Before the addition of hydrochloric acid:We can use the Ka expression to find the percent ionization of the acetic acid solution.Ka = [H+][A-]/[HA]Let x be the concentration of [H+] and [A-] at equilibrium. Then, the Ka expression becomes:1.8 x 10^-5 = x^2 / 0.1 - x Assuming x is small compared to 0.1, we can approximate:1.8 x 10^-5 x^2 / 0.1x^2 1.8 x 10^-6x 1.34 x 10^-3 MPercent ionization = x / initial concentration * 100Percent ionization 1.34 x 10^-3 / 0.1 * 100 1.34%The pH of the solution can be calculated using the formula:pH = -log10[H+]pH -log10 1.34 x 10^-3 2.87After the addition of 0.05 M hydrochloric acid:The strong acid HCl will react with the acetic acid CH3COOH to form its conjugate base CH3COO- and water.CH3COOH + HCl CH3COO- + H2OSince HCl is a strong acid, it will react completely with the acetic acid. The limiting reactant is HCl, so the reaction will consume 0.05 moles of acetic acid and produce 0.05 moles of acetate ion CH3COO- .New concentrations:[CH3COOH] = 0.1 - 0.05 = 0.05 M[CH3COO-] = 0.05 MNow, we need to find the new [H+] concentration. We can use the Ka expression again:1.8 x 10^-5 = [H+][CH3COO-] / [CH3COOH]Let x be the new [H+] concentration. Then:1.8 x 10^-5 = x 0.05 / 0.05x 1.8 x 10^-5 MPercent ionization after the addition of HCl = 1.8 x 10^-5 / 0.05 * 100 0.036%The pH of the solution after the addition of HCl can be calculated using the formula:pH = -log10[H+]pH -log10 1.8 x 10^-5 4.74In summary, the percent ionization of the acetic acid solution decreases from 1.34% to 0.036% after the addition of 0.05 M hydrochloric acid, and the pH of the solution changes from 2.87 to 4.74.