This course of analytical chemistry is distinated to students of chemistry, Pharmacy and biology. It is concerning the equilibrium of solubility and precipitation of salts. The constants of this equilibrium and effecting factors are given with some examples demonstrations

1. Chapter III, Solubility and
precipitation

2. 2
Solubility Equilibria
Solubility: • When something is considered soluble, that means that the attraction between the polar water molecules and
ions is stronger than the attraction between the two ions. If this is the case, a precipitate will not form and the ions will
remain in solution.
• When something is considered insoluble, that means that the attraction between the two ions is stronger than the attraction
between the polar water molecules and the two ions separately. If this is the case, a precipitate will form.
• Solubility depends on the nature of the solute and nature of Solvents
Solubility of solute depends on the temperature, particle size and the nature of the solvent.
Some salts or hydroxides are poorly soluble in water: AgCl, PbCl2, BaSO4, Al(OH)3, Ca(OH)2,
Ag(OH)…….. The ionic species in solution will be in the solid phase, we say that the solution is
saturated (two phases: liquid-solid).

3. 3
• Solubility product (Ksp) – equilibrium constant; has only one value for
a given solid at a given temperature.
• Solubility – an equilibrium position.
Bi2S3(s) 2Bi3+(aq) + 3S2–(aq)
2 3
3+ 2
sp = Bi S 
   
   
K
Precipitation
KSP AND SOLUBILITY
Bismuth (III) sulfide Bi2S3
Generally, for a saturated solution of a
slightly soluble ionic compound with
formula MpXq: Qsp = [Mn+]p[Xz-]q = Ksp

4. WRITING ION-PRODUCT EXPRESSIONS FOR SLIGHTLY SOLUBLE
IONIC COMPOUNDS
SOLUTION:
PROBLEM
:
Write the ion-product expression for each of the following:
(a) magnesium carbonate (b) iron(II) hydroxide
(c) calcium phosphate (d) silver sulfide
PLAN: Write an equation which describes a saturated solution. Take note of the unusual behavior of the
sulfide ion produced in (d).
Ksp = [Mg2+][CO3
2-]
(a) MgCO3(s) Mg2+(aq) + CO3
2-(aq)
Ksp = [Fe2+][OH-]2
(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)
Ksp = [Ca2+]3[PO4
3-]2
(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO4
3-(aq)
(d) Ag2S(s) 2Ag+(aq) + S2-(aq)
S2-(aq) + H2O(l) HS-(aq) + OH-(aq)
Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-
(aq)
Ksp = [Ag+]2[HS-][OH-]

6. 3.
The solubility of a substance is the maximum amount of the substance that can dissolve in a given amount of solvent
at a given temperature SATURATED.
 Once a solution has been saturated with a substance, the addition of more of the substance will simply cause this
extra material to accumulate in undissolved form.
SATURATION exists when:
• Some undissolved material is present, and
• Equilibrium exists between the dissolved ions and the undissolved solid.
SOLUBILITY = equilibrium concentration of the substance in solution at a given temperature or the concentration
of a saturated solution. The solubility is often expressed as MOLAR SOLUBILITY where the units are M
A solution is shown to be saturated by writing an equation showing the substance in equilibrium with its aqueous ions.
Ag2SO4(s) 2Ag+ + SO4
2–
(aq)
a) Solid Ag2SO4 dissociating into ions (DISSOLVING REACTION).
Ag2SO4(s) 2Ag+
(aq) + SO4
2–
(aq)
b) Ag+
and SO4
2–
ions come together and form Ag2SO4 (CRYSTALLIZATION REACTION).
2Ag+
(aq) + SO4
2–
(aq) Ag2SO4(s)
 Initially, there are few ions in solution and the dissolving reaction predominates ,The crystallization reaction speeds up as ion
concentration increases. Eventually, the rate of the dissolving reaction equals the rate of the crystallization reaction and
equilibrium is reached.

