1. The Handbook of Chemistry and Physics says that PbBr 2 is soluble in water to the tune of 8.441 g per kg of water at 25 C. The molar mass of PbBr 2 is 367 g mol 1. (a) What is the ionic strength of an aqueous solution saturated with PbBr 2? (b) What does the Debye-Hückel Limiting Law predict γ ± to be for this solution? (c) The Handbook also says that K = 6.6 10 6 for PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq). What does this number predict γ ± to be? 2. Co and Ni have very similar standard reduction potentials at 298 K: Ni 2+ (aq) + 2e Ni(s) E o red = 0.250 V Co 2+ (aq) + 2e Co(s) E o red = 0.277 V (a) What is the equilibrium constant for Ni 2+ + Co Ni + Co 2+? (b) For the cell Co(s) Co 2+ (aq, a = 1) Ni 2+ (aq, a = 0.01) Ni(s) with ion activities as shown, what is the spontaneous net reaction that would occur if current was allowed to flow between the metal electrodes? (The notation in the cell schematic above represents a salt bridge that connects the two solutions.) 3. A common electrochemical cell used for halide ion determinations consists of a Ag(s) Ag + (aq) half cell connected through a salt bridge to a half cell called a calomel electrode. The calomel half cell uses solid mercury(i) chloride (Hg 2 Cl 2 (s), which is calomel) and a saturated KCl solution in contact with a small amount of elemental mercury, Hg(l), which is in contact with a chemically inert wire to which electrical contact is made. The half reactions and their reduction potentials at 298 K for this cell are Ag + (aq) + e Ag(s) E o red = 0.799 V 1 2 Hg 2 Cl 2(s) + e Hg(l) + Cl (aq, saturated KCl) E red = 0.242 V Note that the calomel reduction potential is listed as E red instead of E o red because the potential, 0.242 V, is based on a saturated solution of Cl (aq) in which [Cl ] does not change. This means the cell emf will depend only on [Ag + ] and these two reduction potentials. The net cell reaction is Hg(l) + Cl (aq) + Ag + (aq) 0.5 Hg 2 Cl 2 (s) + Ag(s). (a) If such a cell exhibits an emf of +0.380 V, what must [Ag + ] be? (b) The Ag Ag + half cell normally contains Ag + (aq) and an inert anion, typically NO 3 (aq). If a solution containing Cl (aq) is added to this half cell, insoluble AgCl(s) precipitates. What is the cell emf if enough Cl (aq) is added to establish [Cl ] = 0.01 mol kg 1, given also that the solubility product for AgCl is K sp = 1.8 10 10. (You may approximate all activity coefficients for ions to be 1, just to simplify the problem a bit.) 4. BaSO 4 (s) is not very soluble in water. Let s look into that fact here. (a) Calculate the solubility product equilibrium constant K sp for BaSO 4 from the following molar free energies of formation, expressed for your computational convenience as ΔG f o /RT at 298 K.
