To calculate the new equilibrium concentration of F- ions, we first need to find the initial concentrations of Al3+ and F- ions in the solution. We can use the Ksp expression for AlF3:Ksp = [Al3+] * [F-]^3Given Ksp = 2.0 x 10^-23, we can set up the equation:2.0 x 10^-23 = [Al3+] * [F-]^3Since AlF3 is in equilibrium, the stoichiometry of the reaction shows that for every 1 mol of Al3+ ion, there are 3 mol of F- ions. Therefore, we can express the concentration of F- ions as 3 times the concentration of Al3+ ions:[F-] = 3 * [Al3+]Now, we can substitute this expression into the Ksp equation:2.0 x 10^-23 = [Al3+] * 3 * [Al3+] ^3Solve for [Al3+]:2.0 x 10^-23 = 27 * [Al3+]^4[Al3+]^4 = 2.0 x 10^-23 / 27[Al3+]^4 = 7.41 x 10^-25[Al3+] = 4.64 x 10^-7 MNow we can find the initial concentration of F- ions:[F-] = 3 * [Al3+][F-] = 3 * 4.64 x 10^-7 M[F-] = 1.39 x 10^-6 MNext, we need to account for the addition of 0.001 mol of NaF. Since NaF is a common ion with F-, it will increase the concentration of F- ions in the solution. Assuming the volume of the solution is 1 L, the increase in F- concentration due to the addition of NaF is:[F-] = 0.001 mol / 1 L = 0.001 MNow, we can find the new equilibrium concentration of F- ions:[F-]new = [F-]initial + [F-][F-]new = 1.39 x 10^-6 M + 0.001 M[F-]new = 0.00100139 MTherefore, the new equilibrium concentration of F- ions is approximately 0.00100139 M.