Ionic equilibrium part 2

More solute
dissolves
No more solute
dissolves
Unstable crystals
form
Unsaturated Saturated Super-saturated
Ionic Equilibrium (Part 2)
Dr. Damodar Koirala
Amarsingh Model Secondary School
Pokhara Nepal
1

Content
 Part I (Previous Part)
 Degree of ionization
 Dissociation constant
 Arrhenius concept
 Ostwald dilution law
 Acid-base concept
2 Dr. Damodar Koirala | koirala2059@gmail.com
 Part II (This Part)
 Hydrolysis of salt
 Solubility product
 Common ion effect
 Buffer solution

Strong acids and strong bases
Strong Acids Strong Bases
perchloric acid (HClO4) lithium hydroxide (LiOH)
hydrochloric acid (HCl) sodium hydroxide (NaOH)
hydrobromic acid (HBr) potassium hydroxide (KOH)
hydrobromic acid (HBr) potassium hydroxide (KOH)
hydroiodic acid (Hl) calcium hydroxide (Ca(OH)2)
nitric acid (HNO3) strontium hydroxide (Sr(OH)2)
sulphuric acid (H2SO4) barium hydroxide (Ba(OH)2)
 All other are considered weak acids or weak bases

Terminologies
 Solution: homogeneous mixture of solute and solvent
 Unsaturated solution: a solution in which more solute can
be dissolved
 Saturated solution: a solution in which all solute are
dissolved and any addition of solute will not get dissolved.
dissolved and any addition of solute will not get dissolved.
 Supersaturated solution: a solution prepared by dissolving
more solute to the saturated solution while heating it.When
the solution is let to cool down, crystals are produced.

Hydrolysis of salt
 Salt hydrolysis is a reaction in which one of the ions from
salt reacts with water, forming either acidic or basic solution
 So the acidic/basic nature of the aqueous solutions of some
salts can be explained by studying hydrolysis
 There are four different types of salts
 There are four different types of salts
Salt of strong acid + strong base, eg: NaCl
Salt of strong acid + weak base, eg: NH4Cl
Salt of weak acid + strong base, eg: CH3COONa
Salt of weak Acid + weak base, eg: CH3COONH4

1. Salts of strong acid and strong base
 Consider,
 The cations of the strong bases and the anions of the
strong acids do not get hydrolyzed.
HCl + NaOH  NaCl + H2O
S.A S.B Salt
strong acids do not get hydrolyzed.
 Therefore the salts of this category do not show any
acid-base behavior and are neutral.
Na+ + H2O  No hydrolysis
Cl- + H2O  No hydrolysis

2. Salts of strong acid and weak base
 Consider,
 The anion (from strong acid) does not get hydrolyzed but
the cation (from weak base) does get hydrolyzed.
HCl + NH4OH NH4Cl + H2O
S.A W.B Salt
 Since it generates H+ ions, the aqueous solution is acidic
in nature.
NH4
+ + H2O  NH4OH + H+
Cl- + H2O  No hydrolysis

3. Salts of weak acid and strong base
 Consider,
 The anion (from strong acid) does not get hydrolyzed but
CH3COOH + NaOH CH3COONa + H2O
W.A S.B Salt
 Since it generates OH- ions, the aqueous solution is basic
in nature.
Na+ + H2O  No hydrolysis
CH3COO- + H2O  CH3COOH + OH-

4. Salts of weak acid and weak base
 Consider,
 Both, the anion (from weak acid) and the cation (from weak
base) get hydrolyzed.
CH3COOH + NH4OH CH3COONH4 + H2O
W.A S.B Salt
base) get hydrolyzed.
 Both H+ and OH- ions are produced. Solution could be acidic,
basic or neutral depending on the relative strength of the
weak acid and the weak base. Usually considered as neutral.
NH4
+ + H2O  NH4OH + H+
CH3COO- + H2O  CH3COOH + OH-

Summary of hydrolysis
 The aqueous solution of the salt of strong acid and strong base
is neutral.The pH of solution = 7. (cancels out each other)
 The aqueous solution of the salt of strong acid and weak
base is acidic.The pH of solution is less than 7. (strong acid
dominates)
 The aqueous solution of the salt of weak acid and strong
base is basic.The pH of solution is more than 7. (strong
base dominates)
 The aqueous solution of the salt of weak acid and weak base is
unknown. It depends on the strength of acid and base. Usually,
it is considered to be neutral solution (cancels out each other)

Problem based on hydrolysis of salt
1. Why is the aqueous solution of FeCl3 acidic ?
2. Explain the fact that aqueous solution of Na2CO3 is basic
while the aqueous solution of NaCl is neutral ?
3. Predict whether the aqueous solution of CuSO acidic, basic
3. Predict whether the aqueous solution of CuSO4 acidic, basic
or neutral.
4. Predict whether the aqueous solution of CaCl2 acidic, basic
or neutral.

