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Problem Set #12, Chem 340, Fall 2013 – Due Wednesday, Dec 4, 2013 Please Show All Work for Credit to Hand In: Ionic Equilibria: –4 1

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Problem Set #12, Chem 340, Fall 2013 – Due Wednesday, Dec 4, 2013 Please Show All Work for Credit to Hand In: Ionic Equilibria: –4 1 Problem Set #12, Chem 340, Fall 2013 – Due Wednesday, Dec 4, 2013 Please show all work for credit To hand in: Ionic equilibria: –4 1. Calculate the value of m± in 5.0 10 molal solutions of (a) KCl, (b) Ca(NO3) 2, and (c) ZnSO4. Assume complete dissociation. a) KCl 1 vv v m= v v m 41 mm = 1 5.0 10 molkg b) Ca(NO3)2 1 vv v m= v v m 11 mm= 4 33 4 5.0 104 molkg 1 7.9 10 4 molkg 1 c) ZnSO4 1 vv v m= v v m 1 mm= 1 2 5.0 1041 molkg 2. Calculate ±, and a±for a 0.0325 m solution of K4Fe(CN) 6 at 298 K. m 2 21 2 2 I v z v z m z m z 22 1 I 4 0.0325 mol kg1 4 2 0.0325 mol kg 1 0.325 mol kg 1 2 I ln 1.173zz 1.173 4 0.325 2.6749 mol kg1 0.069 1 1 vv v 45 1 1 m= v v m 4 0.0325 mol kg 0.099 mol kg m a 0.099 0.069 0.0068 m –3 3. Chloroacetic acid has a dissociation constant of Ka = 1.38 10 . (a) Calculate the degree of dissociation for a 0.0825 m solution of this acid using the Debye–Hückel limiting law. (b) Calculate the degree of dissociation for a 0.0825 m solution of this acid that is also 0.022 m in KCl using the Debye–Hückel limiting law. –3 a) Ka = 1.38 10 0.0825 m 22 mm m 3 Ka 1.38 10 mm0.0825 when = 1 m21.1385 10 4 1.38 10 3 m m21.38 10 3 m 1.1385 10 4 0 2 1.38 103 1.38 10 3 4 1 1.1385 10 4 m 0.0100 molkg1 2 m Im2 0.0100 molkg1 2 ln 1.173 1 0.0100 0.1173 0.8893 when = 0.8893 m21.43956 10 4 1.7449 10 3 m m21.7449 10 3 m 1.43956 10 4 0 2 1.7449 103 1.7449 10 3 4 1 1.43956 10 4 m 0.0112 molkg1 2 m Im2 0.0112 molkg1 2 ln 1.173 1 0.0112 0.1239 Same value, done! 0.8835 when = 0.8835 m21.4587 10 4 1.7681 10 3 m m21.7681 10 3 m 1.4587 10 4 0 2 1.7681 103 1.7681 10 3 4 1 1.4587 10 4 m 0.0112 molkg1 2 I 0.0112 molkg1 ln 1.173 1 0.0112 0.1239 0.0112 The degre of dissociation is 100% 13.6% 0.0825 b) + K Cl 1 1 zz1 1 1 I 0.022 molkg from the KCl alone. We first calculate the activity coefficient using this value. 0.022 I 1 0.022 molkg1 2 ln 1.173 1 0.022 0.1740 0.8403 We next calculate the concentration of the ions produced through the dissociation of the acid in the solution. m2 0.84032 1.38 103 0.0825 m m2 1.9544 103 0.0825 m m21.6124 10 4 1.9544 10 3 m m21.9544 10 3 m 1.6124 10 4 0 2 1.9544 103 1.9544 10 3 4 1 1.6124 10 4 m 0.01176 molkg1 2 The ionic strength is the sum of that due to the chloroacetic acid and the KCl or 1 I = 0.022 + 0.012 = 0.034 molkg . We recalculate using this value for the ionic strength. ln 1.173 1 0.034 0.2163 0.8055 and recalculate m m2 0.80552 1.38 103 giving 0.0825 m 1 m 0.01222 molkg Given the fact that the concentration of KCl is only known to 2 significant figures, this value of m is sufficiently close to the previous value that a further iteration is not 0.01222 14.8%. necessary. The degree of dissociation is 0.0825 4. The principal ions of human blood plasma and their molar concentrations are m 0.14 m, m 0.10 m, m 0.025 m. Calculate the ionic strength Na Cl HCO 3 of blood plasma. Note: obviously something missing, pos-neg imbalance 1 2 1 2 2 2 I mi zi 0.14 m1 0.10 m1 0.025 m1 0.1325 m 2 i 2 5. The oxidation of NADH by molecular oxygen occurs in the cellular respiratory system: + + O2 (g) + 2NADH(aq) + 2H (aq) 2H2O(l) + 2NAD (aq) Using the information in Table 9.7, calculate the standard Gibbs energy change that results from the oxidation of NADH