# "Hydroxylamine", HONH_2 is a weak Bronsted base with K_b=10^-8. What is pH for a 0.024*mol*L^-1 solution?

Apr 21, 2017

We need the stoichiometric equation............and get $p H \cong 9.$

#### Explanation:

$H O N {H}_{2} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s H O N {H}_{3}^{+} + H {O}^{-}$

And then we write the equilibrium expression:

${K}_{b} = {10}^{-} 8 = \frac{\left[H O N {H}_{3}^{+}\right] \left[H {O}^{-}\right]}{\left[H O N {H}_{2}\right]}$,

And we make the usual approximations, and invoke $x$ as the moles of hydroxylamine that associate,

${K}_{b} = {x}^{2} / \left(0.024 - x\right)$

And if $\text{0.024>>>} x$, then $x \cong \sqrt{{K}_{b} \times 0.024}$

i.e. $x \cong \sqrt{{10}^{-} 8 \times 0.024}$

${x}_{1} = 1.55 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$, which is indeed small compared to $0.024$, but let us recycle that value back into the equation......

${x}_{2} = 1.55 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$,

So, from the equilibrium expression, we now know $x = \left[H {O}^{-}\right] = 1.55 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$.

How does this help us, as we were asked to measure $p H$?

Well in aqueous solution under standard condtions, $p H + p O H = 14$, thus $p H = 14 - p O H = 14 + {\log}_{10} \left\{1.55 \times {10}^{-} 5\right\} \cong 9$. This is slightly basic, as we would expect for a weak Bronsted base.