help ! i’m dissolving. using complex algebra to calculate concentrations when multiple equilibria...
TRANSCRIPT
Help !I’m
dissolving
Using complex algebra to calculateconcentrations when multiple equilibria are in play
We can look for two more pieces of info:Charge Balance: Electroneutrality of the
solution; the sum of the positive charge in solution equals the sum of negative charges in the solution.
Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.
We can look for two more pieces of info:Charge Balance: Electroneutrality of the solution;
the sum of the positive charge in solution equals the sum of negative charges in the solution.
Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.
11
[H+] + 2[Ca2+] = [OH-] + [F-] (4)
2[Ca2+] = [F-] + [HF] (5)
How does the solubility of CaF2 depend on pH?
12
CaF2(s) Ca2+ + 2F- Ksp = [Ca2+][F-]2 = 3.910-11 (1)
F-+H2O HF + OH- 11105.1
][F
][HF][OHKb
(2)
H2O H+ + OH- Kw = [H+][OH-] = 1.010-14 (3)
Five unknowns: [Ca2+], [F-], [HF], [H+], and [OH-]Three equations mean we need more equations…
We can look for two more pieces of info:Charge Balance: Electroneutrality of the solution;
the sum of the positive charge in solution equals the sum of negative charges in the solution.
Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.
[H+] + 2[Ca2+] = [OH-] + [F-] (4)
2[Ca2+] = [F-] + [HF] (5)
Ksp = [Ca2+][F-]2 = 3.910-11 (1)
11105.1
][F
][HF][OHKb
(2)
Kw = [H+][OH-] = 1.010-14 (3)
(4)
(5)
We need to substitute to get things in terms of Ca2+ or H+
[H+] + 2[Ca2+] = [OH-] + [F-]
2[Ca2+] = [F-] + [HF]
15
11105.1
][F
][HF][OHKb
Ksp = [Ca2+][F-]2 = 3.910-11
Kw = [H+][OH-] = 1.010-14
2[Ca2+] = [F-] + [HF]
Combine eq. 2 and eq. 5, we have…
][2][
][][ 2
Ca
OH
FKF b
][1
][2][
2
OHK
CaF
b
(B)
Combine (B) and eq. 1, we have…
][1
][2][
2
OHK
CaF
b
(B)
2
22
][1
][2][
OHK
CaCaK
bsp
31
2
2
]1
4][
[OH
KKCa bsp
(C)
It’s much simpler if we can consider the pH fixed (how would we do that?)
If we can fix pH…
pH = 3.00 [H+] = 1.0×10-3 M
Kw
[HF-] =1.5[F-]
Kb
[F-] =0.80[Ca2+]Mass
[Ca2+] = 3.9×10-4 M Ksp
[OH-] = 1.0×10-11 M
18
The [Ca] removed from marble stone (largely dissolution of CaCO3) increases as the [H+] of acid rain increases.
Applications of coupled equilibria in the modeling of environmental problems
CaCO3(s) + 2H+(aq)
Ca2+(aq) + CO2(g) + H2O(l)
SO2(g) + H2O(l) H2SO3(aq)
oxidation H2SO4(aq)
www.chem.wits.ac.za/chem212-213-280
Deposits include CaSO4•2H2O (gypsum), which accumulates creating a black residue.
http://pubs.usgs.gov/gip/acidrain/5.html
19
Total [Al] as a function of pH in 1000 Norwegian lakes.
Acid rain also releases Al, Hg, and Pb into the environment.
www.chem.wits.ac.za/chem212-213-280