by molecular oxygen. The standard reduction potential for nicotine adenine dinucleotide is NAD aq H aq 2e NADH aq E 0.320V Using the half cell reactions: E 0.320 V Oxidation: 2 NADH 2 NAD 2 H 4 e 0 E 0.815 V Reduction: O2 4H 4 e 2 H2O 0 Ecell = Ered + Eox = 1.135 V ΔG n F Δφ 496485 C mol-1 1.135 V 438.0 kJ mol-1 reaction 6. Determine Ksp for AgBr at 298.15 K using the electrochemical cell described by Ag s AgBr s Br aq, a Ag aq, a Ag s . The half cell and overall Br Ag reactions are AgBr + e– → Ag + Br– Eº = +0.07133 V Ag → Ag+ + e– Eº = –0.7996 V AgBr → Ag+ + Br– Eº = –0.72827 V nE 0.729 V logK 12.310 10 sp 0.05916 0.05916 V K 4.90 1013 sp 7. For a given overall cell reaction, S 17.5 J mol1 K1 and H 225.0 kJ mol1. R R ° E Calculate E and . T P GHTS 225000 J mol1 298.15 K 17.5 J K 1 mol 1 E RRR 2.38 V nF nF 1 96485 Cmol1 11 E SR 17.5 J K mol 41 1 = 1.81 10 V K TP nF 1 96485 Cmol 8. The following data have been obtained for the potential of the cell PtsH g, f 1 atmHCl aq, mAgClsAgs as a function of m at 25°C. 2 m (molkg–1) E (V) m (mol kg–1) E (V) m (mol kg–1) E (V) 0.00100 0.597915 0.0200 0.43024 0.500 0.27231 0.00200 0.54425 0.0500 0.38588 1.000 0.23328 0.00500 0.49846 0.100 0.35241 1.500 0.20719 0.0100 0.46417 0.200 0.31874 2.000 0.18631 ° Calculate E and ± for HCl at m = 0.00100, 0.0100, 0.100, and 1.000. + – Cell reaction: 2AgCl(s) + H2(g) 2Ag(s) + 2H (aq) + 2Cl (aq) RT2 RT E E ln a a E ln a a 2FF H Cl H Cl a a a2 2 m 2 H Cl 22RT m RT EE ln ln F m F In the low concentration limit we can use the Debye-Hückel result mm ln 0.50926log10 1.172614ln mm Therefore, for dilute solutions 22RT m RT m EEln 1.172614 F m F m 2RT m m E ln Using this result, a plot of Fm (y axis) vs. m (x axis) will have an intercept of E . We use the data up to m = 0.100, as the Debye-Hückel model is not valid for more concentrated solutions. m E m m m 0.001 0.031623 0.57915 0.224212 0.002 0.044721 0.54425 0.224909 0.005 0.070711 0.49846 0.226203 0.010 0.1 0.46417 0.227531 0.020 0.141421 0.43024 0.229218 0.050 0.223607 0.38588 0.231943 0.100 0.316228 0.35241 0.234090 The data in the table is graphed below. The best fit line gives a value for E of 0.223665 V. Given we can now find from Fm ln EE ln RT m E 0.223655 V ln mm 0.001 –0.010654 0.989 0.010 –0.0755112 0.927 0.100 –0.203036 0.816 9. A fuel cell develops an electric potential from the chemical reaction between reagents supplied from an outside source. What is the cell potential of a cell fuelled by (a) hydrogen and oxygen, (b) the combustion of butane at 1.0 bar and 298 K? E 10. Consider the cell, Zn(s)|ZnCl2 (0.0050 mol kg−1)|Hg2Cl2(s)|Hg(l), for which the − 2+ cell reaction is Hg2Cl2(s) + Zn(s)→2 Hg(l) + 2 Cl (aq) + Zn (aq). o 2+ o Given that E (Zn ,Zn) = −0.7628 V, E (Hg2Cl2,Hg) = +0.2676 V, and that the cell potential is +1.2272 V, (a) write the Nernst equation for the cell. Determine (b) the standard cell potential, (c) ΔrG, ΔrGo, and K for the cell reaction, (d) the mean ionic activity and activity coefficient of ZnCl2 from the measured cell potential, and (e) the mean ionic activity coefficient of ZnCl2 from the Debye–Hückel limiting law. (f ) −1 Given that (∂Ecell /∂T)p = −4.52 × 10−4 V K , calculate ΔrS and ΔrH. 11. The standard potentials of proteins are not commonly measured by the methods described in this chapter because proteins often lose their native structure and function when they react on the surfaces of electrodes. In an alternative method, the oxidized protein is allowed to react with an appropriate electron donor in solution.
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