1. Solubility Product Expression
• Silver chloride is so insoluble in water (.0.002
g/L) that a saturated solution contains only
about 1.3 x 10-5 moles of AgCl per liter of
water

2. 1.3 x 10-5 moles of AgCl
• AgCl(s) <-> Ag+(aq) + Cl-(aq)
• Ag+ = 107. amu
• Cl- = 35.45 amu== 143 amu
• 143.34 g/mol

3. AgCl(s) <-> Ag+(aq) + Cl-(aq)

4. • (Water isn't included in the equilibrium
constant expression because it is neither
consumed nor produced in this reaction, even
though it is a vital component of the system.)

5. • The [Ag+] and [Cl-] terms represent the
concentrations of the Ag+ and Cl- ions in moles
per liter when this solution is at equilibrium

6. • The third term [AgCl] is more ambiguous. It
doesn't represent the concentration of AgCl
dissolved in water because we assume that
AgCl dissociates into Ag+ ions and Cl- ions
when it dissolves in water.

7. • It can't represent the amount of solid AgCl in
the system because the equilibrium is not
affected by the amount of excess solid added
to the system.

8. • The [AgCl] term has to be translated quite
literally as the number of moles of AgCl in a
liter of solid AgCl.

9. • The concentration of solid AgCl can be
calculated from its density and the molar mass
of AgCl.

11. • This quantity is a constant, however. The
number of moles per liter in solid AgCl is the
same at the start of the reaction as it is when
the reaction reaches equilibrium

12. • Since the [AgCl] term is a constant, which has
no effect on the equilibrium, it is built into the
equilibrium constant for the reaction.
• [Ag+][Cl-] = Kc x [AgCl]

13. [Ag+][Cl-] = Kc x [AgCl]
• This equation suggests that the product of the
equilibrium concentrations of the Ag+ and Cl-
ions in this solution is equal to a constant.
• [Ag+][Cl-] = Kc x [AgCl]
• [Ag+][Cl-] = Kc x [1]
• [Ag+][Cl-] = Kc

14. • Since this constant is proportional to the
solubility of the salt, it is called the solubility
product equilibrium constant for the reaction,
or Ksp.

15. solubility product equilibrium
constant
• Ksp = [Ag+][Cl-] = 100
• [50 moles][50 moles-]= 100 get a solid
• [25moles][75 moles-]= no solid

16. • Ksp = [Ag+][Cl-]
• When= 1.3 x 10-5 moles of AgCl is reached

17. raised to a power equal
• The Ksp expression for a salt is the product of
the concentrations of the ions, with each
concentration raised to a power equal to the
coefficient of that ion in the balanced
equation for the solubility equilibrium

18. The Relationship Between Ksp And the
Solubility of a Salt
• Ksp is called the solubility product because it is
literally the product of the solubilities of the
ions in moles per liter. The solubility product
of a salt can therefore be calculated from its
solubility, or vice versa.

19. AgBr crystals that do not absorb light
• Photographic films are based on the sensitivity
of AgBr to light. When light hits a crystal of
AgBr, a small fraction of the Ag+ ions are
reduced to silver metal. The rest of the Ag+
ions in these crystals are reduced to silver
metal when the film is developed. AgBr
crystals that do not absorb light are then
removed from the film to "fix" the image

20. solubility of AgBr in water
• Example: Let's calculate the solubility of AgBr
in water in grams per liter, to see whether
AgBr can be removed by simply washing the
film.

21. AgBr(s) <-> Ag+(aq) + Br-(aq)
• When the combined concentrations of the
anion and cation reach this-the precipitate will
form
• Ksp = [Ag+][Br-] = 5.0 x 10-13

22. can't be solved for two unknowns
• One equation can't be solved for two
unknowns the Ag+ and Br- ion concentrations.
We can generate a second equation, however,
by noting that one Ag+ ion is released for
every Br- ion. Because there is no other source
of either ion in this solution, the
concentrations of these ions at equilibrium
must be the same.

23. • [Ag+] = [Br-]
• Substituting this equation into the Ksp
expression gives the following result.
• [Ag+]2 = 5.0 x 10-13
• Taking the square root of both sides of this
equation gives the equilibrium concentrations
of the Ag+ and Br- ions.

24. .
• [Ag+] = [Br-] = 7.1 x 10-7M
• Once we know how many moles of AgBr
dissolve in a liter of water, we can calculate
the solubility in grams per liter.

25. .
• The solubility of AgBr in water is only 0.00013
gram per liter. It therefore isn't practical to try
to wash the unexposed AgBr off photographic
film with water.

26. • Solubility product calculations with 1:1 salts
such as AgBr are relatively easy to perform. In
order to extend such calculations to
compounds with more complex formulas we
need to understand the relationship between
the solubility of a salt and the concentrations
of its ions at equilibrium.

27. Confirmation of Silver
• (Ag+)
• Save 5 ml of this solution immediately-just
for safe keeping

28. I need the unknown number from
Samantha and Tyler
• Save 5 ml of this solution immediately-just
for safe keeping
• This number should match your locker
• Silver , mercury and lead
• Drop as chlorides
• Heat separate the lead

29. • Silver forms a soluble complex ion with
aqueous ammonia. The presence of silver is
confirmed by dissolving any remaining solid
residue in 6 M NH3 (aq) and then re-
precipitating the chloride by freeing the silver
ion from the complex ion using 6 M acid

30. • Add ~5 drops of 6 M NH3 (aq) to the solid from
A, keeping your test tube near the inlet of the
fume exhaust vent. The solid should dissolve,
but if any precipitate remains, centrifuge and
proceed using only the centrifugate

31. • . Add 6 M HNO3 to the solution until the
solution is acidified, using pH indicator paper
to test for acidification. A white precipitate
(AgCl) confirms the presence of Ag+. The Cl-
needed for precipitation will be present from
the prior dissolution of AgCl

32. • Lead (Pb2+)
Lead chloride is almost three times more
soluble in hot water than cold. One may use
this as a basis for separating it from silver
chloride. The presence of lead is then
confirmed by precipitation of yellow lead
chromate.

33. confirmed by precipitation of yellow
lead chromate.