q2fall2004_key4324

q2fall2004_key4324 - Caveat: be watchful, wary and wise....

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Quiz 2 EMA 4324, Stability of Materials, LecSec 3009 KEY Friday, September 17, 2004 Thermodynamic data for the JRA - H 2 O system are presented on the following table. 1. [5 points] Calculate the numerical value of the solubility product for JRA(OH) 2 in pure water at 25 o C. *************************************************************************** JRA(OH) 2 (c) = JRA +2 (s) + 2OH - ; G r o = 2*(-35595) - 16142 - (-120054) = 32722 cal/mol K sp = exp[- G r o /RT] = exp[-32722/1.098*298] = 1.00x10 -24 2. [5 points] The solubility of (JRA)SO 4 in water at 25 o C is 14.3 g per 100 g H 2 O. What solution pH would be required to precipitate enough JRA(OH) 2 to decrease the JRA +2 concentration to 10 -6 M? *************************************************************************** 1.00x10 -24 = [10 -6 ][OH - ] 2 ; [OH - ] = [10 -24 /10 -6 ] 1/2 = 10 -9 M; [H + ] = K w /[OH - ] = 10 -14 /10 -9 = 10 -5 ; pH = -log[H + ] = 5 3. [5 points] What is the maximum solution pH above which JRA cannot be oxidized [to JRA +2 ] in a deaerated solution [no air] containing a JRA +2 concentration of 10 -6 M? [Instructor’s
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Unformatted text preview: Caveat: be watchful, wary and wise. ...] *************************************************************************** G r o = 0 - (-16142) = 16142 cal/mol;E o (JRA +2 /JRA) = - G r o /nF = -16142/2*23060 = -0.350 v E(JRA +2 /JRA) = -0.350 + 0.0296log[10-6 ] = -0.528 v E(H + /H 2 ) = -0.0592pH = E(JRA +2 /JRA) = -0.528; pH max = -0.528/(-0.0592) = 8.91 The problem, of course, is that this pH is far greater than that required to maintain a metal ion concentration of 10-6 M; as the pH increases, [JRA s +2 ] decreases, with E(JRA s +2 /JRA c ) always staying below E(H + /H 2 ). [note to Brandon: deduct no more than one point if some recognition of the problem is not recognized] H +-56,690 H 2 O-35,595 OH--H(JRA)O 2-1-120,054 JRA(OH) 2 (c)-16,142 JRA +2 (s) JRA(c) Free Energy of Formation, G o , cal/mol, 25 o C substance F [Faraday's constant] = 23060 cal/volt-equivalent 2.303RT/F = 0.0592; R = 1.987 cal/mol* o K...
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