Calculate the \(\mathrm{pH}\) at the equivalence point for titrating $0.200 \mathrm{M}\( solutions of each of the following bases with 0.200 \)M$ HBr: \((\mathbf{a})\) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\).

At the equivalence point, the moles of base will be equal to the moles of acid. Since the given bases are strong bases that react completely with HBr, we can predict the species formed as follows: a) NaOH + HBr -> NaBr + H2O b) NH2OH + HBr -> NH3 + H2O c) C6H5NH2 + HBr -> C6H5NH3+ + Br-

The concentration of the species formed depends only on the concentration of the base and acid used in the titration because of their equal molar concentrations at the equivalence point. For a), the concentration of NaBr is 0.200 M. For b), the concentration of NH3 is 0.200 M. For c), the concentration of C6H5NH3+ is 0.200 M.

For the pH calculation, we need to consider the different equilibria. a) Since the formation of NaBr involves a strong base and a strong acid, there is no hydrolysis, and the pH at the equivalence point is 7.00. b) For NH3, we must consider the ammonia equilibrium: NH3 + H2O <-> NH4+ + OH- We apply the Kb for NH3: Kb = [NH4+][OH-] / [NH3] = 1.8 x 10^-5 Since the system is at the equivalence point and [NH3] = [NH4+], we can express Kb in terms of x (the concentration of OH- ions): Kb = x^2 / (0.200 - x) Solving the quadratic equation for x, we get: x ≈ 5.9 x 10^-4 M Then, the pOH can be calculated: pOH = - log10(5.9 x 10^-4) ≈ 3.23 Finally, the pH is obtained by subtracting pOH from 14: pH = 14 - pOH ≈ 10.77 c) For C6H5NH3+, the acid dissociation constant (Ka) of its conjugate acid (C6H5NH2) is 2.5 x 10^-10. Submitting a mass balance, we obtain the following equation for the acid-base equilibrium of aniline and its conjugate acid: 2.5 x 10^-10 = x*(0.200 + x) / (0.200 - x) Solving the quadratic equation for x yields: x ≈ 2.2 x 10^-11 M The H3O+ ion concentration (x) can be directly used to calculate the pH: pH = - log10(2.2 x 10^-11) ≈ 10.66 In summary, the pH at the equivalence point for each titration is: a) NaOH: pH = 7.00 b) NH2OH: pH ≈ 10.77 c) C6H5NH2: pH ≈ 10.66