Friday october 1
chem 127
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background:
exam on Wednesday (evening)
sunday, 7 PM Overman (meet by the lab; west door access)
web site added information, including a 2003 exam II and review
notes
added remark: the page of problems I distributed last
Friday (copy on the Web with solutions.) These were NOT intended to be typical
of exam problems. Many of the questions were deliberately complicated so we
could focus on the task of dissecting a question, finding what tools we can
bring to bear (equilibrium tools), and how we could attack the problem. On this
coming exam, I’d like the problems to be much more direct.
not a new topic, been using examples since Ch 13
salt ----water---100% ---> ions
PbCl2 --> Pb2+ + 2Cl-
AgCl ---> Ag+ + Cl-
CuS ---> Cu2+ + S=
Ksp is the equilibrium constant for solubility process
Ksp = [Pb2+ ] [Cl-]2
Ksp = [Ag+] [Cl-]
Ksp= [Cu2+ ] [ S=]
(Technically this is valid for all ionic compounds, but results are
only good for salts with relatively low solubility.)
The reverse is basically a discussion of precipitation
No new questions, equations, methods
same core equilibrium problems
But some examples appear frequently.
Throw a salt in water, how much will dissolve ?
Ag2SO4 ----> 2Ag+ + SO42-
Classic Equilibrium, but easier than most
initial concentrations -- zero
change -- s and 2s (both same sign)
s = # moles that dissolved (per liter)
reactant (solid salt)-- doesn't enter
calculation
equilibrium values are 2s for Ag+ and s for SO42--
2s = [Ag+]
s= [SO42-]
Ksp = (2s)2 (s) or Ksp = 4 s3
s = { Ksp/4} = conc of salt in solution.
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Common Ion Version--
how much ____ will dissolve in _____
Molar Na ______?
Common Ion Version-- how much Ca(OH)2 will dissolve in 0.50 Molar
NaOH? (Ksp= 4.7 x 10-6)
Ca(OH)2 ---> Ca2+ + 2 OH-
s= # moles that dissolve
Initial [Ca2+]=0 so final [Ca2+] = s
Initial [OH- ] = 0.50 so final [OH-] = 0.50+2s
Ksp = [Ca2+ ][OH-]2
4.7 x 10-6= s * (0.50 +2s)2 ... solve for s
0 = -4.7 x 10-6 +0.25 s + 2 s2 + 4s3 ...cubic eq (ugh)
could we cheat ? approximate
small Ksp so solubility in water is small
common ion, so s will c quite small here
let's approximate 0.50 + x as 0.50
now
4.7 x 10-6= s * (0.50 +2s)2 = 0.25 s
s = 4.7 x 10-6/ 0.25 = 1.87 x 10-5
a great approximation:
0.50 and 0.50+2*1.87 x 10-5 are nearly the same?
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The other major solubility problem is really the reverse
will a precipitate form if we mix cation and anion ?
We start with a little fiction...
we mix some solutions, but turn off equilibrium process
we find the initial concentrations
then we turn equilibrium back on
do concentrations in Q exceed Ksp (if so, ppt)
then, if requested, we solve for final equil concentrations
we add 1 ml of 5.0 M NaOH to 1.0 liters of 0.0020 M Mg2+
will we get Mg(OH)2 to precipitate?
minor cheating... let's assume volume remains 1.0 liters
initial [Mg2+] = 0.0020
initial [OH- ]=(0.001 L * 5.0 mol/L) / 1.001 L= 0.0050 M
Q = [Mg2+ [OH-]2 = (2x10-3) (5 x 10-3)2 = 50 x 10-9
Ksp (tables) = 5.6 x 10-12
yes, we get precipitate to form
in fact, Q>>Ksp & we need massive reduction in Q.
if limiting reagent idea is used Mg2+ could be totally removed and OH-
would still be 0.0010M, so we should expect most of the Mg2+ to be
consumed.
if x is the extent of precipitation (conc change)
[Mg2+] initially 0.0020 and finally 0.0020-x
[OH-] initially 0.0050 and finally 0.0050 - 2x
Ksp =[Mg2+] [OH-]2
5.6 x 10-12 = (0.0020-x ) (0.0050 - 2x)2
another ugly cubic equation , but mathematically, one step from a solution
this actually is easy to approximate
since almost all Mg2+ is consumed,
the remaining [OH-] = 0.0050 – 2*0.002 = 0.0010
of 5.6 x 10-12 = [Mg+2] * (0.0010)2
[Mg2+] at equil = 5.6 x 10-6 (tiny enough to validate our approximations)
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The last common solubility problem is associated with another simultaneous equilibrium...
example... what is the solubility of Silver Acetate if
we manage to keep the pH at 3.00 ? (a pH buffer)
(similar examples-- will CdS ppt if we supply H2S at pH 5.0?)
calculation is more complex (very poor exam question)
I hope you can follow, and see that this is valid. I don’t expect that you will figure out how to do this on your own or that your would now be able to solve a similar problem if it showed up.
1) Ksp = [Ag+] [Ac-]
2) Ka = [H+] [Ac-] / [HAc]
if H+ conc is large, [Ac-] is kept low
3) stoichiometry links these ... [Ag+] = [Ac-] + [HAc]
If x = final Ag+ concentration, eq 1 becomes
[Ac-] = Ksp /x
and eq 3 becomes
x = (Ksp/x) + [H+] (Ksp/x) /Ka
x2 = Ksp + [H+]/ Ka right side = known numbers
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Closely related
why AgCl (in photographic film) dissolves in "fixer"
sodium thiosulfate, even though Ksp is tiny.
reaction with thiosulfate keeps [Ag+]free very tiny
can even dissolve in conc NaCl, forming AgCl2- ion
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Review ... types of exam questions and type of problems