BaSO 4 (s) Ba 2+ 2 (aq) SO 4 (aq) ΔG o f /RT 549.53 226.21 300.34 (b) Now calculate K sp from another line of reasoning. Experimentally, one finds 0.246 mg (or 1.054 10 6 mol) of BaSO 4 dissolves in 0.1 L of water at 298 K. What K sp does this fact predict, assuming that all activity coefficients are 1? (c) Your answer in part (b) should be close to that in (a), but a bit larger. Why are the answers different, and why is the part (b) answer larger? 5. Several areas of contemporary research rely on a technique know as time-of-flight mass spectrometry. This device lets a brief pulse of a gas sample into a vacuum chamber, into which the gas molecules fly on straight-line paths without collision until they reach a detector that (often via electron bombardment) turns them into gaseous cations that can be detected as an electrical current. Suppose a puff of a mixture of He (mass 4 g mol 1 ) and Ar (mass 40 g mol 1 ) is injected into such a device via a small hole equipped with a fast-acting shutter. Suppose as well that the gas starts at 300 K. If the detector is 0.125 m away from the hole, at what times after the shutter opens will the He and Ar detector signals be largest? (Assume the shutter opens and closes much more quickly than the times it takes either gas to reach the detector.) 6. Solid antimony is placed in an evacuated cell which is heated to 763 K. Antimony vapor is allowed to effuse from this cell into a vacuum chamber through a pinhole of area 4 10 3 cm 2. The vapor pressure inside the cell is 1.75 10 6 atm. The cell is observed to change mass at the rate 2.46 10 7 g s 1. Is the principle species in antimony vapor Sb atoms, Sb 2, Sb 3, or what? 7. The measured diffusion coefficient for N 2 gas at 1 atm and 80 C is 2.87 10 5 m 2 s 1, and at 1 atm and 78.5 C, D = 0.946 10 5 m 2 s 1. (a) If you had only the 80 C value and our hard-sphere theory of transport phenomena, what would you predict the 78.5 C value to be? (b) What would you predict the N 2 viscosity coefficient, η, to be at 80 C and 1 atm? (c) Now consider the diffusion constant of gaseous ethylene, C 2 H 4. Compared to the N 2 value at some T and P, would the ethylene value be larger or smaller at that same T and P, and why? 8. Robert Boyle did the following experiment in 1660. He observed the damping time of a pendulum in a vessel at two different air pressures. If a pendulum takes 15 min to come to rest swinging in a gas at 1 atm, will it come to rest faster or slower in a gas at 0.1 atm, and why? 9. The K/Na phase diagram is shown below. Key temperatures are shown above the diagram lines, and the composition is expressed in atomic percent (which is mole fraction times 100) with pure K on the left and pure Na on the right.
(a) Note the short vertical line at 66.6% Na that ends at 6.92 C. It represents a stoichiometric compound Na 2 K. Unlike most compounds, though, Na 2 K(s) does not melt at one sharp temperature. What does the phase diagram say happens to Na 2 K(s) at 6.92 C? (b) A liquid Na/K solution at the eutectic composition is cooled to 12.62 C. What happens at this temperature, i.e., what phases are in equilibrium?
Solutions 1. The Handbook of Chemistry and Physics says that PbBr 2 is soluble in water to the tune of 8.441 g per kg of water at 25 C. The molar mass of PbBr 2 is 367 g mol 1. (a) What is the ionic strength of an aqueous solution saturated with PbBr 2? (b) What does the Debye-Hückel Limiting Law predict γ ± to be for this solution? (c) The Handbook also says that K = 6.6 10 6 for PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq). What does this number predict γ ± to be? (a) At saturation, we can find the analytic molality of dissolved PbBr 2 and thus of the dissolved ions quite easily: 8.441 g kg 1 m PbBr2 = 367 g mol 1 = 0.0230 mol kg 1 = m + m = 2m + = 0.0460 mol kg 1 and from this, we find the ionic strength, I: I = 1 2 22 0.0230 mol kg 1 + ( 1) 2 0.0460 mol kg 1 = 0.0690 mol kg 1 (b) The D-H limiting law next tells us (z + = 2, z = 1) γ ± = exp z + z 