Hydroxylamine, HONH₂, readily forms salts such as hydroxylamine hydrochloride which are used as antioxidants in soaps. Hydroxylamine has K_b = 9.1 × 10^-9. What is the pH of a 0.025 M HONH₂ solution? A) 2.90 B) 4.82 C) 9.18 D) 9.91 E) 11.10

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To solve this problem, we need to use the acid dissociation constant expression (Ka) for the ionization of HONH₂: HONH₂ + H2O ⇌ H3O+ + ONH_b- Ka = [H3O+][ONH_b-] / [HONH₂] Since the concentration of HONH₂ is much greater than the concentration of H3O+ or ONH_b-, we can assume that the concentration of HONH₂ is approximately equal to the initial concentration (0.025 M) and use the approximation [HONH₂] ≈ 0.025 M. Next, we can use the fact that the concentration of H3O+ is equal to the concentration of OH- in a neutral solution (pH = 7) and write the expression for the ion product constant (Kw): Kw = [H3O+][OH-] = 1.0 × 10^-14 Since Kw = Ka × Kb (where Kb is the base dissociation constant), we can solve for Kb: Kb = Kw / Ka = 1.0 × 10^-14 / 9.1 × 10^-9 = 1.1 × 10^-6 Now we can write the expression for the base dissociation constant (Kb) for the ionization of ONH_b-: ONH_b- + H2O ⇌ OH- + HONH₂ Kb = [OH-][HONH₂] / [ONH_b-] Since the initial concentration of ONH_b- is negligible compared to the concentration of HONH₂, we can assume that [HONH₂] ≈ [OH-] and write: Kb ≈ [HONH₂]2 / [ONH_b-] We can now use the Kb value to solve for [OH-]: Kb = [HONH₂]2 / [ONH_b-] 1.1 × 10^-6 = (x)2 / 0.025 x = 9.89 × 10^-4 Finally, we can use the [OH-] value to calculate the pH: pH = -log[H3O+] pH = -log(1.0 × 10^-14 / 9.89 × 10^-4) pH = 9.18 Therefore, the best choice is C, and the explanation is as above.

Hydroxylamine, HONH2, Readily Forms Salts Such as Hydroxylamine Hydrochloride Which

Hydroxylamine, HONH₂, Readily Forms Salts Such as Hydroxylamine Hydrochloride Which