When 0.1 moles of HCl is added to 1 liter of a 0.1 M solution of HA, the HCl will dissociate completely, as it is a strong acid. This will result in an increase in the concentration of H3O+ ions in the solution. Let's calculate the new concentration of H3O+ ions and the degree of ionization of the weak acid HA.Initial concentration of HA = 0.1 MInitial concentration of H3O+ from HA dissociation = xInitial concentration of A- = xHCl dissociates completely, so the concentration of H3O+ from HCl = 0.1 MTotal concentration of H3O+ = [H3O+]_total = x + 0.1The dissociation of HA can be represented as:HA <=> H3O+ + A-The Ka expression for this reaction is:Ka = [H3O+][A-] / [HA] Since the initial concentration of HA is 0.1 M, and x moles of HA dissociate:[HA] = 0.1 - xSubstituting the values in the Ka expression:1.8 x 10^-5 = x + 0.1 x / 0.1 - x To simplify the equation, we can assume that x is very small compared to 0.1, so 0.1 - x 0.1:1.8 x 10^-5 = 0.1x /0.1Solving for x:x = 1.8 x 10^-5So, the initial concentration of H3O+ from HA dissociation is 1.8 x 10^-5 M. The total concentration of H3O+ after adding HCl is:[H3O+]_total = x + 0.1 = 1.8 x 10^-5 + 0.1 0.1 MThe degree of ionization of HA can be calculated as the ratio of the concentration of dissociated HA x to the initial concentration of HA 0.1 M :Degree of ionization = x / 0.1 = 1.8 x 10^-5 / 0.1 = 1.8 x 10^-4The addition of HCl has increased the concentration of H3O+ ions in the solution to approximately 0.1 M, and the degree of ionization of the weak acid HA is 1.8 x 10^-4.