Chapter 5 - Homework solutions

Transcription

1 Chapter 5 - Homework solutions Q Ex 1,2,3,4,5,7,9,13,18,20,21,23, 27,29 and Prob. 3,4,6,7 1) Li Be B C N O F acidic Na Mg Al Si P S Cl amphoteric K Ca Ga Ge As Se Br basic Rb Sr In Sn Sb Te I Cs Ba Tl Pb Bi Po At 2) [Co(NH 3 ) 5 OH] 2+ SO 4 2- MeO - HPO 4 2- SiO(OH) 3 - S 2-4) 5) NH 2 - > HS - > F - > I - i) O 2- - too strong CO OK ClO 4 -, NO too weak ii) ClO too weak HSO 4 -, NO 3 -- OK 7) HAsO 4 2- has one As=O unit. Using Pauling s rules the pka for H 3 AsO 4 should be ~ 8-5 = 3 The pka for HAsO 4 2- should be ~ = 13. There is reasonable agreement between Pauling s rules and the experimental pka.

2 9) a) Fe(OH 2 ) 6 3+ is the stronger acid as the higher charge on the metal makes it easier to loose a proton. b) Al(OH 2 ) 3+ 6 is more acidic as the metal center is smaller making it easier to loose a proton. c) Si(OH) 4 is more acidic again due to the size of the central atom. d) HClO 4 is more acidic as this acid has more Cl=O units. e) HMnO 4 will be the stronger acid as it has more M=O units. f) H 2 SO 4 is the stronger acid as there are more X=O units in it. 13) Mo, Si, B, As form polyanions Al forms polycations According to the text Ti and Cu also form polycations. I am not aware of any important examples of such species. 18) Question not very clear - I wish I had not set this! a) SO 3 and H + are acids, H 2 O and HSO 4 - are bases. The reaction can be regarded as the formation of a complex or an acid base displacement. b) The reaction is an acid base displacement reaction. The soft acid Hg 2+ competes for and wins the soft base from the soft acid B 12 + c) The reaction is an example of complex formation with Cl - acting as a Lewis base and SnCl 2 acting as a Lewis acid. d) The process is an acid base displacement reaction with AsF 3 acting as a base and SbF 5 acting as an acid. e) Ethanol dissolving in pyridine is perhaps best regarded as complex formation with the pyridine acting as an acceptor and the ethanol (OH proton) acting as a donor in a hydrogen bond. (When discussing hydrogen bonds, the molecule with the hydrogen attached is the hydrogen bond donor even though it is acting as a Lewis acid).

3 20) a) <1 b) >1 as S in SO 2 is soft and P is soft H is H-O- is hard. c) <1 as Hg is soft and I is softer than Cl d) >1 Ag + is soft and CN - is softer base than Cl - 21) 23) N binds to B in BF 3 and P binds to BH 3 P is softer than N and B in BH 3 is softer than B in BF 3. a) Acetone appears to be a base with similar properties to DMSO on the basis of the E and C parameters with DMSO being a marginally better donor. b) DMS is a much better soft base than DMSO but a worse hard base. DMSO appears to be acting primarily as a donor through the oxygen atom rather than the sulfur on the basis of the above observations. 26) a) A hard acidic solvent would favor displacement of Cl - by I - at an acidic center. For example water. b) A hydrogen bond donor, for instance water or perhaps an alcohol would interact with the nitrogen lone pair on the amine and mask it. The arsine would not hydrogen bond very strongly and it s basicity would be enhanced relative to that of the amine. c) A hard donor solvent that is not likely to loose a proton may be effective at enhancing the acidity of Ag + relative to Al 3+. For example methanol may bind to Al 3+ and reduce it s acidity but it is not likely to strongly bind to Ag +. d) A solvent that will not bind to the iron center competitively with Cl -. For example concentrated HCl. Alternatively a solvent like DMSO will not coordinate strongly to Fe 3+ allowing the formation of FeCl ) Donor solvents would react with I 2 + and Se 8 + so a very poor donor solvent is required. A good Lewis acid as the solvent allows the formation of stable noncoordinating anions like SbF 6 -. S 4 2- and Pb 9 4-

4 are basic species if the solvent were acidic it would react with ions. By having a basic solvent the solvent will not attack the ions and will stabilize the counter cation.

5 29) Hg 2+ is a very soft acid. The exclusive formation of sulfide ores is a reflection of this. Zn 2+ is not as soft as Hg 2+ consequently it can be found in ores with harder counter ions as well as sulfides (Zn 2+ is borderline on the hard/soft acid scale). Prob 3) CO 2 + H 2 O > H 2 CO 3 M 2 SiO 4 + 2H 2 CO > 2MCO 3 + SiOH 4 nsioh > nsio 2 + 2H 2 O? Silicic acid can condense to form all sorts of polysilicates they do not have to be SiO 2. The above scheme shows the removal of CO 2 by reaction with water followed by silicates. Prob 4) a) Fe(NO 3 ) 3 + 3H 2 O > 3HNO 3 + Fe(OH) 3 b) Formula weight of Fe(NO 3 ) 3.9H 2 O is ~ g So mols of Fe 3+ are added to the solution Fe 3+ + H + /3 ~ In a 100 L system [Fe 3+ ] + [H + ]/3 = But [Fe 3+ ]/[H + ] 3 = 10 4 So 3[H + ] 3 x [H + ] = 0.49 [H + ] must be significantly less than one for the above equation to be true so ignore the [H + ] term and solve.

6 [H + ] ~ M (check the approximation by back substitution) ph ~ 1.6 [Fe 3+ ] ~ M Ignored Fe(H 2 O) 5 OH 2+ and Fe(H 2 O) 4 OH 2 +

7 Prob 6) As the solution is conducting it must contain ions. As AlCl 3 is a Lewis acid it will be able to act as a chloride ion acceptor. Species such as AlCl 4 - may be present along with say AlCl 2 (NCMe) 4 + or Al 3+ (MeCN) 6 In the case of these species the Lewis base MeCN displaces the Lewis base Cl - from the AlCl 3 and the Cl - is the picked up by AlCl 3 acting as a Lewis acid. Prob 7) The initial data suggests that six coordinate Fe 3+ is reddish in the ligand field provided by chloride and the tetrahedral species is yellow. Both sets of equilibria appear to explain the observations. However, it seems unlikely that the solvent would ionize and act as a good chloride donor and much more probable that the solvent would coordinate the metal center via oxygen. Consequently, the second equilibrium reactions is a probably a better description of what happens in solution.