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Unformatted text preview: CE/ESR 479/579
PSet 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 8.7OE05 8.7OE—05 K+ 0.5 39.1 1.28E05 1.28E05 Mg2+ 0.3 24.3 1.23E05 2.47E05 1.25E04
CI 1.5 35.5 4.23E05 4.23E05 $042— 2 96 2.08E—05 4.17E—05 NO3— 0.5 62 8.06E06 8.06E—06 —1.07EO4 1.82E05
Hence not accurate or complete 1.48E05 —1.48E~05 3. (10) [4 pts total]
a) [2 pt] 7 Given [H2CO3*] = 105 M and pH = 4, 7, 10, hence {H+} = 104, 10—7, 10—10
Find [HCO3—] so use mass action law: _] _____ 10~63 [H2C03*] 10‘6'3[H2CO3*] __ 10“” [HCO31 = {If} 10”, pH lHCO3—I 4 107'3 = 5e08 M
7 104'3 = 56—05 M
10 ' 10'1'3 = 5802 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 GammaO (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 gamma1 = —0.15 y1[HC03]{H*} _ 10‘“ 10‘“ ____:__:106.15 70[H2C03*] 71 10—0” Solve for [HCO3] as above but with corrected log K = —6. 1 5
pH IHC03— .
4 107'15 = 7e08 M 10""15 = 7e~05 M 10"'15 = 7e02 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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 Spring '09
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