To solve this problem, we will use the Henderson-Hasselbalch equation:pH = pKa + log [A-]/[HA] where pH is the final pH of the solution, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base NH3 , and [HA] is the concentration of the conjugate acid NH4+ .First, we need to find the pKa of the NH3/NH4+ buffer system. The Kb base dissociation constant of NH3 is 1.8 x 10^-5. We can find the Ka acid dissociation constant of NH4+ using the relationship:Ka * Kb = Kwwhere Kw is the ion product of water 1.0 x 10^-14 . Ka = Kw / Kb = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10Now we can find the pKa:pKa = -log Ka = -log 5.56 x 10^-10 = 9.25Next, we need to find the initial moles of NH3 and NH4+ in the solution. Since the volume of the solution is constant, we can use the initial concentrations to find the moles:moles of NH3 = 0.1 M * volumemoles of NH4+ = 0.1 M * volumeNow, when we add 0.01 moles of HCl, it will react with NH3 to form NH4+:NH3 + HCl NH4+ + Cl-moles of NH3 final = 0.1 M * volume - 0.01 molesmoles of NH4+ final = 0.1 M * volume + 0.01 molesSince the volume is constant, we can find the final concentrations by dividing the final moles by the volume:[NH3] final = 0.1 M * volume - 0.01 moles / volume[NH4+] final = 0.1 M * volume + 0.01 moles / volumeNow we can use the Henderson-Hasselbalch equation to find the final pH:pH = pKa + log [NH3] final / [NH4+] final pH = 9.25 + log 0.1 M * volume - 0.01 moles / volume / 0.1 M * volume + 0.01 moles / volume Since the volume is constant, we can simplify the equation:pH = 9.25 + log 0.1 M * volume - 0.01 moles / 0.1 M * volume + 0.01 moles Now we can plug in the values and solve for the final pH:pH = 9.25 + log 0.1 * volume - 0.01 / 0.1 * volume + 0.01 Since the volume is constant and we are only interested in the change in pH, we can ignore the volume term:pH = 9.25 + log 0.1 - 0.01 / 0.1 + 0.01 pH = 9.25 + log 0.09 / 0.11 pH 9.25 - 0.10pH 9.15So, the final pH of the solution after adding 0.01 moles of HCl is approximately 9.15.