7. • The solubility is low so most will not dissolve
• What does dissolve will dissociate into ions
PbCl2 (s) Pb2+(aq) +2Cl-
(aq)
• The solid is in equilibrium with the dissolved ions
• The equilibrium expression is:
Ksp =[Pb2+][Cl-
]2
• Ksp is an approximation used to estimate solubility
• Measures the extent to which a substance will dissolve in water
• Larger Ksp =higher solubility

9. Calculating Ksp from Solubility
The molar solubility of CaF2 at 35°C is 1.24x10-3 M.
(a) What is the solubility of CaF2 in g/L? A: 0.0968g/L
(b) What is Ksp at this temperature? A: 7.63x10-
9
Initial Solid 0 0
Change Solid +x +2x
Equilibrium +x +2x
Ksp = [Ca++][F-]2
s = 1,24.10-3 M = n/V = m/MV Molar mass of CaF2 = 78 g/Mol
m = s.Mv or v = 1l
m = 1,24.10-3.78 .1l = 0,0968g/l
X =s = 1,24.10-3 M that means [Ca++] = s = 1,24.10-3 M
and
[F-] = 2.2S = 2.1,24.10-3 M
Ksp = [Ca++][F-]2 = s.(2s)2 = 4S3
= 4.1,24.10-3 = 7,62.10-9 M
Pks = 8,11

10. Comparing Molar Solubilities vs. Ksp
Compound
BaSO4
Mg3(AsO4)2
Ksp
1.1.10-10
2.10-20
Molar Solubility
1.0x10-5M
5.0x10-5M
Molar Solubility Comparison:
Mg3(AsO4)2 ( ma g n e s i u m a r s e n a t e ) molar solubility is 5 times greater than
BaSO4(Barium sulfate)
Ksp Comparison
BaSO4 has a Ksp that is 109 times greater than Mg3(AsO4)2
Be careful using Ksp directly to compare solubilities
• Number of ions present also matters
• Can only directly compare Ksp if # ions produced is identical

11. Precipitation
1. Reaction between calcium nitrate and carbonic acid: (species
exist largely undissociated in solution)
Molecular form:
Complete ionic:
Net Ionic:
As seen above, CaCO3 is the precipitate.

12. As seen above, FePO4 is the precipitate.
Complete ionic:
Molecular form
2. Reaction between iron(III) chloride and phosphoric acid:
Net Ionic:
3. Reaction between potassium nitrate and sodium iodide:
Molecular form:
Complete ionic:
Net Ionic:
No precipitate forms and all species remain in aqueous solution due to all species being soluble
None because no precipitate forms and nothing can be removed as spectators.

13. Molecular form:
4 Reaction between sodium hydroxide and copper(II) sulfate:
Complete ionic:
Net Ionic:
As seen above, Cu(OH)2 is the precipitate

14. Ksp AND THE REACTION QUOTIENT Q (CONDITIONS OF PRECIPITATION)
DETERMINING WHETHER PRECIPITATION OCCURS
To know if there is precipitation, we calculate the ionic product assuming the total reaction and then compare it to
the KS ƒ
what concentrations of ions are required to begin the precipitation of an insoluble salt.
1. Q < Ksp, the system is not at equil. (unsaturated)
2. Q = Ksp, the system is at equil. (saturated) 3.
3. Q > Ksp, the system is not at equil. (supersaturated) Precipitates form when the solution is supersaturated!!!
Problem 1 Predicting whether a precipitate will form
A common laboratory method for preparing a precipitate is to mix solutions of the component ions.
Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

15. PLAN: Write out a reaction equation to see which salt could form. Look up the Ksp values in a
table. Treat this as a reaction quotient, Q, problem and calculate whether the
concentrations of ions are > or < Ksp. Remember to consider the final diluted solution
when calculating concentrations.
SOLUTION:
CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2 x 10-11
mol Ca2+ = 0.100 L(0.30 mol/L) = 0.030 mol [Ca2+] = 0.030 mol/0.300 L = 0.10 M
mol F- = 0.200 L(0.060 mol/L) = 0.012 mol
[F-] = 0.012 mol/0.300 L = 0.040 M
Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6 x 10-4
Q is >> Ksp, thus CaF2 will precipitate.