1.173 kg 1/2 mol 1/2 I 1/2 = 0.540 (c) If K sp = 6.6 10 6, then we can write 2 K = a Pb 2+a Br = γ ± m 3 ± = γ 3 ± m+ m 2 or γ ± = K m + m 2 1/3 = 0.514 2. Co and Ni have very similar standard reduction potentials at 298 K: Ni 2+ (aq) + 2e Ni(s) E o red = 0.250 V Co 2+ (aq) + 2e Co(s) E o red = 0.277 V (a) What is the equilibrium constant for Ni 2+ + Co Ni + Co 2+? (b) For the cell Co(s) Co 2+ (aq, a = 1) Ni 2+ (aq, a = 0.01) Ni(s) with ion activities as shown, what is the spontaneous net reaction that would occur if current was allowed to flow between the metal electrodes? (The notation in the cell schematic above represents a salt bridge that connects the two solutions.) (a) We first find E o from the standard reduction potentials (based on the net reaction as written with Ni 2+ reduced and Co oxidized): E o = E o red(reduction 1/2-rxn) Ered o (oxidation 1/2-rxn) = 0.250 V ( 0.277 V) = 0.027 V and from this, we calculate K, the equilibrium constant, noting that the net reaction is a two-electron redox reaction: n = 2 (and recalling that RT/F = 25.693 mv at 298 K) K = e ΔG R o /RT = e + nfeo /RT = e 2(0.027 V)/(0.025693 V) = 8.2 (b) The full Nernst equation here is
E = E o RT nf ln a Co 2+ a Ni 2+ = 0.027 V 0.025693 V ln 1 2 0.01 = 0.032 V and, because this value is negative, we see that the spontaneous process in this cell would be the reverse of the original net reaction, i.e., Ni + Co 2+ Ni 2+ + Co. 3. A common electrochemical cell used for halide ion determinations consists of a Ag(s) Ag + (aq) half cell connected through a salt bridge to a half cell called a calomel electrode. The calomel half cell uses solid mercury(i) chloride (Hg 2 Cl 2 (s), which is calomel) and a saturated KCl solution in contact with a small amount of elemental mercury, Hg(l), which is in contact with a chemically inert wire to which electrical contact is made. The half reactions and their reduction potentials at 298 K for this cell are Ag + (aq) + e Ag(s) E o red = 0.799 V 1 2 Hg 2 Cl 2(s) + e Hg(l) + Cl (aq, saturated KCl) E red = 0.242 V Note that the calomel reduction potential is listed as E red instead of E o red because the potential, 0.242 V, is based on a saturated solution of Cl (aq) in which [Cl ] does not change. This means the cell emf will depend only on [Ag + ] and these two reduction potentials. The net cell reaction is Hg(l) + Cl (aq) + Ag + (aq) 0.5 Hg 2 Cl 2 (s) + Ag(s). (a) If such a cell exhibits an emf of +0.380 V, what must [Ag + ] be? (b) The Ag Ag + half cell normally contains Ag + (aq) and an inert anion, typically NO 3 (aq). If a solution containing Cl (aq) is added to this half cell, insoluble AgCl(s) precipitates. What is the cell emf if enough Cl (aq) is added to establish [Cl ] = 0.01 mol kg 1, given also that the solubility product for AgCl is K sp = 1.8 10 10. (You may approximate all activity coefficients for ions to be 1, just to simplify the problem a bit.) (a) The Nernst equation here is (with E o based on the net reaction in which Ag + is reduced and Hg is oxidized) E = E o RT nf ln 1 a Ag + = (0.799 V 0.242 V) 0.025693 V 1 ln 1 a Ag + = 0.380 V and we solve for the silver activity: a Ag + = 0.00102 or [Ag + ] 1.02 mm with [Ag + ] in mm units which is an appropriate approximation because the activity is so small, making the activity coefficient close to 1 and the molality concentration approximately equal to a molarity concentration. (b) This is easy. If [Cl ] = 0.01 mol kg 1, then [Ag + ] = K sp /[Cl ] = 1.8 10 8 mol kg 1 so that
