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.)

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.

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

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.

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.