To solve this problem, we need to find the equilibrium constant Ka for the dissociation of the weak acid HA. We are given the following information:1. 0.10 moles of HA are dissolved in 1.00 L of water.2. The pH of the solution is 4.20.3. The acid is 10% ionized.First, let's write the dissociation equation for the weak acid HA:HA H + ASince the acid is 10% ionized, we know that 10% of the 0.10 moles of HA have dissociated into H and A ions. Therefore, the concentration of H ions is:[H] = 0.10 moles 0.10 = 0.010 molesNow, we can use the pH formula to find the concentration of H ions:pH = -log[H]4.20 = -log[H]To find [H], we need to take the inverse log antilog of both sides:[H] = 10^-4.20 6.31 10^-5 MSince the concentration of H ions is 0.010 moles, and we calculated [H] to be 6.31 10^-5 M, we can now find the concentration of A ions, which is equal to the concentration of H ions:[A] = [H] = 6.31 10^-5 MNow, we can find the concentration of the undissociated HA:[HA] = 0.10 moles - 0.010 moles = 0.090 molesSince the volume of the solution is 1.00 L, the molar concentration of HA is:[HA] = 0.090 moles / 1.00 L = 0.090 MFinally, we can calculate the equilibrium constant Ka using the equilibrium expression:Ka = [H][A] / [HA]Ka = 6.31 10^-5 M 6.31 10^-5 M / 0.090 M Ka 4.42 10^-6 So, the equilibrium constant Ka for the dissociation of the weak acid HA is approximately 4.42 10^-6 .