precipitation equilibria. solubility product ionic compounds that we have learned are insoluble in...
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Precipitation Equilibria
Solubility Product Ionic compounds that we have learned are
insoluble in water actually do dissolve a tiny amount.
We can quantify the solubility using the equilibrium expression or solubility product.
Solubility Product Example: What is the solubility of silver
chloride in pure water? Ksp = 1.8x10-10
Write the equilibrium expression:
[Ag+][Cl-] = Ksp
=1.8x10-10
why are we ignoring the AgCl??
-(aq)(aq)(s) ClAgAgCl
Solubility Product Use (ICE)
Ag+ Cl-
I 0 0
C +x +x
E x x
-(aq)(aq)(s) ClAgAgCl
Solubility Product Plug the equilibrium (E) values into the
equilibrium expression:[x][x] = x2= 1.8x10-10
x = 1.3x10-5
[Ag+] = 1.3x10-5M, and [Cl-] = 1.3x10-5M
Solubility Product Example: What is the solubility of lead
iodide in pure water? Ksp = 7.1x10-9
Write the equilibrium expression:
[Pb2+][I-]2 = 7.1x10-9
-(aq)
2(aq)2(s) 2IPbPbI
Solubility Product Use (ICE)
Pb2+ I-
I 0 0
C +x +2x
E x 2x
-(aq)
2(aq)2(s) 2IPbPbI
Solubility Product Plug the equilibrium (E) values into the
equilibrium expression:[x][2x]2 =2x3= 7.1x10-9
x = 1.2x10-3
[Pb2+] = 1.2x10-3M, then [I-] = ??
Solubility Therefore, the solubility of PbI2 is:
1.2x10-3 mol/L
That is, 1.2x10-3 moles of PbI2 will dissolve in 1L of water.
Or, multiply by the MW of PbI2 to find that:(1.2x10-3mol/L)(461.0g/mol) = 0.55g/L
The Common Ion Effect What if there is already an ion dissolved in
the water that is common with the ionic compound?
For example: What is the solubility of silver chloride in a solution that contains 2.0x10-
3M Cl-?
-(aq)(aq)(s) ClAgAgCl
The Common Ion Effect Write the equilibrium expression:
[Ag+][Cl-] = 1.8x10-10
Use the ‘ICE’ method:
Ag+ Cl-
I 0 2.0x10-3
C +x +x
E x 2.0x10-3 +x
The Common Ion Effect Plug the equilibrium (E) values into the
equilibrium expression:[x][2.0x10-3 +x] = 1.8x10-10
Solve:x2 + 2.0x10-3x – 1.8x10-10 = 0
x = 9.0x10-8
Solubility of AgCl is 9.0x10-8 mol/L vs. 1.3x10-5 mol/L when no common ion was present!
Another ExampleAdd 10.0mL of 0.20M AgNO3 to 10.0mL of
0.10M NaCl. How much Cl- will remain in solution?
First, this is a limiting reagent problem:
-(aq)(aq)(s) ClAgAgCl
Since we are combining two solutions, find moles:
Since we know that AgCl is insoluble, the amount of Ag+ remaining in solution is:
2.0x10-3 – 1.0x10-3 = 1.0x10-3moles[Ag+] = 0.050M
-3--
3-
Cl molesx100.1Cl
Ag moles2.0x101000mL
(10mL) moles20.0Ag
To determine [Cl-], simply use ICE and the equilibrium expression:
[0.050+x][x] = 1.8x10-10
ignore x in the Ag+ termx = 3.6x10-9
[Cl-] = 3.6x10-9MWhat is [Ag+] at equilibrium?
Ag+ Cl-
I 0.050 0
C +x +x
E 0.050+x x
So, how do we tell when a ppt will form?
We use, P (analagous to Q)If P > Ksp, ppt will formIf P< Ksp, no ppt will formIf P=Ksp, solution is saturated but no ppt yet
Solubility rules we used earlier work onlyWhen the concentration is 0.1 mol or greater
Dissolving ppts
Many methods are used to make water-insoluble ionic solids ionize
Most commonlyH+ is used to react with basic anions
a strong acid, often HCl, is usedworks on virtually all carbonatesmany sulfides
NH3 or OH- is used to react with metal cationsUse K = Ksp x Kf (2 steps)
Qualitative AnalysisObjective is to
separate and identify cations present in an “unknown” solution
Use ppt reactions to divide the ions into 4 groups
Then bring ions into solution, separate and identify
Groups for Qualitative AnalysisGroup I
Cations that form insoluble chlorides: Ag, Pb, Hg2
Group IICations that form
insoluble sulfidesH2S (toxic and stinky)
at pH 5 used: Cu, Bi, Hg, Cd, Sn, Sb
Group IIICations that from more
soluble sulfidesDon’t ppt at ph 5 but do
at pH 9: Al, Cr, Co, Fe, Mn, Ni, Zn
Group IVSoluble chlorides and
sulfideAlkaline earth (Mg, Ca,
Ba) ppt as carbonatesAlkali metal (Man, K) can
be identified with flame tests