Q19.35P - Chemistry: Molecular Nature Of Matter And Change

Question

Choose specific acid-base conjugate pairs to make the following buffers: (a) $\mathbf{\text{pH}}\mathbf{\approx }\mathbf{\text{3.5}}$ ; (b) $\mathbf{\text{pH}}\mathbf{\approx }\mathbf{5}\mathbf{\text{.5}}$
. (See Appendix C.)

Step-by-Step Solution

Answer

1. ${\text{HCOCOOH/HCOCOO}}^{\text{ - }}$, ${\text{HOCH}}_{\text{2}}{\text{CH(OH)COOH/HOCH}}_{\text{2}}{\text{CH(OH)COO}}^{\text{ - }}$ or ${\text{CH}}_{\text{3}}{\text{COOC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COOH/CH}}_{\text{3}}{\text{COOC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COO}}^{\text{ - }}$.
2.  $\text{HOOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{COO}}^{\text{ - }}\text{/OOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{COO}}^{\text{2 - }}$, ${\text{HOOCCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}{\text{/OOCCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{2 - }}$ or ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{ + }}{\text{/C}}_{\text{6}}{\text{H}}_{\text{5}}\text{N}$.

Step 1: Buffer solution

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.

Step 2: Subpart (a)

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.

 ${\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}$

Where,  $\left[{\text{A}}^{\text{ - }}\right]\text{ = [HA]}$

So  $\text{log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\text{ = log1 = 0 so}...$

 ${\text{pH = pK}}_{\text{a}}$

Calculate ${\text{K}}_{\text{a}}$.

 $$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ {\mathrm{K}}_{\mathrm{a}}={10}^{-\mathrm{pH}} \\ ={10}^{-3.5} \\ {\mathrm{K}}_{\mathrm{a}}=3.2\times {10}^{-4} \end{gathered}$$

As, ${\text{HCOCOOH/HCOCOO}}^{\text{ - }}$, with a ${\text{K}}_a=3.5\times {10}^{-4}$,  ${\text{HOCH}}_{\text{2}}{\text{CH(OH)COOH/HOCH}}_{\text{2}}{\text{CH(OH)COO}}^{\text{ - }}$, with a ${\mathrm{K}}_{\mathrm{a}}=2.9\times {10}^{-4}$, and ${\text{CH}}_{\text{3}}{\text{COOC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COOH/CH}}_{\text{3}}{\text{COOC}}_{\text{6}}{\text{H}}_{\text{4}}{\text{COO}}^{\text{ - }}$, with a ${\mathrm{K}}_{\mathrm{a}}=3.6\times {10}^{-4}$ are the closest pairs. We can also search for ${\text{K}}_{\text{b}}$.

Solve for ${\text{pK}}_{\text{b}}$ first, then ${\text{K}}_{\text{b}}$.

 $$\begin{gathered} {\mathrm{pK}}_{\mathrm{w}}={\mathrm{pK}}_{\mathrm{b}}+{\mathrm{pK}}_{\mathrm{a}} \\ {\mathrm{pK}}_{\mathrm{b}}={\mathrm{pK}}_{\mathrm{w}}-{\mathrm{pK}}_{\mathrm{a}} \\ =14-3.5 \\ {\mathrm{pK}}_{\mathrm{b}}=10.5 \\ {\mathrm{pK}}_{\mathrm{b}}=-{\mathrm{logK}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{b}}={10}^{-{\mathrm{pK}}_{\mathrm{b}}} \\ ={10}^{-10.5} \\ {\mathrm{K}}_{\mathrm{b}}=3.2\times {10}^{-11} \end{gathered}$$

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

Step 3: Subpart (b)

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.

 ${\text{pH = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}$

Where,  $\left[{\text{A}}^{\text{ - }}\right]\text{ = [HA]}$

So $\text{log}\frac{\left[{\text{A}}^{\text{ - }}\right]}{\text{[HA]}}\text{ = log1 = 0 so}...$

 ${\text{pH = pK}}_{\text{a}}$

Calculate ${\text{K}}_{\text{a}}$.

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ {\mathrm{K}}_{\mathrm{a}}={10}^{-\mathrm{pH}} \\ ={10}^{-5.5} \\ {\mathrm{K}}_{\mathrm{a}}=3.2\times {10}^{-6} \end{gathered}$$

 

As, $\text{HOOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{COO}}^{\text{ - }}\text{/OOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{COO}}^{\text{2 - }}$, with a ${\mathrm{K}}_{\mathrm{a}}=3.8\times {10}^{-6}$, and${\text{HOOCCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}{\text{/OOCCH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{2 - }}$, with a ${\mathrm{K}}_{\mathrm{a}}=2.3\times {10}^{-6}$, are the closest pairs. We can also search for ${\text{K}}_{\text{b}}$.

Solve for ${\text{pK}}_{\text{a}}$ first, then ${\text{K}}_{\text{b}}$.

 $$\begin{gathered} {\mathrm{pK}}_{\mathrm{w}}={\mathrm{pK}}_{\mathrm{b}}+{\mathrm{pK}}_{\mathrm{a}} \\ {\mathrm{pK}}_{\mathrm{b}}={\mathrm{pK}}_{\mathrm{w}}-{\mathrm{pK}}_{\mathrm{a}} \\ =14-5.5 \\ {\mathrm{pK}}_{\mathrm{b}}=8.5 \\ {\mathrm{pK}}_{\mathrm{b}}=-{\mathrm{logK}}_{\mathrm{b}} \\ {\mathrm{K}}_{\mathrm{b}}={10}^{-p{K}_b} \\ ={10}^{-8.5} \\ {\mathrm{K}}_{\mathrm{b}}=3.2\times {10}^{-9} \end{gathered}$$

Therefore, ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{ + }}{\text{/C}}_{\text{6}}{\text{H}}_{\text{5}}\text{N}$ is the closest pair, with a ${\mathrm{K}}_{\mathrm{b}}=1.7\times {10}^{-9}$.