Answer :
[tex]Our \ goal \ here\ is \ to \ find \ an \ acid-base \ conjugate \ pair \ that \ have \a \\\$pK_a-p H$. This is because we assume that the acid-base conjugate pair have an equal concentration to be considered an effective buffer.[/tex]
[tex]$p H-p K_a+\log \frac{\left[A^{-}\right]}{[H A]}$[/tex]
[tex]$Where $\left[A^{-}\right]-[H A]$, so $\log \frac{\left[A^{-}\right]}{[H A]}-\log 1-0_{\text {so ... }}$$[/tex]
pH-pKa
[tex]$Solve for $K_a$.$[/tex]
[tex]$\begin{aligned}p H-p K_a &--\log K_a \\K_a &-10^{-p H} \\&-10^{-3.5} \\K_a &-3.2 \times 10^{-4}\end{aligned}$[/tex]
[tex]$The nearest pairs are $\mathrm{HCOCOOH} / \mathrm{HCOCOO}^{-}$with a $\mathrm{K}_{\mathrm{a}}-3.5 \times 10^{-4}$,[/tex]
[tex]$\mathrm{HOCH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} / \mathrm{HOCH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{COO}^{-}$with a $\mathrm{K}_{\mathrm{a}}-2.9 \times 10^{-4}$, and[/tex]
[tex]$\mathrm{CH}_3 \mathrm{COOC}_6 \mathrm{H}_4 \mathrm{COOH} / \mathrm{CH}_3 \mathrm{COOC}_6 \mathrm{H}_4 \mathrm{COO}^{-}$with a $\mathrm{K}_a-3.6 \times 10^{-4}$. We can also check for the $\mathrm{K}_b$. First, solve for $p K_b$ then the $K_b$ -[/tex]
[tex]$\begin{aligned}p K_w &-p K_b+p K_a \\p K_b &-p K_w-p K_a \\&-14-3.5 \\p K_b &-10.5 \\p K_b &--\log K_b \\K_b &-10^{-p K_b} \\&-10^{-10.5} \\K_b &-3.2 \times 10^{-11}\end{aligned}$[/tex]
[tex]There \ are\ no \ pairs \ near \ this \ $K_b$ value.[/tex]
What is a conjugate acid and base pair?
An acid-base pair that differs by one proton is referred to as a conjugate pair. A conjugate acid-base pair is a pair of substances that can both absorb and donate hydrogen ions to one another. A proton is added to the compound to create the conjugate acid, and a proton is taken out to create the conjugate base. When a proton is supplied to a base, a conjugate acid is created, and vice versa when a proton is taken away from an acid, a conjugate base is created.
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