E = E o RT nf ln 1 a Ag + = 0.177 V 0.025693 V ln 1 = 0.281 V 1 1.8 10 8 4. BaSO 4 (s) is not very soluble in water. Let s look into that fact here. (a) Calculate the solubility product equilibrium constant K sp for BaSO 4 from the following molar free energies of formation, expressed for your computational o convenience as ΔG f /RT at 298 K. BaSO 4 (s) Ba 2+ 2 (aq) SO 4 (aq) ΔG o f /RT 549.53 226.21 300.34 (b) Now calculate K sp from another line of reasoning. Experimentally, one finds 0.246 mg (or 1.054 10 6 mol) of BaSO 4 dissolves in 0.1 L of water at 298 K. What K sp does this fact predict, assuming that all activity coefficients are 1? (c) Your answer in part (b) should be close to that in (a), but a bit larger. Why are the answers different, and why is the part (b) answer larger? (a) The K sp net reaction is BaSO 4 (s) Ba2+ (aq) + SO 4 2 (aq) so that ΔGR o is given by so that ΔG o R RT = ΔG f o (Ba 2+ ) + ΔG f o 2 (SO 4 ) ΔG f o (BaSO 4 ) = 22.98 RT RT RT K sp = e ΔG R o /RT = e 22.98 = 1.047 10 10 (b) Here, we turn the solubility information into ion concentrations at saturation: [Ba 2+ ] = [SO 4 2 ] = 1.054 10 6 mol 0.1 L = 1.054 10 5 M and write K sp = [Ba 2+ ][SO 4 2 ] = (1.054 10 5 ) 2 = 1.11 10 10. The reason this value is a bit larger than the value based on free energy of formation values is easy to see: we have assumed activity coefficients of 1, but in fact they will be a bit less that 1 (the Debye-Hückel limiting law predicts γ ± = 0.970) so that the real activities will be a bit less than the molar concentrations, dropping K sp a bit. In fact, with this γ ± value, we change K sp to (1.11 10 10 )(0.970) 2 = 1.045 10 10, spot on the value from part (a)! 5. Several areas of contemporary research rely on a technique know as time-of-flight mass spectrometry. This device lets a brief pulse of a gas sample into a vacuum chamber, into which the gas molecules fly on straight-line paths without collision until they reach a detector that (often via electron bombardment) turns them into gaseous cations that can be detected as an electrical current.
Suppose a puff of a mixture of He (mass 4 g mol 1 ) and Ar (mass 40 g mol 1 ) is injected into such a device via a small hole equipped with a fast-acting shutter. Suppose as well that the gas starts at 300 K. If the detector is 0.125 m away from the hole, at what times after the shutter opens will the He and Ar detector signals be largest? (Assume the shutter opens and closes much more quickly than the times it takes either gas to reach the detector.) The signal will be largest when the greatest number of molecules happen to fly by the detector. These molecules will be moving at the most probable speed, v mp so that v mp = 2k B T m = (128.95 m s 1 ) T/K M/g mol 1 = 1116 m s 1 for He 353.1 m s 1 for Ar Their time to travel 0.125 m will be t = 0.125 m v mp = 112 μs for He 354 μs for Ar While microsecond times seem like they would be no time at all, it is easy to record events happening on this time scale (and even briefer scales) with modern electronics. 6. Solid antimony is placed in an evacuated cell which is heated to 763 K. Antimony vapor is allowed to effuse from this cell into a vacuum chamber through a pinhole of area 4.00 10 3 cm 2 in a very thin cell wall. The vapor pressure inside the cell is 1.75 10 6 atm. The cell is observed to change mass at the rate 2.46 10 7 g s 1. Is the principle species in antimony vapor Sb atoms, Sb 2, Sb 3, or what? We have a cell at temperature T and (unknown) vapor pressure P which has a small circular hole of radius r through which mass Δm is lost in time t. The rate of loss of molecules is given by dn/dt = Z w A where A = πr 2, the hole area, and Z w is the rate of collisions with the wall: Z w = P 2πmk B T 1/2 Thus, the total number of molecules lost in time t is just ΔN = Z w At, and thus the mass loss is just this number times the mass per molecule: Δm = mpat 2πmk B T 1/2 which can be solved for P (but we know P here) or rearranged to give the mass loss rate: Δm = m 1/2 PA t 2πk B T Here, we know the rate and seek the mass of effusing particles, m. Converting our input data into base SI quantities gives us
