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FT_SS04_2008 - CE/ESR 479/579 P-Set 4 SOLUTION SET 1(7...

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Unformatted text preview: CE/ESR 479/579 P-Set 4 SOLUTION SET 1. (7) Easiest way is just look at the verb. if the process is proceeding (present participle for you grammarians) then the system is not at equilibrium, If the Verb is a "state" (like "is") then could be at equilibrium A. "components...are leaching” so must be a kinetic process B. "salt...is dissolving” so must be kinetic process C. "There is a concentration...” Sounds like an equilibrium state 2. (9) V54 ELeaiAanguT/{Aery : Chm3¢ 50.00%? m Cg CL W/L mg/L MW mol/L eq/L} Sums Difference pH H+ 5.01507 5.01507 Na+ 2 23 8.7OE-05 8.7OE—05 K+ 0.5 39.1 1.28E-05 1.28E-05 Mg2+ 0.3 24.3 1.23E-05 2.47E-05 1.25E-04 CI- 1.5 35.5 4.23E-05 -4.23E-05 $042— 2 96 2.08E—05 -4.17E—05 NO3— 0.5 62 8.06E-06 -8.06E—06 —1.07E-O4 1.82E-05 Hence not accurate or complete 1.48E-05 —1.48E~05 3. (10) [4 pts total] a) [2 pt] 7 Given [H2CO3*] = 10-5 M and pH = 4, 7, 10, hence {H+} = 10-4, 10—7, 10—10 Find [HCO3—] so use mass action law: _] _____ 10~63 [H2C03*] 10‘6'3[H2CO3*] __ 10“” [HCO3-1 = {If} 10-”, pH lHCO3—I 4 107'3 = 5e-08 M 7 104'3 = 56—05 M 10 ' 10'1'3 = 58-02 M b) [2 pt] To correct for seawater at I = 0.5 M rewrite mass action law in its proper from of aactivities and then convert to concentrations with gamma factors {Hc03—}{H+} :10-.. 2 71 [Hem—1 {Ha {Hzcofi} , yoiHZCOm Gamma-O (for uncharged species is 1.0 (no charge, no correction) and get garnma—l (correction for monovalent ion) from Table 1—2 in text. Find for I = 0.5, —log gamma = 0.15, or log gamma-1 = —0.15 y1[HC03-]{H*} _ 10‘“ 10‘“ ____:__:10-6.15 70[H2C03*] 71 10—0” Solve for [HCO3-] as above but with corrected log K = —6. 1 5 pH IHC03—| . 4 107'15 = 7e-08 M 10""15 = 7e~05 M 10"'15 = 7e-02 M 4e (12) [2 pts Totl] 3) [1 Pt] 0.693 71/2 = T k : 0.693 = 0.693 : 0'058yr_1 11/2 12)” The only trick here is to see that question is asking about the molar ratio of tritium to its product after 25 yr, NOT the ratio of tritium to the original amount of tritium (C/CO): 3H—>3He 3H(t):3HO e-(o,058yr")r and amount of helium— 3 produced equals the difference between original tritium and the tritium remaining: 3He(t)=3Ho—3H(t) 3H0) _ 3H0) 3Hem 3Ho~3H(z‘) Looks like the algebra is easier if we flip this ratio and simplify: 3He(t) _ 3H0—3H(t) _ 3H0 3H0 1 Molar ratio = 3H0) 3H(t) " 3H<t> fl 3H0 {mossy—U; e—(o,053yr")z Notice the amount of original tritium is irrelevant, which is to be expected since that quantity not given. Cam’ol 3He(t)_ 1 _1=__1.__—1=3.26 3H0) _ e‘(o.osgyrt')(25yr) 0.235 3 H“) :__1_=o_31 3He(t) 3.26 If you did not see your way through the algebra, another way to solve this is to assume some arbitrary initial amount of tritium, decay it for 25 yr, determine the amount of hellumi3- by the difference between the original and the decayed tritium, and then set up the ratio of tritium remaining to helium produced. 5 (if) a”? germ'a 14mg {4%w) Kw afgxmw’ M éggm {3%)} {gggfimlg (Am 2%; Wait}: 2m¢3¢i§m Qt. W x .5; mm” ’ Eifirj [fi‘yfifj d? MW ‘CEAEOE: [Hgorg f: 5.39;,» (Tea? {W} - :1» ...
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