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Unformatted text preview: 254 Aquatic Chemistry
Problem Set #1 Solution Set
2.3, 3.4, 3.6, 3.7, 3.8 Problem 2.3 ' In concentrated ferric iron solutions, numerous species may appear as intermediates before iron
precipitates as Fc(OH)3(s). What is the total ferric iron concentration in mgfL as Fe if the
concentrations of Fe“, Fe(OH)2*, Fe(0H);, and Fez(OH)2‘*are, respectively, 1.0x] 0‘3 M. 8.9x 10"1 M, 4.9x10’T M, and 1.2x10'3 M. Purpose of problem
Units conversion and stoichiometry Relevant section(s) of text
Sections 2.4.2 and 2.6 and Table 2.2 Solution
The total concentration as Fe is: {Fe} (mol/L) = (1 eq Fe/mol)[Fe3‘} + (1 eq Fe/mol)[Fe(OII)2+] + (1 eq Fclmol)[Fc(OH)f]  + (2 eq Fe/mol)[Ee3(OH)2“’]
= (1 eq Felrnol)(1.0><10"3 mol/L) + (1 eq Fe/mol)(8.9><10"‘ moi/L) + (1 eq Fe/mol)(4.9><10‘7 moi/L) + (2 eq Fe/tnoi)(l.2)<10‘3 molfL)
=4.3><10'3 Incl/L / ' [Fe] ( rug/L) = [Fe in tnoI/LKMW of Fe in g/moi)( 1000 mg/g}
= (’1.3><10‘3 moi/L)(55.85 g/mol)(1000 mg/g)
= 240 mg/L The total iron concentra tion is 240 nag/L as Fe Problem 3.4 _
Consider the reaction: HSO4“ = 5042' + H". Is the reaction endothermic or exothermic if all
concentrations are 1 M? Does the reaction proceed spontaneously as written if all concentrations are 1 M? For H8041 H"; = — 887.3 kamolﬁf = 132 J/molJ’K, and G; = —756.0 lei/mot. For
804213; 2 —909.2 kJ/mol,§; = 20.1 J/mol—“K, and G} = —744.6 kI/rnol. For H“: H} =
kJ/molﬁ, : 0 J/mol»°K, and G“! I O kJ/mol. Purpose of problem
Use of thermodynamic equations Relevant section(s) of text
Section 3.7 Solution Since alt concentrations are l M, you can use thermodynamic properties evaluated at standard
state. To determine if the reaction is endothermic or exothermic, calculate AH"m assuming all
concentrations are 1M: Aanm=§a 1 +§FH+“£t—£a flu3'04 49,336: = _9092 + 0 _ = ‘21.9 Heat is released (AQp= AH?” < 0) and therefore the reaction is eitothermic if all
concentrations are l M. To determine if the reaction proceeds spontaneously, calculate A011,, assuming all concentrations
are 1M: AG; =35“. + iii”. — aims —744.6 + o m (— 756.0) = +1 1.4'ldfmol A62” > 0 and therefore the reaction does not proceed spontaneously as written (assuming
that all concentrations are l Problem 3.6
Calculate K for the reaction in Problem 3.4. (At equilibrium. you cannot assume all species
concentrations are 1 M.) At what pH are the equilibrium activities of H804“ and 8042' equal? Purpose of problem I
Test understanding of equilibrium and use of thennodynamic equations Relevant section{s) of text
Section 3.9 ' Solution
From eq. 3.18: K : exp(—AG"M/RT)
From Problem 3.4: AG?“ a +1 1.4 kJ/mol
Thus, at 25“C ("4 298°K):
K : exp[—(11.4 klinlol)/(8.3l4><10"3 kJ/mole—“K)(298°K)] m 1.9X10'2
Also: K = {3042' } {If}! {HSOA’} Thus, {8042'} = {F130;} when {IF} 2K1“ 1.(l><10‘2 or pH = —log{H"} = ~log(1.0><10‘2) : 2.0 Problem 3.7
Calculate K for the reaction in Problem 3.5. (At equilibrium, you cannot assume aii species
concenn‘ations are 0.01 M.) At What range of pH is {NH‘C} > {NFL} at equilibrium? Purpose of problem
Test understanding of equilibrium and use of thermodynamic equations Relevant seetiou{s) of text
Section 3.9 Solution
From eq. 3.18: K = eprAG‘Q/RT)
From Problem 3.5: A61," = +52% kJ/mol
Thus, at 25“C (= 298°K):
K = cum—(52.8 lernol)f(8.3 14X 10‘3‘ kJ/mol.e°K)(298“K)} = 5.6X10‘ "'
Also: .K = {N143} {11"}! {NI14*}
Thus, {NRC} > {NHB} when {H} > K= 5.6><10“”
This occurs when or pH = ~iog{H*} < —log(5.6x 10' '0) or pH < 9.3 (Note: {I—I‘} > K =» log {H*} > iogK a ~log {F} < —iogK, since the inequality sign is reversed
when you multiply both sides of an inequality by a negative number) Problem 3.8
What is the criterion for equilibrium in terms of G? AGm? A0,“? Purpose of problem
Test understanding of equilibrium Relevant section(s) of text
Sections 3.8 and 3.9 Solution
From the text: 6' is minimized at equilibrium
AGm = 0 at equilibrium (since G is minimized)
A61“ = RTanat equilibrium ...
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 Fall '11
 JenniferJay
 Chemistry, Thermodynamics, aquatic chemistry, eq Fe/mol

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