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