As mentioned in class briefly, the Archean Ocean was relatively iron-rich, much
ID: 997783 • Letter: A
Question
As mentioned in class briefly, the Archean Ocean was relatively iron-rich, much more than today. Let's assume for the moment that the amount of Fe2^+ in the Archean Ocean was controlled by one of two minerals: siderite (FeCO_3) or mackinawite (FeS). Minteq will not be much help here...go pencil/paper. a. Given conditions listed below, which of these two minerals would have controlled the Fe^2+ concentration in the Archean Ocean if the system were at equilibrium? Ignore activity corrections, and assume 25 degree C. b. What would have been the equilibrium [Fe2^+] under those conditions? Ignore activity corrections, and assume 25 degree C. c. What would have been the concentration of total Fe ([Fe]_T)?Explanation / Answer
first we need see the eqution and balancing
FeCO3 we have FeCO3(s) -> Fe+2 + CO3-2 Ka = [CO3][Fe+2]
FeS we have FeS(s) -> Fe+2 + S-2 Kb = [ S-2][Fe+2]
now need see which species are present in solution and which is more favored in the equilibrium
for CO3-3 we have
CO2g <--> CO2l K1 H2CO3 <--> HCO3- + H+ K2 HCO3- <--> CO3-2 + H+ K3 now we rewritte all species of carbonates in function of CO3-2 to see their dependence on pH
K3 = [CO3-2][H+]/[HCO3- ] =) [CO3-2] = K3[HCO3- ]/[H+]
K2 = [HCO3- ][H+]/[H2CO3 ] as [H2CO3 ] = [HCO3- ][H+]/K2
Ct = [CO3-2] + [H2CO3 ] + [HCO3- ] =) and as
[H2CO3 ] = [HCO3- ][H+]/K2 and [CO3-2] = K3[HCO3- ]/[H+]
Ct = K3[HCO3- ]/[H+] + [HCO3- ][H+]/K2 + [HCO3- ]
in my initial conditions i have alcalinity [HCO3- ] = 4.5mM and Ct = 12.5mM K2 = 5x10-7 K3 = 5x10-11 [H+] = 1x10-3mM
Ct =5x10-11[4.5mM- ]/[1x10-3mM ] + [4.5mM- ][1x10-3mM ]/5x10-7 + [4.5mM ] =
Ct = 9x103mM = 9M as Ct = 12.5mM in our initial condition the different is as FeCO3(s)
for S-2
HS <--> H+ + S-2 Ka = 1.2x10-13
H2S <---> H+ + HS Kb = 1x10-7
H2S(g) <---> H2S Kc = 0.1
Ka = [S-2][H+]/[HS] =) [HS ] = [S-2][H+]/Ka
Kb = [ HS][H+]/[H2S] as [HS ] = [S-2][H+]/Ka replacing Kb = [S-2][H+]/Ka [H+]/[H2S] rewritting
Kb = [S-2][H+][H+] / [H2S]Ka =) [H2S] = [S-2][H+]2 / KbKa
H2S(g) =[H2S]/Kc as [H2S] = [S-2][H+]2 / KbKa replacing H2S(g) =[S-2][H+]2 / KbKaKc
St = H2S(g) + [H2S] +[HS ] + [S-2] =) St = [S-2][H+]2 / KbKaKc +[S-2][H+]2 / KbKa +[S-2][H+]/Ka + [S-2] =)
St = [S-2] ([H+]2 / KbKaKc +[H+]2 / KbKa +[H+]/Ka + 1) =
St = [S-2] (1x10-3mM 2 / 1x10-7.2x10-13 0.1 +1x10-3mM 2 / 1x10-71.2x10-13 +1x10-3mM /1.2x10-13 + 1) =
St = [S-2] 8.3x1016 whit a [S-2] 1x10-5mM St = St =1x10-5mM 8.3x1016 =) St = 8.3x108M
a) the FeS will be the predominant cristal form
b)
Fe(OH)3 + 3H <-> Fe+2 + 3H2O K1 = 1x1031 K1 = [Fe+2]/[H+]3 =) [Fe+2] = K1[H+]3
Fe(OH)2 + 2H <-> Fe+2 + 2H2O K2 = 4x1020 K2 = [Fe+2]/[H+]2 =) [Fe+2] = K2[H+]2
Fe(OH)+ H <-> Fe+2 + H2O K3 = 3.x109 K3 = [Fe+2]/[H+] =) [Fe+2] = K3[H+]
FeCO3(s) -> Fe+2 + CO3-2 Ka = [CO3][Fe+2] =) [Fe+2] = Ka/[CO3]
FeS(s) -> Fe+2 + S-2 Kb = [ S-2][Fe+2] =) [Fe+2] = Kb/[ S-2]
c) [Fe+2] = K1[H+]3 + K2[H+]2 + K3[H+] + Ka/[CO3] + Kb/[ S-2]
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