox State = (Group # & 8) = (7 & 8) = &1 d. Na (group 1) Max Ox State = Group# = +1 Min Ox State = (Group # & 8) = (1 & 8) = &7 Never achieved! e. Se (group 6) Max Ox State = Group# = +6 Min Ox State = (Group # & 8) = (6 & 8) = &2 f. Ca (group 2) Max Ox State = Group# = +2 Min Ox State = (Group # & 8) = (2 & 8) = &6 Never Achieved! g. C (group 4) Max Ox State = Group# = +4 Min Ox State = (Group # & 8) = (4 & 8) = &4 h. B (group 3) Max Ox State = Group# = +3 Min Ox State = (Group # & 8) = (3 & 8) = &5 52. (+) (+7) (&2) (+6) (&2) (+6) (&1) a. Cs Cl O 4 b. S O 3 c. U F6 (+4) (&1) (+) (+5) (&2) (+3) (+6) (&2) d. Xe F 4 e. K N O 3 f. Al2 (S O )4 3 (+1) (&1) (+1) (+6) (&2) (+8) (&2) g. Cs H h. Na Cr O i. Os O 2 2 7 4 (&3) (+1) (+4) (&2) (+2) (&2) (0) j. (N H )4 2 C O 3 k. Ba Se l. S8 (+2) (+6) (&2) (+5) (&2) (+2) (&1)!!! + m. Mg 3 U O6 n. N O 2 o. Sr O 2 (+8) (&2) (+6) (&2) (&3) (+1) p. Xe O &4 q. S O &2 r. As H 6 2 7 3 53. a. The Ox. St. of Mn increases from +4 in MnO to +2 in MnO and th4s the Mn lost 3 electrons during the reaction. Therefore, Mn underwent an oxidation. b. The Ox. St. of Bi decreases from +5 in BiO to 33 in Bi and thus the Bi gained 2 electrons during the reaction. Therefore, Bi underwent a reduction. c. The Ox. St. of S increases from +4 in SO t2 +6 in SO an3 thus the S lost 2 electrons during the reaction. Therefore, S underwent an oxidation. d. TheOx.St.ofClincreasesfrom+1inOCl to+5inClO andthustheCll3st4electronsduring the reaction. Therefore, Cl underwent an oxidation. e. The Ox. St. of N decreases from +4 in N O2to4+1 in N O 2nd thus the N gained3 electrons each during the reaction. Therefore, N underwent a reduction. f. The Ox. St. of O decreases from &1 in H O 2o2&2 in H O a2d thus the O gained 1 electron each during the reaction. Therefore, O underwent a reduction. 11 54. As mentioned in problem 43, the oxidizing agent (or oxidant or oxidizer) is the chemical that gains electronsfromtheotherchemical. Thereducingagent(orreductantorreducer)isthechemicalthat loses electrons to the other chemical. & a. TheCl g2inselectronstoformCl andthereforetheCl istheoxid2zingagent(oroxidizer). The & Br loses electrons to form Br a2d therefore is the reductant (or reducer). b. The oxidation state of N decreases from +4 in NO to 23 in NH . Thi3 is the gain of electrons and therefore NO i2 the oxidizing agent. The oxidation state of H increases from 0 in H to 21 in NH 3nd H O.2This is the loss of electrons and therefore H is 2he reducing agent. c. The oxidation state of Mn decreases from +7 in MnO &to +2 in Mn . This is the gain of 4 electronsandthereforeMnO istheoxidizingagent. TheoxidationstateofSincreasesfrom+4 4 in SO &2to +6 in SO . This is the loss of electrons and therefore SO &2is the reducing agent. 3 4 3 d. The oxidation state of Cl decreases from +1 in ClO to &1 in Cl . This is the gain of electrons and therefore ClO is the oxidizing agent. The oxidation state of Cr increases from +3 in Cr(H 2) (3H) to3+6 in CrO . T4is is the loss of electrons and thereforeCr(H O) (O2) i3 the 3 reducing agent. & e. The oxidation state of Cl decreases from 0 in Cl 2o &1 in Cl . This is the gain of electrons and therefore Cl2is the oxidizing agent. The oxidation state of Cl increases from 0 in Cl to2+1 in HClO. This is the loss of electrons and therefore Cl 2s the reducing agent. Cl a2ts as both the oxidizing agent and the reducing agent in this case. This is calleda disproportionation reaction. 55. Cu(H O) +2(aq) + 2 e & xxv Cu (s) + 6 H O ( R) (Reduction ½ Reaction) 2 6 2 2 Cl (aq) xxv Cl (g) (Oxidation ½ Re
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