Choose specific acid-base conjugate pairs to make the following buffers:

(a)$\left[{\text{H}}_3{O}^{+}\right]\mathbf{\approx }\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{\text{M}}$ ;

(b)$\left[O{H}^{-}\right]\mathbf{\approx }\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{\text{M}}$ . (See Appendix C.)

A buffer solution can withstand pH changes when acidic or basic components are added. It can neutralize little amounts of additional acid or base, allowing the pH of the solution to remain relatively constant.

The goal is to find an acid-base conjugate pair with a ${\text{pK}}_{\text{a}}$equal to $\text{pH}$. This is because, in order to be deemed an efficient buffer, we expect that the acid-base conjugate pair has equal concentration.

As, $\text{pH = - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$and ${\text{pK}}_{\text{a}}{\text{= - logK}}_{\text{a}}$.

So, role="math" localid="1663352784139" ${\text{K}}_{\text{a}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$. Find a pair of numbers with ${\mathrm{K}}_{\mathrm{a}}=1\times {10}^{-9}$.

As, ${\text{HBrO/BrO}}^{\text{-}}$is the closest pair, with a ${\mathrm{K}}_{\mathrm{a}}=2.3\times {10}^{-9}$. We can also look for the ${\text{K}}_{\text{b}}$. Calculate ${\text{K}}_{\text{b}}$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}} \\ =\frac{1\times {10}^{-14}}{1\times {10}^{-9}} \\ {\mathrm{K}}_{\mathrm{b}}=1\times {10}^{-5} \end{gathered}$$

Therefore, ${\text{NH}}_{\text{4}}^{\text{+}}{\text{/NH}}_{\text{3}}$is the closest pair, with a ${\mathrm{K}}_{\mathrm{b}}=1.76\times {10}^{-5}$.

The goal is to find an acid-base conjugate pair with a ${\text{pK}}_{\text{a}}$equal to $\text{pH}$. This is because, in order to be deemed an efficient buffer, we expect that the acid-base conjugate pair has equal concentration.

Since ${\text{pK}}_{\text{a}}\text{= pH}$, ${\text{pK}}_{\text{b}}\text{= pOH}$, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{=K}}_{\text{a}}$, and $\left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{b}}$, we can start by looking for an acid-base conjugate pair with ${\mathrm{K}}_{\mathrm{b}}=3\times {10}^{-5}$from the prior problem.As, ${\text{HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{3}}^{\text{+}}{\text{/HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}$is the closest pair, having a ${\mathrm{K}}_{\mathrm{b}}=3.2\times {10}^{-5}$. We can also keep an eye out for the ${\text{K}}_{\text{a}}$. Make a solution for ${\text{K}}_{\text{a}}$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \\ =\frac{1\times {10}^{-14}}{3\times {10}^{-5}} \\ {\mathrm{K}}_{\mathrm{a}}=3.3\times {10}^{-10} \end{gathered}$$

Therefore, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{OH/C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}^{\text{-}}$is the closest pair, with a ${\mathrm{K}}_{\mathrm{a}}=1.0\times {10}^{-10}$.