Terminologies
 Sparingly soluble salt : salts that are not completely soluble
in water even at larger extend of dilution
 Ionic product (I.P) of a salt: the product of concentration
of ions of salt in any given solution
 Solubility product (Ksp) of an electrolyte: the product
 Solubility product (Ksp) of an electrolyte: the product
of concentration of ions in its saturated solution
 Solubility of a salt: the greatest possible product of
concentration (molarity) and each concentration terms raised
to a power equal to the corresponding coefficient in balanced
chemical reaction

Solubility product: expression
 Let BA be binary salt that is sparingly soluble in water.Then,
BA(s) BA (solution) B+ + A-
 Applying the law of mass action,
K = [B+] [A-]
K* [BA(s)] = [B+] [A-]
 The concentration of solid salt [BA(s)] is constant. By
convention it is unity. So, [BA(s)] = 1
K = [B+] [A-]
[AB(s)]

 Hence, [B+] [A-] = constant = Ksp
 The constant Ksp is called the solubility product of salt BA.
 The above equation signifies that the product of concentration
of B+ and A- of a salt is constant independent of the individual
of B+ and A- of a salt is constant independent of the individual
concentration of B+ and A- ions when temperature is constant
 If S is the solubility of BA, then [B+] = S and [A-] = S
 Ksp = [B+] [A-] = S * S = S2
 Here, unit of Ksp is (mol/L)2

 Salt of BA type: ( For example AgCl, CaSO4)
 The solubility equilibrium can be represented as
BA(s) A-(aq) + B+(aq) and Ksp = [A-] [B+]
 Let S mol/L be the solubility then
[A-] = [B+] = S
 Substituting the values in the expression of Ksp
 Ksp = S * S = S2
 Here, unit of Ksp is (Mol/L)2

 Salt of BA2 type: ( For example CaF2).
BA2(s) 2A-(aq) + B2+ (aq) and Ksp = [B2+] [A–]2
[B2+] = S and [A–] = 2S
 Ksp = S * (2S)2 = 4 S3

 Salt of B2A type: ( For example Ag2CrO4).
B2A (s) A2-(aq) + 2B+(aq) and Ksp = [B+]2 [A2-]
[B+] = 2S and [A2-] = S
 Ksp = (2S)2 * S = 4 S3

 Salt of B3A2 type: ( For example Ca3(PO4)2).
B3A2 (s) 2A3-(aq) + 3B2+(aq) and Ksp = [B2+]3 [A3–]2
[B2+] = 3S and [A3–] = 2S
 Ksp = (3S)3 * (2S)2 = 108 S5

Salt [A-] [B+] Ksp Ksp Ksp unit
AB S S [A-] [B+] S . S M2
A2B 2S S [A-]2 [B2+] 22. S2. S M3
AB2 S 2S [A2-] [B+]2 22. S. S2 M3
A2B3 2S 3S [A3-]2 [B2+]3 22. 33. S2. S3 M5
A3B2 3S 2S [A2-]3 [B3+]2 33.22.S3.S2 M5
AxBy xS yS [Ay-]x [Bx+]y xx. yy. Sx. Sy M(x+y)

Solubility product principle
 By comparing the solubility product and ionic product of a salt
solution, it is possible to predict whether the precipitation will
occur or not.
 The precipitation of sparingly soluble salt takes place only
when the ionic product exceeds the solubility product.This
when the ionic product exceeds the solubility product.This
principle is called solubility product principle.
Ksp > IP , the solution is unsaturated, ppt does not occur
Ksp = I.P , the solution is saturated, ppt does not occur
Ksp < IP , the solution is super-saturated, ppt occurs

