Question ba54f

Aug 24, 2015

$\left(\left[{\text{B"])/(["BH}}^{+}\right]\right) = 0.0281$

Explanation:

There are two ways in which you can approach this problem, one using the base dissociation constant and the solution's $p O H$, and the other one using the acid dissociation constant and the solution's pH.

I'll show you how to solve it using ${K}_{b}$ and $\left[{\text{OH}}^{-}\right]$, and you try the other approach as practice. So, you know that you're dealing with a weak base that has the base dissociation constant, ${K}_{b}$, equal to $8.91 \cdot {10}^{- 6}$.

The equilibrium dissociation of the weak base, which I'll call $\text{B}$ for simplicity, looks like this

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

By definition, the base dissciation constant is equal to

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

Use the blood's pH to determine what the $p O H$ is

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

$p O H = 14 - 7.4 = 6.6$

The concentration of the hydroxide ions present in solution will be

$\left[{\text{OH}}^{-}\right] = {10}^{- p O H}$

["OH"^(-)] = 10^(-6.6) = 2.5 * 10^(-7)"M"#

Rearrange the equation for ${K}_{b}$ to get

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

$\frac{\left[{\text{B"])/(["BH"^(+)]) = (["OH}}^{-}\right]}{K} _ b = \frac{2.5 \cdot {10}^{- 7}}{8.91 \cdot {10}^{- 6}} = \textcolor{g r e e n}{0.0281}$