Hydroxylamine, HONH₂, readily forms salts such as hydroxylamine hydrochloride which are used as antioxidants in soaps. Hydroxylamine has K_b of 9.1 * 10¯⁹. What is the pH of a 0.025 M HONH₂ solution?

Question

Quizplus · Accepted Answer

HONH$_2^+$ is the conjugate acid of HONH, which is a weak base. Therefore, we need to use the Kb expression to solve for the pH.

Kb = [NH$_2$OH][H$^+$]/[HONH]

Since HONH is a monoprotic acid, [H$^+$] = [HONH] and we can simplify:

Kb = [NH$_2$OH][H$^+$]/[H$^+$] = [NH$_2$OH]

9.1 x 10$^{-9}$ = [NH$_2$OH]$^2$/[HONH]

[NH$_2$OH] = $\sqrt{(9.1 	imes 10^{-9})	imes 0.025}$ = 1.41 x 10$^{-5}$ M

pH = 14 - pOH = 14 - 8.85 = 5.15

However, this is the pH of the solution if we had pure HONH. When we dissolve HONH in water, it reacts with water to form NH$_2$OH$^+$ and OH$^-$.

HONH + H$_2$O $ightleftharpoons$ NH$_2$OH$^+$ + OH$^-$

Since this is an acid-base reaction, we can use the Kw expression to find the concentrations of NH$_2$OH$^+$ and OH$^-$:

Kw = [NH$_2$OH$^+$][OH$^-$] = 1.0 x 10$^{-14}$

Since [NH$_2$OH$^+$] = [HONH] and [OH$^-$] = x,

x$	imes$(1.41 x 10$^{-5}$) = 1.0 x 10$^{-14}$

x = 7.09 x 10$^{-10}$

pOH = -log(7.09 x 10$^{-10}$) = 9.15

pH = 14 - pOH = 14 - 9.15 = 4.85

Therefore, the best answer is C.

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

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