Numerical: Solubility and solubility product
 The solubility of CaF2 in water at 18C is 2.05X10-4 mol/litre.
Calculate its solubility product .
 The solubility of CaCO3 is 0.0305 g/L. Calculate its solubility
product.
 0.00143 gram of AgCl dissolve in one liter of water at 25C to
 0.00143 gram of AgCl dissolve in one liter of water at 25C to
form a saturated solution.What is the solubility product of the
salt ? (Ag = 108, Cl = 35.5)
 The solubility product constant of BaSO4 in water at 25C is
1*10-10 mol2 L-2. Calculate the solubility in g/l

Common ion effect
 The degree of ionization of a weak electrolyte is suppressed
(decreased) by the addition of strong electrolyte having a
common ion is know as common ion effect.
 Eg: when NH4Cl (strong electrolyte) is added to NH4OH the
common ion NH4+ provided by NH4Cl suppress the
ionization of NH OH.
4 4
ionization of NH4OH.
 As a result, the concentration of OH- ion will decrease
otherwise the constant Ksp will change
 Therefore, NH4+ and OH- will combine to give the solid
NH4OH(s) and the solubility of NH4OH will decrease.

Application: Common ion effect
 It is used for the precipitation of group II metals (Cu2+, Hg2+,
Pb2+,Ar3+, Sb3+, Bi3+, Cd3+) in qualitative salt analysis
 Used for the precipitation of group IIA metal ions (Fe2+, Fe3+,
Al3+, Cr3+) in qualitative salt analysis
 Why should we use HCl in the precipitation of group II
 Why should we use HCl in the precipitation of group II
 What is the use of NH4Cl for the precipitate of Group IIIA
metal ion

 A sample of AgCl was treated with 5 ml of 2M Na2CO3
solution to produce Ag2CO3.The remaining solution contained
0.003 gm of Cl- per litre. Calculate the solubility product of
AgCl (Ksp of Ag2CO3 = 8.2 X 10-12)
 At 25 C the solubility of BaSO4 is 0.00233 g/l. Calculate the
solubility product of the salt. Calculate the solubility of BaSO
Numerical: Common ion effect
solubility product of the salt. Calculate the solubility of BaSO4
in a solution of (NH4)2SO4 containing 13.2 g/L at 25C.
 The solubility of AgCl in water at 298K is 1.43X10-3 g/L.
Calculate its solubility in 0.5M KCl solution

Buffer solution
 A solution which retain its pH value when stored for ong
time is known as buffer solution
 Even addition of small amount of strong acid or strong
base, its pH does not alter or change
 There are two type:
 Acidic buffer: it is the mixture of weak acid and a salt of the
same acid with a strong base .The pH of such buffer is less
than 7. Example: the solution of acetic acid with sodium acetate
 Basic buffer: it is the mixture of weak base and a salt of the
same base with a strong acid.The pH of such buffer is greater
than 7. Example: the solution of ammonium hydroxide with
ammonium chloride

Mechanism of acidic buffer
 Consider an acidic buffer solution prepared by mixing acetic
acid (weak acid) and CH3COONa ( salt of weak acetic acid
and a strong base NaOH) i.e. CH3COOH + CH3COONa
 Addition of small amount of HCl (strong acid) :
Extra H+ combine with acetate ion from salt to form
unionized acetic acid. Change in pH is minimized
unionized acetic acid. Change in pH is minimized
 Addition of small amount of NaOH (strong base):
Extra OH- ions react with H+ ion from acetic acid to form
neutral water.The change is pH is minimized
H+ + CH3COO-  CH3COOH
OH- + H+  H2O

Mechanism of basic buffer
 Consider a basic buffer solution prepared by mixing NH4OH
(weak base) and NH4Cl ( salt of weak base and a strong acid
HCl) i.e. NH4Cl + NH4OH
 Addition of small amount of HCl (strong acid) :
Extra H+ combine with OH- ion from NH4OH to form
neutral water. Change in pH is minimized
neutral water. Change in pH is minimized
 Addition of small amount of NaOH (strong base):
Extra OH- ions react with NH4+ ion salt to form weak base.
The change is pH is minimized
OH- + NH4+  NH4OH
H+ + OH-  H2O

Ionic equilibrium part 2

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Ionic equilibrium part 2