Final answer:
The concentration of all species in a 0.220 M C6H5NH3Cl solution can be calculated using the Kb value for C6H5NH2. The concentrations of the solute species at equilibrium are [C6H5NH3+] = [OH-] = 5.8 x 10^-6 M.
Step-by-step explanation:
The concentration of all species in a 0.220 M C6H5NH3Cl solution can be calculated using the given Kb value for C6H5NH2. First, we can find the concentration of C6H5NH2 at equilibrium using the Kb expression. Assuming x is << 0.220, we can simplify the equation and solve for x. Then, we can calculate the equilibrium concentrations of C6H5NH3+ and OH- by setting them equal to the concentration of C6H5NH2, as they all have the same concentration. Thus, the concentrations of the solute species at equilibrium are [C6H5NH3+] = [OH-] = 5.8 x 10^-6 M.