# Question 5f837

Feb 11, 2016

$1.0 \cdot {10}^{- 8}$

#### Explanation:

Start by writing the balanced chemical equation for this hydrolysis reaction

${\text{B"_text((aq])^(+) + "H"_2"O"_text((l]) rightleftharpoons "BOH"_text((aq]) + "H}}_{\textrm{\left(a q\right]}}^{+}$

The **equilibrium constant, ${K}_{e q}$, for this reaction is defined as

${K}_{e q} = \left(\left[\text{BOH"] * ["H"^(+)])/(["B"^(+)] * ["H"_2"O}\right]\right)$

Now, because water's concentration is assumed to be constant, you can rewrite this equation as

$\overbrace{{K}_{e q} \cdot \left[{\text{H"_2"O"])^(color(purple)(=K_h)) = (["BOH"] * ["H"^(+)])/(["B}}^{+}\right]}$

The left-hand side of the equation is equal to the hydrolysis constant ,${K}_{h}$, so you can say that

K_h = color(blue)((["BOH"])/(["B"^(+)])) * ["H"^(+)]" " " "color(red)("(*)")

You are told that the base dissociation constant, ${K}_{b}$, for this reaction is equal to $1.0 \cdot {10}^{- 6}$. The balanced chemical equation for the partial dissociation of the base looks like this

${\text{BOH"_text((aq]) rightleftharpoons "B"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

By definition, ${K}_{b}$ will be equal to

${K}_{b} = \left(\left[\text{B"^(+)] * ["OH"^(-)])/(["BOH}\right]\right)$

Rearrange this to get

${K}_{b} / \left(\left[\text{OH"^(-)]) = (["B"^(+)])/(["BOH}\right]\right)$

color(blue)((["BOH"])/(["B"^(+)])) = (["OH"^(-)])/K_b

Plug this into equation $\textcolor{red}{\text{(*)}}$ to get

K_h = (["OH"^(-)])/K_b * ["H"^(+)]#

${K}_{h} = \frac{\left[{\text{OH"^(-)] * ["H}}^{+}\right]}{K} _ b$

You also know that the ion product constant for water, ${K}_{W}$, equal to ${10}^{- 14}$ at room temperature, is defined as

${K}_{W} = \left[{\text{H"^(+)] * ["OH}}^{-}\right]$

This means that you have

${K}_{h} = {K}_{W} / {K}_{b}$

Plug in your value to get

${K}_{h} = {10}^{- 14} / \left(1.0 \cdot {10}^{- 6}\right) = \textcolor{g r e e n}{1.0 \cdot {10}^{- 8}}$