Question

Choose specific acid-base conjugate pairs to make the following buffers: (a)$\mathbf{\text{pH}}\mathbf{\approx }\mathbf{\text{4.5}}$; (b)$\mathbf{\text{pH}}\mathbf{\approx }\mathbf{\text{4.5}}$. (See Appendix C.)

Short answer

(a)$\text{HOOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{COOH/HOOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{COO}}^{\text{-}}$or${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}^{\text{+}}{\text{/C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$

(b)${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{\text{/H}}_{\text{2}}{\text{AsO}}_{\text{4}}^{\text{-}}$or${\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}^{\text{-}}{\text{/HPO}}_{\text{3}}^{\text{2-}}$

Step-by-step solution

Define buffer

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.

Explanation

(a) 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.

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

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

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

${\text{pH=pK}}_{\text{a}}$

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

$\begin{array}{c}{\text{pK}}_{\text{a}}{\text{=-logK}}_{\text{a}}\\ {\text{K}}_{\text{a}}{\text{=10}}^{\text{-pH}}\\ {\text{=10}}^{\text{-4.5}}\\ {\text{K}}_{\text{a}}{\text{=3.2×10}}^{\text{-5}}\end{array}$

With a , ${\text{K}}_{\text{a}}{\text{=3.8×10}}^{\text{-5}}$the closest pair is.$\text{HOOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}\text{COOH/HOOC}{\left({\text{CH}}_{\text{2}}\right)}_{\text{4}}{\text{COO}}^{\text{-}}$ We can also search for.${\text{K}}_{\text{b}}$

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

$\begin{array}{c}{\text{pK}}_{\text{w}}{\text{=pK}}_{\text{b}}{\text{+pK}}_{\text{a}}\\ {\text{pK}}_{\text{b}}{\text{=pK}}_{\text{w}}{\text{-pK}}_{\text{a}}\\ \text{=14-4.5}\\ {\text{pK}}_{\text{b}}\text{=9.5}\\ {\text{pK}}_{\text{b}}{\text{=-logK}}_{\text{b}}\\ {\text{K}}_{\text{b}}{\text{=10}}^{{\text{-pK}}_{\text{b}}}\\ {\text{=10}}^{\text{-9.5}}\\ {\text{K}}_{\text{b}}{\text{=3.2×10}}^{\text{-10}}\end{array}$

Therefore,,${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}^{\text{+}}{\text{/C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ with a${\text{K}}_{\text{b}}{\text{=4.0×10}}^{\text{-10}}$, is the closest pair.

Explanation

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

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

Where, data-custom-editor="chemistry" $\left[{\text{A}}^{\text{-}}\right]\text{=[HA]}$

Sodata-custom-editor="chemistry" $\text{log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}}\text{=log1=0so}\mathrm{...}$

data-custom-editor="chemistry" ${\text{pH=pK}}_{\text{a}}$

Calculate.data-custom-editor="chemistry" ${\text{K}}_{\text{a}}$

data-custom-editor="chemistry" $\begin{array}{c}{\text{pK}}_{\text{a}}{\text{=-logK}}_{\text{a}}\\ {\text{K}}_{\text{a}}{\text{=10}}^{\text{-pH}}\\ {\text{=10}}^{\text{-7.0}}\\ {\text{K}}_{\text{a}}{\text{=1×10}}^{\text{-7}}\end{array}$

As, data-custom-editor="chemistry" ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{\text{/H}}_{\text{2}}{\text{AsO}}_{\text{4}}^{\text{-}}$, with a , data-custom-editor="chemistry" ${\text{K}}_{\text{a}}{\text{=1.1×10}}^{\text{-7}}$and, data-custom-editor="chemistry" ${\text{H}}_{\text{2}}{\text{PO}}_{\text{3}}^{\text{-}}{\text{/HPO}}_{\text{3}}^{\text{2-}}$with a, data-custom-editor="chemistry" ${\text{K}}_{\text{a}}{\text{=1.7×10}}^{\text{-7}}$are the closest pairs. We can also search for .data-custom-editor="chemistry" ${\text{K}}_{\text{b}}$

Solve fordata-custom-editor="chemistry" ${\text{pK}}_{\text{b}}$ first, then .data-custom-editor="chemistry" ${\text{K}}_{\text{b}}$

data-custom-editor="chemistry" $\begin{array}{c}{\text{pK}}_{\text{w}}{\text{=pK}}_{\text{b}}{\text{+pK}}_{\text{a}}\\ {\text{pK}}_{\text{b}}{\text{=pK}}_{\text{w}}{\text{-pK}}_{\text{a}}\\ \text{=14-7}\\ {\text{pK}}_{\text{b}}\text{=7}\\ {\text{pK}}_{\text{b}}{\text{=-logK}}_{\text{b}}\\ {\text{K}}_{\text{b}}{\text{=10}}^{{\text{-pK}}_{\text{b}}}\\ {\text{=10}}^{\text{-7}}\\ {\text{K}}_{\text{b}}{\text{=1×10}}^{\text{-7}}\end{array}$

Therefore, there are no pairs that are close to thisdata-custom-editor="chemistry" ${\text{K}}_{\text{b}}$ value.