16. 1. Determining Precipitation Conditions A solution is prepared by adding 750.0 mL of 4.00 × 10−3 M
Ce(NO3)3 to 300.0 mL of 2.00 × 10−2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 × 10−10) precipitate from this
solution?
2. A solution is prepared by mixing 150.0 mL of 1.00 × 10-2 M Mg(NO3)2 and 250.0 mL of 1.00 × 10−1 M
NaF. Calculate the concentrations of Mg2+ and F− at equilibrium with solid MgF2 (Ksp = 6.4 × 10−9 )
PROBLEM 2

17. 42
FACTORS AFFECTING SOLUBILITY:
Common-Ion Effect
• One of the ions in the compound is also part of another compound present in the
solution
pH
• Presence of hydroxide (OH-) or hydronium ions (H3O+)
Complexation
• Formation of coordinate bonds with solvent or other molecules present in solution

18. 43
COMMON-ION EFFECT
The extent of ionization of a weak electrolyte is decreased by the addition of a strong electrolyte
that has an ion in common with the weak electrolyte.
Equilibrium process – presence of ions shifts process back to reactants.
Ca(OH)2(s) Ca2+(aq) +2OH-
(aq)
Add extra Ca2+ from CaCl2
Shift equilibrium left
Will affect pH if solution is acidic or basic
-in this case shift toward products decreases
concentration of OH- in solution.
Ex: A solution is made with 1.0M CaCl2 and 2.0M Ca(OH)2

19. THE EFFECT OFA COMMON ION ON SOLUBILITY
PbCrO4(s) Pb2+(aq) + CrO4
2-(aq) PbCrO4(s) Pb2+(aq) + CrO4
2-(aq)
CrO4
2- added

20. Example of practice
1. Calculate the pH of a solution containing 0.085M nitrous acid (HNO2; Ka =4.5x10-
4) and 0.10M potassium
nitrite (KNO2). Answer pH =3,42
2. The Ksp of Mn(OH)2 is 1.6x10-13. Calculate the molar solubility of Mn(OH)2 in:
A. water Answer : 3.4x10-5M (use the Ksp expression)
B. A solution that contains 0.020M NaOH A: 4.0x10-10 M ( use the Le Châteliers principale ( common-ion effect:
Write the RICE table and put the concentrations at equilibrium
A. Compare the solubility of Mn(OH)2 in these solutions
A: 85,000 times more soluble in water ( use the ration between the two solubilities)
3. How much is the solubility of lead (II) chloride changed in the presence of 0.85M NaCl? Ksp =1.6x10-5
Answer : More than 700 times less soluble than in water : use the same methode in question B

21. pH effect on Solubility
• The solubility of an ionic solute may be greatly affected by pH if an acid-base reaction also occurs as the solute dissolves.
• In other words, some salts will not dissolve well in pure water, but will dissolve in an acid or a base.
• If the anion (A-) of the salt/precipitate is that of a weak acid, the salt/precipitate will dissolve more when in a strong acid (H+
ions will form HA with A-)
• However, if the anion of the precipitate is that of a strong acid, adding a strong acid will have no effect on the precipitate
dissolving more.
Example How would the addition of HCl affect the solubility of PbCl2?
Cl- is the conjugate base of a strong acid, thus it is a weak base.
It will not react with H+ ions, so there is no effect.
How would the addition of HCl affect the solubility of FeS?
S2- is a strong base (conjugate base of weak acid)
Thus, it will react with H+ ions to form H2S,
This will shift the equilibrium to make more FeS dissolve!