1. Dissociation Constant or Ionization Constant for Weak Acids & Bases When a weak acid, HX, is placed into water some of it reacts with water to form H 3 O + and X - ions. An equilibrium is established, in which the dissociation reaction HX + H 2 O  H 3 O + + X - occurs at the same rate as the association reaction, H 3 O + + X -  HX + H 2 O.

2. We describe this situation by writing the reaction equation as: HX (aq) + H 2 O (l)  H 3 O + (aq) + X - (aq) When the law of chemical equilibrium is applied, the mass action expression we get is: Keq = [H 3 O + ] [X - ] [HX] Since Keq expresses the tendency for the acid to dissociate, it is often called the “dissociation constant” or “ionization constant” for the weak acid. symbolically: Kdiss or Ka  weak acids Kdiss or Kb  weak bases

3. NOTE: 1. If Ka, Kb, Kdiss > 10: The substance is essentially 100% dissociated for our purposes. This will happen with strong acids & bases like HCl or NaOH. 2. If Ka, Kb, Kdiss < 10: Dissociation is less extensive & the acid or base is said to be weak. 3. A knowledge of Kdiss for a weak acid or base is useful to chemists since it allows the [H 3 O + ] & pH of the solution to be calculated.

4. Ka Problems – Type I Finding the % dissociation & Kdiss given concentration If the pH of a 0.100 M solution of CH 3 COOH is 2.86, find a) % diss b) Ka

5. Type II: Finding concentrations, % dissociation when given Ka or Kdiss Given a 0.900 M solution of a weak acid, HX, whose Ka is 4.5 x 10 -4, find the [H 3 O + ], pH & % dissociation.

6. Type II with a twist Find the [H 3 O + ] & pH of a solution which was made up to be 0.100 M HOCN (cyanic acid) & 0.0500 M in NaOCN (sodium cyanate). Assume complete dissociation. Ka = 3.47 x 10 -4