The equilibrium product constant
 A salt is considered soluble if it dissolves in water to
give a solution with a concentration of at least 0.1 M at
room temperature.
 A salt is considered insoluble if the concentration of an
aqueous solution is less than 0.0001 M at room
temperature.
 Salts with solubilities between 0.0001 M and 0.1 M are
considered to be slightly soluble.
 Can a particular
we want some things to
be in solution and some
things to settle… chlorine
is nice to have out of
water, but we don’t want
lead to settle.
compound be added to a
lake to kill organisms
/plants /fish, yet not ppt
to the bottom?
#5 will be homework
 1. Write the dissociation of
each salt, so that you know
want the initial amount is.
 2. Calculate the
concentration of the PbCl2
 3. Look at the Ksp chart
and see if the
concentration is stronger
(the power of 10 is > than
the Ksp – which means less
negative.)
 1. Pb(NO3)2  Pb+2 + 2NO3FeCl2  Fe+2 + 2Cl 2. Ksp of the lead chloride =
[Pb+2] [2Cl-]2 = [.01] [2 x .06]2
*you need to have .06 since the original
dissociation was 2 x .03M…
This calculates out to [.01] [.0144] =
= .00014 = 1.44 x 10-4
The Ksp = 1.7 x 10-5,
so there will be a ppt since the
answer is stronger
There will be answers to it tomorrow in class.
 An experiment calls for 22.5g of Lead (II) Chloride
dissolved in 500 mL of distilled water. Is this
possible?
 Calculate the Ksp at room temperature and compare to
the actual.
 What could you do to dissolve this salt if 22.5 g saturates
the solution?
 The formula for lead(II) chloride is : PbCl2
 The formula mass is
207 + 35.5 + 35.5 = 278g
 22.5g x 1 mole = .0809 moles
278g
 The concentration or molarity is:
0809mo. = .162M
.500L
Ksp comes from the dissociation equation which is…
PbCl2  Pb+2 + 2Cl-1
Ksp = [.162]1 [ 2 x .162]2 = (.162) (.105)
= .0170 = 1.70 x 10 -2
Why does Hoffmann
Always make one
More step
Of work?
 There is one final step that we need to recognize and
understand….regarding the ions that have a coefficient greater that 1 in the
balanced equation.
 Similar to Keq, we need to square or cube the concentration; but there is
one more step. (What’s new??)
 We must recognize that the coefficient NOT ONLY is acknowledged in the
superscript, but is also recognized as an amount of twice or three times the
number of ions as follows:
The concentration of each of the three
Parts is .162
So the Keq would actually be:
(.162) (2 x .162)2 =
Pb+2
.0170
-1
Cl
This Is a much greater
-1
Cl
Value than the .00425
Try this qualitative problem:
 If sodium bromide were added to the
beaker, what would settle out?
 If this(these) ppt. were separated from the
solution of the other ions and at this point,
and OH- were added to the solution, what
would settle?
 After separation of this(these) ppt.
potassium carbonate is added to the
solution. What settles?
CuBr
Mg(OH)2
Fe(OH)2
BaCO3
A paragraph fill in the blank using vocabulary terms
correctly.
2. A qualitative precipitate in a f low chart problem.
3. A quantitative Ksp problem similar to yesterday’s work
with given concentrations like #5 and #6.
4. A quantitative Ksp problem that you need to calculate if
given concentrations
1.
https://www.youtube.com/watch?v=u1oPVBPU0wk
https://www.youtube.com/watch?v=u1oPVBPU0wk
(go to 6 min on the second one)
 http://www.scienceiscool.org/ksp/tutorials.html
 http://chemed.chem.purdue.edu/genchem/topicrevie
w/bp/ch18/ksp.php
 http://www.chemteam.info/Equilibrium/Calc-Ksp-
FromMolSolub.html