First, we need to find the ionization constant Ka at 25C using the given pKa value:pKa = -log Ka 4.76 = -log Ka Ka = 10^-4.76 1.74 x 10^-5Now, we will use the ionization constant Ka at 50C, which is given as 1.32 x 10^-5.The ionization reaction for acetic acid CH3COOH is:CH3COOH CH3COO- + H+Let x be the concentration of CH3COO- and H+ ions at equilibrium. Then, the concentration of CH3COOH at equilibrium will be 0.1 - x .The ionization constant Ka expression is:Ka = [CH3COO-][H+]/[CH3COOH]At 50C, we have:1.32 x 10^-5 = x x / 0.1 - x We can assume that x is very small compared to 0.1, so we can approximate:1.32 x 10^-5 x^2 /0.1Now, we can solve for x:x^2 = 1.32 x 10^-5 * 0.1x^2 = 1.32 x 10^-6x 1.15 x 10^-3Now, we can find the percent ionization:Percent ionization = x / initial concentration * 100Percent ionization = 1.15 x 10^-3 / 0.1 * 100Percent ionization 1.15%So, the percent ionization of a 0.1 M solution of acetic acid at 25C, if its ionization constant Ka changes to 1.32 x 10^-5 when the temperature is raised to 50C, is approximately 1.15%.