To determine the equilibrium constant, Kc, we need to use the reaction quotient, Qc, and the initial concentrations of the reactants and products. The reaction is:I2 aq + 2S2O3^2- aq + 2H+ aq 2I^- aq + S4O6^2- aq At equilibrium, Qc = Kc. We are given the initial concentrations of the reactants:[I2] = 0.02 M[S2O3^2-] = 0.05 M[H+] = 0.1 MWe are also given that Qc = 0.04 at equilibrium. To find Kc, we need to determine the equilibrium concentrations of all species involved in the reaction. We can do this by setting up an ICE Initial, Change, Equilibrium table:| | I2 | S2O3^2- | H+ | I^- | S4O6^2- ||---------|----|---------|----|-----|---------|| Initial | 0.02 | 0.05 | 0.1 | 0 | 0 || Change | -x | -2x | -2x | 2x | x || Equilibrium | 0.02-x | 0.05-2x | 0.1-2x | 2x | x |Now, we can write the expression for Qc:Qc = [I^-]^2 * [S4O6^2-] / [I2] * [S2O3^2-]^2 * [H+]^2 At equilibrium, Qc = Kc, so we can substitute the equilibrium concentrations from the ICE table into the expression:0.04 = 2x ^2 * x / 0.02-x * 0.05-2x ^2 * 0.1-2x ^2 Now, we need to solve this equation for x, which represents the change in concentration for each species. Since the equation is quite complex, we can make an assumption that x is small compared to the initial concentrations, so we can simplify the equation:0.04 4x^3 / 0.02 * 0.05^2 * 0.1^2 Now, solve for x:x^3 0.04 * 0.02 * 0.05^2 * 0.1^2 / 4x^3 5e-9x 1.71e-3Now that we have the value of x, we can find the equilibrium concentrations of all species:[I2] = 0.02 - x 0.01829 M[S2O3^2-] = 0.05 - 2x 0.04658 M[H+] = 0.1 - 2x 0.09316 M[I^-] = 2x 0.00342 M[S4O6^2-] = x 0.00171 MNow, we can plug these equilibrium concentrations back into the Kc expression:Kc = [I^-]^2 * [S4O6^2-] / [I2] * [S2O3^2-]^2 * [H+]^2 Kc = 0.00342^2 * 0.00171 / 0.01829 * 0.04658^2 * 0.09316^2 Kc 0.04So, the equilibrium constant Kc for the reaction between iodine and thiosulfate ions in aqueous solution at 25C is approximately 0.04.