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

In the titration of a strong acid with a strong base, the pH at equivalence point is equal to 7. Since NaOH is a strong base, and HBr is a strong acid, the pH at the equivalence point will be: pH = 7

Let's write the equation for the titration reaction: NH2OH + HBr → NH3OH+ + Br- Calculate the Kb of hydroxylamine (NH2OH) from the given Ka of its conjugate acid (NH3OH+). The relationship is Kb * Ka = Kw, where Kw is 1.0×10^-14. \(K_{b} = \frac{K_{w}}{K_{a}} = \frac{1 \times 10^{-14}}{1.1 \times 10^{-9}} = 9.09091 \times 10^{-6}\) Now, let's write the equation for the reaction of hydroxylamine with water: \(NH_{2}OH + H_{2}O \rightleftharpoons NH_{3}OH^{+} + OH^{-}\) At the equivalence point, the moles of NH2OH and HBr are equal, which means all NH2OH has been converted to its conjugate acid NH3OH+. However, since the volumes are equal, the concentration of NH3OH+ remains 0.200 M. Use an ICE table and solve for the concentration of OH-. NH2OH + H2O ↔ NH3OH+ + OH- Initial 0 - 0.200 0 Change -x - +x +x Equilibrium -x - 0.200+x x \(K_{b} = \frac{[NH_{3}OH^{+}][OH^{-}]}{[NH_{2}OH][H_{2}O]}\) \(9.09091 \times 10^{-6} = \frac{(0.200+x)x}{(-x)}\) Now we can assume that x << 0.200, and ignore x in the denominator. \(9.09091 \times 10^{-6} = \frac{0.200x}{(-x)}\) x = [OH-] = 1.818×10^-6 Now, calculate the pOH: pOH = -log(1.818×10^-6) = 5.74 Finally, calculate the pH: pH = 14 - pOH = 14 - 5.74 = 8.26

Similar to Step 2 with hydroxylamine, write the titration equation and the base dissociation constant expression: C6H5NH2 + HBr → C6H5NH3+ + Br- \(K_{b} = \frac{K_{w}}{K_{a}} = \frac{1 \times 10^{-14}}{2.5 \times 10^{-5}} = 4 \times 10^{-10}\) Reaction with water: \(C_{6}H_{5}NH_{2} + H_{2}O \rightleftharpoons C_{6}H_{5}NH_{3}^{+} + OH^{-}\) In this case, at the equivalence point, all C6H5NH2 has been converted to C6H5NH3+, and the concentration of C6H5NH3+ is also 0.200 M. Now, use an ICE table and solve for the concentration of OH-. C6H5NH2 + H2O ↔ C6H5NH3+ + OH- Initial 0 - 0.200 0 Change -x - +x +x Equilibrium -x - 0.200+x x \(K_{b} = \frac{[C_{6}H_{5}NH_{3}^{+}][OH^{-}]}{[C_{6}H_{5}NH_{2}][H_{2}O]}\) \(4 \times 10^{-10} = \frac{(0.200+x)x}{(-x)}\) Ignore x in the denominator again. \(4 \times 10^{-10} = \frac{0.200x}{(-x)}\) x = [OH-] = 2×10^-9 Now, calculate the pOH: pOH = -log(2×10^-9) = 8.70 Finally, calculate the pH: pH = 14 - pOH = 14 - 8.70 = 5.30

The pH at the equivalence point for titrating the three bases with HBr are: a) Sodium Hydroxide: pH = 7 b) Hydroxylamine: pH = 8.26 c) Aniline: pH = 5.30