1. Ionic equilibrium Lec.9

2. Introduction Ionization: a process in which unionized molecules are changed into ions. Types of electrolytes: Strong electrolytesWeak electrolytes complete ionizationpartial ionization All Salts, strong acids & strong basesWeak acids and weak bases HCl H + +Cl - NaOH Na + +OH - Na Cl Na + + Cl - CH 3 COOH CH 3 COO - + H + NH4OH NH 4 + + OH - No equilibrium take places,so no equilibrium constant The reaction stop when reached equilibrium(IONIC Equilibrium ), so it has equilibrium constant (K)

3. Where: K a = acid ionization constant K b = base ionization constant Ostwald Law(1888): Described the relation between the extent of ionization (α) and concentration (C) mole/ liter for solutions. Weak electrolyte dissociates in water to ions. Where : [A - ], [B + ] : concentrations of the ions produced [AB] :the undissociated form at equilibrium state. K : ionization or dissociation constant Suppose V = volume Concentration = number of moles/ volume (L) No.of moles

4. In case of very weak electrolyte α = very small value and can be neglected i.e. (1- α) = 1 Where : C =1/V Very weak electrolyte weak electrolyte Degree of Ionization

5. Determination of (α) Experimentally: Determine the value of( v ) : equivalent conductance with different weak electrolyte concentrations and calculated K in each case. K was always constant and independent of the dilution only in case of weak electrolytes. Determine the value of( o )= equivalent conductance at infinite dilution. If λ v ↑ α ↑ ---------- λ v α (α) α= v / o

6. Example (1):

7. The acid can be considered very weak, [α ‹‹‹‹ 1] Concentration of hydrogen ion:

8. The Ion Product of Water Water is very slightly ionized ( Very weak electrolyte ) H 2 O is very weak electrolyte [H 2 O] = constant value [H 2 O]*K=constant value [H 2 O]*K=KW= ion product of water [H + ] * [OH - ]= KW= 10 -14 at constant temperature

9. For example: at samples of H 2 O: 1- When [H + ] = 10 -5 > 10 -7 (acidic solution) 2-When [H + ] = 10 -9 < 10 -7 ( alkaline solution) 4- When [OH - ] =10 -11 <10 -7 (acidic solution)

10. The Hydrogen Ion Exponent:( pH) pH is defined as the negative exponent of 10 which gives the hydrogen ion concentration, pH= - log 10 [H + ] pOH = - log [OH - ] [H + ][OH - ] = K w =10 -14 -log [H + ]-log[OH - ]=-log10 -14 p H +p OH =14

11. Example (1): What is the pH value of: Pure water. HCl (0.01 mole/L). NaOH (0.001 mole/L).

12. Example (2): The ionization constant of acetic acid at 25 0 C is 1.82x10 -5. Calculate the pH of 0.1 mole/ L acid. Another method of calculating [H+]: According to Ostwald’s dilution law, pH= -log [H+]= 2.87

13. Example (3): If the pH value of 0.1 mole/L acetic acid is 2.872, calculate the ionization constant of the acid. Answer C=0.1pH= 2.872K=? pH= -log [H + ] 2.872 = -log [H + ] [H + ]= 1.353x10 -3