# Question #d40e3

Apr 17, 2015

Start with the balanced chemical equation for the equilibrium reaction that exists when hydroxylamine reacts with water

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

Use the solution's pH to determine the pOHm which will allow you to calculate the concentration of the hydroxide ions

$p {H}_{\text{sol" = 14 - pOH => pOH = 14 - pH_"sol}}$

$p O H = 14 - 10.11 = 3.89$

The concentration of hydroxide ions will be

$\left[O {H}^{-}\right] = {10}^{- p O H} = {10}^{3.89} = 1.29 \cdot {10}^{- 4} \text{M}$

Since you have $1 : 1$ mole ratios between all the species of interest, the initial concentration of hydroxylamine will decrease by the same amount the concentration of hydroxide increased.

At equilibrium, you'll have

$\left[N {H}_{3} O {H}^{+}\right] = \left[O {H}^{-}\right] = 1.29 \cdot {10}^{- 4} \text{M}$

$\left[N {H}_{2} O H\right] = {\left[N {H}_{2} O H\right]}_{0} - \left[O {H}^{-}\right] = 0.15 - 1.29 \cdot {10}^{- 4} = \text{0.14987 M}$

By definition, the base dissociation constant, ${K}_{b}$, will be equal to

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

${K}_{b} = \frac{1.29 \cdot {10}^{- 4} \cdot 1.29 \cdot {10}^{- 4}}{0.14987} = \textcolor{g r e e n}{1.1 \cdot {10}^{- 9}}$