Question

A 0.1 M solution of ammonia (NH3) has a pH of 11.64 at room temperature. If sodium hydroxide (NaOH) is added to this solution such that the concentration of OH- ions becomes 0.01 M, what will be the new pH of the solution? (Assume that the dissociation of water can be ignored)

Answer 1

To find the new pH of the solution, we first need to determine the new concentration of OH- ions after adding NaOH. Since NaOH is a strong base, it will dissociate completely in water, and the concentration of OH- ions will be the sum of the concentrations from NH3 and NaOH.First, we need to find the concentration of OH- ions from the NH3 solution. We can do this using the pOH, which is related to the pH by the following equation:pOH = 14 - pHFor the NH3 solution, the pOH is:pOH = 14 - 11.64 = 2.36Now, we can find the concentration of OH- ions from the NH3 solution using the following equation:[OH-] = 10^-pOH [OH-] = 10^-2.36  = 4.57 x 10^-3  MNow, we can find the total concentration of OH- ions after adding NaOH:[OH-]_total = [OH-]_NH3 + [OH-]_NaOH = 4.57 x 10^-3  M + 0.01 M = 0.01457 MNow, we can find the new pOH of the solution:pOH_new = -log10 [OH-]_total  = -log10 0.01457  = 1.836Finally, we can find the new pH of the solution:pH_new = 14 - pOH_new = 14 - 1.836 = 12.164So, the new pH of the solution after adding NaOH is 12.164.