# Question 1809d

Oct 2, 2015

${K}_{b} = 3.7 \cdot {10}^{- 6}$

#### Explanation:

Start by using the known pH of the solution to find the molarity of the hydrogen ions, ${\text{H}}^{+}$. This will help you find the acid dissociation constant, ${K}_{a}$, for quinine's conjugae acid, ${\text{QH}}^{+}$.

So, you know that quinine's conjugate acid, ${\text{QH}}^{+}$, will ionize in aqueous solution to give

${\text{OH"_text((aq])^(+) -> "Q"_text((aq]) + "H}}_{\textrm{\left(a q\right]}}^{+}$

The concentration of the conjugate acid is simply the ratio between the number of moles you have and the volume of the solution

["QH"^(+)] = "0.23 moles"/"1.0 L" = "0.23 M"

The concentration of ${\text{H}}^{+}$ will be

["H"^(+)] = 10^(-"pH") = 10^(-4.58) = 2.63 * 10^(-5)"M"

Since the dissociation reactrion produces equal numbers of moles of $\text{Q}$ and ${\text{H}}^{+}$, you get that

["Q"] = ["H"^(+)] = 2.63 * 10^(-5)"M"#

The acid dissociation constant for this equilibrium will thus be

${K}_{a} = \left(\left[{\text{Q"] * ["H"^(+)])/(["QH}}^{+}\right]\right)$

${K}_{a} = \frac{2.63 \cdot {10}^{- 5} \cdot 2.63 \cdot {10}^{- 5}}{0.23} = 2.7 \cdot {10}^{- 9}$

Finally, to get the base dissociation constant, ${K}_{b}$, use the fact that

${K}_{a} \cdot {K}_{b} = {K}_{W} \text{ }$, where

${K}_{W}$ - the water self-ionization constant, equal to ${10}^{- 14}$.

This means that ${K}_{b}$ will be equal to

${K}_{b} = {K}_{W} / {K}_{a} = {10}^{- 14} / \left(2.7 \cdot {10}^{- 9}\right) = \textcolor{g r e e n}{3.7 \cdot {10}^{- 6}}$