m = Δm t 2πk B T 1/2 PA 2 = 2.46 10 10 kg s 1 2πk B (763 K) 1/2 0.1773 Pa 4.00 10 7 m 2 = 7.96 10 25 kg = 4 (the mass of one Sb atom, which is 2.02 10 25 kg) Thus, Sb 4 is the main gas-phase form of Sb at this temperature. (And by the way, the hard part of these experiments is finding a way to accurately measure μg mass changes in a metal cell at 700+ K while also measuring the gas pressure in the cell, all suspended in an ultra-high vacuum chamber!) 7. The measured diffusion coefficient for N 2 gas at 1 atm and 80 C is 2.87 10 5 m 2 s 1, and at 1 atm and 78.5 C, D = 0.946 10 5 m 2 s 1. (a) If you had only the 80 C value and our hard-sphere theory of transport phenomena, what would you predict the 78.5 C value to be? (b) What would you predict the N 2 viscosity coefficient, η, to be at 80 C and 1 atm? (c) Now consider the diffusion constant of gaseous ethylene, C 2 H 4. Compared to the N 2 value at some T and P, would the ethylene value be larger or smaller at that same T and P, and why? (a) Our theory says that, at constant P, D varies with T as T 3/2. Thus, we can write D( 78.5 C = 194.65 K) = D(80. C = 353.15 K) 194.65 3/2 = 1.17 10 5 m 353.15 2 s 1 (b) If we compare our expressions for D and η, we see that ρd = η where ρ is the gas mass density, mn/v: 2 D = v λ 3 η = v λ 3 mn so η = D mn V V = Dρ We can also express the gas mass density in more familiar terms with M the gas molar mass (in kg mol 1 units): ρ = mn V = mnp nrt = MP RT which, for N 2 at 1 atm and 80 C, yields ρ = 0.967 kg m 3 and thus we predict η = Dρ = 2.77 10 5 Pa s. The literature value is 2.01 10 5 Pa s. Not bad! (c) Our expression for D has two simple factors: the mean speed and the mean free path. The mass of ethylene is pretty much identical to that of N 2 (28.05 g mol 1 for C 2 H 4 versus 28.01 g mol 1 for N 2 ), which gives them equal mean speeds. But C 2 H 4 is bigger (has a larger collision cross-section) than N 2, making its mean free path smaller than that of N 2. Thus D(N 2 ) > D(C 2 H 4 ). 8. Robert Boyle did the following experiment in 1660. He observed the damping time of a pendulum in a vessel at two different air pressures. If a pendulum takes 15 min
to come to rest swinging in a gas at 1 atm, will it come to rest faster or slower in a gas at 0.1 atm, and why? This question boils down (no pun intended...) to the question of how viscosity varies with pressure for gases around 1 atm pressure. Our viscosity theory shows that η is independent of P: as the gas density goes up with increasing pressure at constant T, the mean free path goes down at the same rate. More molecules are available to transport momentum at higher pressures, but they carry it over shorter distances. The two effects cancel exactly. Mr. Boyle s pendulum would take the same 15 minutes to come to rest at 0.1 atm as at 1 atm. Gaseous thermal conductivity is independent of pressure (unless the pressure is very low) for the same reason. 9. The K/Na phase diagram is shown below. Key temperatures are shown above the diagram lines, and the composition is expressed in atomic percent (which is mole fraction times 100) with pure K on the left and pure Na on the right. liquid + liquid + Na 2 K + liq. Na 2 K + Na 2 K + (a) Note the short vertical line at 66.6% Na that ends at 6.92 C. It represents a stoichiometric compound Na 2 K. Unlike most compounds, though, Na 2 K(s) does not melt at one sharp temperature. What does the phase diagram say happens to Na 2 K(s) at 6.92 C?
The diagram above has had the various areas labeled to show what phase or phases each represents. On the K side, the area labeled α is a one-phase homogeneous solid solution that is mostly K, and on the Na side, the area labeled β is a similar one-phase solid solution that is mostly Na. At 6.92 C, Na 2 K(s) melts into a liquid solution and a solid β phase. Much like a eutectic point, we have three phases in equilibrium: solid Na 2 K, solid β, and the liquid solution. The temperature is stuck at 6.92 C until all the solid Na 2 K is gone and only liquid + β appears. (b) A liquid Na/K solution at the eutectic composition is cooled to 12.62 C. What happens at this temperature, i.e., what phases are in equilibrium? Here, the eutectic liquid freezes into a mixture of α solid and Na 2 K(s). The temperature is stuck at 12.62 C until all the liquid has solidified into this mixture of α solid and Na 2 K(s).