The addition of a strong acid or base to a solution containing a weak acid or base will affect the ionization of the weak acid or base by changing the equilibrium concentrations of the ions in the solution. This occurs because the strong acid or base will react with the weak acid or base, altering the concentrations of the ions involved in the equilibrium reaction.Let's consider a solution containing a weak acid, acetic acid CH3COOH , and a strong base, sodium hydroxide NaOH . The ionization of acetic acid can be represented by the following equilibrium reaction:CH3COOH aq + H2O l CH3COO- aq + H3O+ aq The equilibrium constant for this reaction, Ka, is given by:Ka = [CH3COO-][H3O+] / [CH3COOH]Now, let's say we add a strong base, NaOH, to the solution. The NaOH will react with the acetic acid as follows:CH3COOH aq + OH- aq CH3COO- aq + H2O l This reaction will consume some of the acetic acid and produce more acetate ions CH3COO- . As a result, the concentration of CH3COOH will decrease, and the concentration of CH3COO- will increase. This will cause the equilibrium to shift to the left, decreasing the concentration of H3O+ ions in the solution.To calculate the resulting change in equilibrium concentrations, we can use an ICE Initial, Change, Equilibrium table. Let's assume we initially have 0.1 M CH3COOH and add 0.05 M NaOH to the solution. CH3COOH + OH- CH3COO- + H2OInitial: 0.1 0.05 0 -Change: -0.05 -0.05 +0.05 -Equilibrium: 0.05 0 0.05 -Now, we can use the Ka expression to find the new equilibrium concentrations:Ka = [CH3COO-][H3O+] / [CH3COOH]For acetic acid, Ka = 1.8 x 10^-5. Plugging in the equilibrium concentrations:1.8 x 10^-5 = 0.05 x / 0.05 Solving for x the concentration of H3O+ ions :x = 1.8 x 10^-5Thus, the addition of the strong base NaOH has decreased the concentration of H3O+ ions in the solution, which in turn affects the ionization of the weak acid, acetic acid.