91Ó°ÊÓ

A buffer is a solution that 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 pH. This is because, in order to be deemed an efficient buffer, we expect that the acid-base conjugate pair has equal concentration.

Since we know that ${\text{pK}}_{\text{a}}\text{= pH}$, ${\text{pK}}_{\text{b}}\text{= pOH}$, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{=K}}_{\text{a}}$, and $\left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{b}}$from the preceding issue, we may begin by looking for an acid-base conjugate pair with ${\mathrm{K}}_{\mathrm{b}}=1×{10}^{-6}$.

As, ${\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{3}}^{\text{2 +}}{\text{/H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{3}}^{\text{+}}$is the closest pair, with a ${\mathrm{K}}_{\mathrm{b}}=3×{10}^{-6}$. We can also look for the ${\text{K}}_{\text{a}}$. Calculate ${\text{K}}_{\text{a}}$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}×{\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \\ =\frac{1×{10}^{-14}}{1×{10}^{-6}} \\ {\mathrm{K}}_{\mathrm{a}}=1×{10}^{-8} \end{gathered}$$

Therefore, ${\text{HClO/HClO}}^{\text{-}}$is the closest pair, with a ${\mathrm{K}}_{\mathrm{a}}=2.9×{10}^{-8}$.

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

Since we know that ${\text{pK}}_{\text{a}}\text{= pH}$, ${\text{pK}}_{\text{b}}\text{= pOH}$, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{=K}}_{\text{a}}$, and $\left[{\text{OH}}^{\text{-}}\right]{\text{=K}}_{\text{b}}$from the preceding issue, we may begin by looking for an acid-base conjugate pair with ${\mathrm{K}}_{\mathrm{a}}=4×{10}^{-4}$.

As, ${\text{CH}}_{\text{3}}{\text{COOC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COOH/CH}}_{\text{3}}{\text{COOC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COOH}}^{\text{-}}$, with a ${\mathrm{K}}_{\mathrm{a}}=3.6×{10}^{-4}$and ${\text{HCOCOOH/HCOCOO}}^{\text{-}}$, with a ${\mathrm{K}}_{\mathrm{a}}=3.5×{10}^{-4}$. We can also look for the ${\text{K}}_{\text{b}}$. Calculate ${\text{K}}_{\text{b}}$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}={\mathrm{K}}_{\mathrm{a}}×{\mathrm{K}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}} \\ =\frac{1×{10}^{-14}}{4×{10}^{-4}} \\ {\mathrm{K}}_{\mathrm{b}}=2.5×{10}^{-11} \end{gathered}$$

Therefore, there are no pairs that are close to this ${\text{K}}_{\text{b}}$value.