HW10_Answer - Chem340 Physical Chemistry for Biochemists...

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Chem340 Physical Chemistry for Biochemists Dr. Yoshitaka Ishii Homework 10 Please solve the following questions. This is a good part of the final Exam. P9.13-14, P9.18, P9.21, P9.24, P9.28-9, P9.32, P9.40, P9.44 P9.48, P9.50 P10.1-10.2, P10.4-10.7, P10.9 P9.13) In the Debye-Hückel theory, the counter charge in a spherical shell of radius r and thickness dr around the central ion of charge + q is given by – q 2 re r dr. Calculate the most probable value of r , r mp , from this expression. Evaluate r mp for a 0.050 m solution of NaCl at 298 K.  2 23 8 0 1 10 ; 4.30 10 m = 43.0 nm mp mp r rr mp mp mp mp r mp dq r e qe qre dr r r      P9.14) Calculate the Debye–Hückel screening length 1/ at 298 K in a 0.00100 m solution of NaCl. For water, the screening length at 298 K in m -1 can be calculated as: 1 8 1 8 r 1 - 10 m 10 04 . 1 m 78.54 997 . 0 0.0010 10 9.211 ε kg mol I/ 10 91 . 2 κ solvent   1 8 1 8 r 1 - 10 m 10 33 . 7 m 78.54 997 . 0 0.0500 10 9.211 ε kg mol I/ 10 91 . 2 κ solvent nm 36 . 1 m 10 36 . 1 κ 1 7
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nm 65 . 9 m 10 65 . 9 κ 1 9 P9.18) Calculate I, ± , and a ± for a 0.0250 m solution of K 2 SO 4 at 298 K. Assume complete dissociation. How confident are you that your calculated results will agree with experimental results?  24 1 2 35 1 K SO 2, 1, 1, 2 0.0250 2 4 0.0750 mol kg 2 ln 1.173 2 0.0750 0.6425 0.523 0.050 0.0250 6.25 10 0.03969 molkg 0.03969 0.5260 0.0209 vzv z I m a m m m a             Not very confident. Figures 10.5 and 10.6 show significant deviations from predicted behavior for I > 0.01 mol kg –1 . P9.21) Dichloroacetic acid has a dissociation constant of K a = 3.32 10 –2 . Calculate the degree of dissociation for a 0.125 m solution of this acid (a) using the Debye–Hückel limiting law and (b) assuming that the mean ionic activity coefficient is one. a) We first consider the case when is given by the Debye–Hückel limiting law. The ionic strength is given by 1 2 0.04992 molkg 2 ln 1.173 1 0.04992 0.2621 0.7694 m Im
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We recalculate the ionic strength and iterate several times. when = 0.7694, 22 2 3.32 10 0.125 m m     23 2 3 1 1 7.0097 10 0.05608 m 0.05608 m 7.0097 10 0 0.05608 0.05608 4 1 7.0097 10 0.06025 molkg 2 0.06025 molkg ln 1.173 1 0.06025 0.2879 0.7498 m m m I       when = 0.7498   2 2 3 1 1 0.05905 0.125 0.05905 m 7.3815 10 0 0.05905 0.05905 4 1 7.3815 10 0.06132 molkg 2 0.06132 molkg ln 0.7479 0.7479 m m m m I when = 0.7479   2 2 3 1 1 0.05935 0.125 0.05935 m 7.4191 10 0 0.05935 0.05935 4 1 7.4191 10 0.06143 molkg 2 0.06143 molkg ln 0.2907 0.7477 m m m m I
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when = 0.7477    2 23 2 3 1 0.05938 0.125 0.05938 m 7.4228 10 0 0.05938 0.05938 4 1 7.4228 10 0.06144 molkg 2 m m m m    This is a sufficiently good result, and we calculate the degree of dissociation to be 0.06144 100% 49%.
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HW10_Answer - Chem340 Physical Chemistry for Biochemists...

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