1. Solubility Product & Common-Ion Effect Acids & Bases
Aqueous Equilibria
Solubility Product & Common-Ion Effect
Acids & Bases

2. Dynamic equilibrium occurs in reversible reactions.
For a reaction: aA + bB  cC + dD
The equilibrium-constant expression is
Kc = [C]c [D]d (Products)
[A]a [B]b (Reactants)
Kp = PCc PDd P – partial pressures PAa PBb
Le Chatelier’s principle: the effects of changes in concentration of reactants or products, pressure (volume) or temperature can be predicted.
The equilibrium constant varies only with temperature.
Catalyst increase rates of forward & reverse rxns.

3. Solubility Equilibria and Solubility Product
Solubility equilibrium is any chemical equilibrium between a solid compound and its dissolved states at saturation.
For example:
CaSO4 (s)  Ca2+ (aq) + SO42- (aq)
Using the equilibrium expression as before:
Kc = [Ca2+(aq)][SO42-(aq)]
{CaSO4(s)} but {CaSO4 (s)} = 1

4. Solubility Equilibria and Solubility Product
Therefore, this reduces to the solubility-product expression:
Ksp = [Ca2+][SO42-] = 4.93 × 10−5
This expression means that for an aqueous solution in equilibrium (saturated) with solid CaSO4, the product of the concentrations of the two ions equals the solubility-product constant, Ksp.
Units for Ksp?

5. Solubility Product
Note that solubility and solubility product are not the same.
The solubility of a substance is the quantity of substance that dissolves in a solvent to form a saturated solution, usually in g / L.
The solubility product is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution.

6. Addition of Ca2+ or F- shifts equilibrium
Common-Ion Effect
The solubility of a substance is affected by:
temperature
presence of other solutes.
So, the presence of Ca2+ (aq) or F- (aq) in a solution reduces the solubility of CaF2:
CaF2 (s)  Ca2+ (aq) + 2F- (aq)
Addition of Ca2+ or F- shifts equilibrium

7. Common-Ion Effect
The common-ion effect refers to the fact that solubility equilibria will shift in response to additions of a common ion – Le Chatelier’s principle.
Ksp does not change when additional solutes are present.

8. Examples
1. Solid silver chromate is added to water at 25C. Some of the solid remains undissolved at the bottom of the flask. The equilibrium solution is analysed and the concentration of Ag+ is determined to be 1.3 x 10-4 M. Assuming that Ag2CrO4 dissociates completely in water, and there are no other important equilibria involving the Ag+ or CrO42-, calculate the Ksp for the compound.
2. Calculate the molar solubility of CaF2 at 25C in a solution that is in M Ca(NO3)2. Ksp (CaF2) = 3.9 x

9. Acid-Base Equilibria Arrhenius Acid:
substance that when dissolved in water increases the concentration of H+ ions.
Example: HCl (g)  H+ (aq) + Cl- (aq)
Arrhenius Base:
increases concentration of OH- ions
H2O

10. Acid-Base Equilibria Brønsted-Lowry Acids:
substance (molecule or ion) that can donate a proton to another substance.
ACID: HCl(g) + H2O(l)  H3O+(aq) + Cl- (aq)
While a base accepts a proton:
BASE: HCl(g) + NH3(g)  NH4Cl(s) (NH Cl-)
Hydronium ion

11. Acid-Base Equilibria Lewis Acid:
substance that can accept a pair of electrons.
Lewis bases are electron-pair donors.
base
acid

12. Acid-Base Equilibria
When an acid goes into solution, a conjugate base is formed:
HA (aq) + H2O (l) ↔ A- (aq) + H3O+ (aq)
HA and A- are the
conjugate acid-base pair.
Water will also “autoionize”:
H2O (l) + H2O (l) ↔ H3O+ (aq) + OH- (aq)
Conjugate
base

13. Acid-Base Equilibria OH- H2O H3O+
conjugate base conjugate acid of OH conjugate acid of H2O conjugate base of H3O+ of H2O
Kw = [H3O+] [OH-] = 1.0 x at 25C (Kw – ionic product of water)
[H3O+] = [OH-] = 1.0 x = 1.0 x 10-7 M
at 25C

14. Strength of Acids and Bases
Acid and base strengths:
Strong acids completely transfer their protons to water leaving no undissociated molecules in aqueous solution (completely dissociates)
Weak acids only partially dissociate in water and therefore exist as a mixture of acid molecules and ions in equilibrium.
A strong acid, e.g., HCl, will almost completely dissociate in water to H3O+ and Cl- ions.

15. Strength of Acids and Bases
While, H2O and acetic acid are weak acids:
CH3COOH (l) ↔ CH3COO- (aq) + H3O+ (aq)
Strong and weak bases:
NaOH (s)  Na+ (aq) + OH- (aq)
NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq)
H2O
H2O

16. Water and the pH Scale pH scale: 1 – 14 pH = -log10[H3O+]
Kw = [H3O+][OH-] = 1.0 x at 25C
At 25 C, [H3O+] = 1.0 x 10-7 M
 pH of water at 25 C = -log (1.0 x 10-7 M) = 7.0
pOH = -log10[OH-] and pH + pOH = 14

17. Water and the pH Scale
0.10 M HCl solution (strong acid) contains 0.10 M H3O+ ions:
 pH of 0.10 M HCl = -log (0.10 M) = 1.0
A 0.10 M NaOH solution (strong base) has what pH?
Kw = [H3O+] [OH-]  [H3O+] = Kw / [OH-]
pH of 0.10 M NaOH = -log (1.0 x / 0.10 M)
= 13

18. Acid and Base Dissociation Constants
For weak acids and bases, an equilibrium exists:
HCl (aq) + H2O (l) ↔ Cl- (aq) + H3O+ (aq)
Ka = [H3O+] [Cl-]
[HCl]
NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq)
Kb = [NH4+ [OH-]
[NH3]

19. Acid and Base Dissociation Constants
Ka – acid dissociation constant
Kb – base dissociation constant
Kw – ionic product of water
Ka x Kb = Kw or pKa + pKb = pKw
pKa = -log10Ka pKb = -log10Kb
pKw = -log10Kw

20. Example
Problem: A student prepared a solution of M acetic acid solution and measured its pH to be Calculate the Ka for acetic acid.
CH3COOH (aq) + H2O(l) H3O+ (aq) + CH3COO- (aq)
Ka = [H3O+] [CH3COO-]
[CH3COOH]
pH = 2.88  -log [H3O+] = 2.88
 [H3O+] = = 1.3 x 10-3 M
Ka = (1.3 x 10-3 M) (1.3 x 10-3 M) = 1.7 x 10-5
(1.0 x 10-1 M)