Answer :
[tex]Our \ goal \ here \ is \ to \ find \ an \ acid-base \ conjugate \ pairs \ that \ has \ a \ $pK_a-p H$. This is because, we assume that the acid-base conjugate pair have equal concentration to be considered an effective buffer.[/tex]
[tex]$p H-p K_a+\log \frac{\left[A^{-}\right]}{[H A]}$Where $\left[A^{-}\right]-[H A]$, so $\log \frac{\left[A^{-}\right]}{\mid H A]}-\log 1-0$ so...$[/tex]
[tex]pH-p K_a[/tex]
[tex]$Solve for $K_a$ -$\begin{aligned}p H-p K_a--\log K_a \\K_a &-10^{-p H} \\&-10^{-5.5} \\K_a &-3.2 \times 10^{-6}\end{aligned}$[/tex]
[tex]The nearest pairs are $\mathrm{HOOC}\left(\mathrm{CH}_2\right)_4 \mathrm{COO}^{-} / \mathrm{OOC}\left(\mathrm{CH}_2\right)_4 \mathrm{COO}^{2-}$ with a $\mathrm{K}_a-3.8 \times 10^{-6}$, and[/tex]
[tex]$\mathrm{HOOCCH}_2 \mathrm{CH}_2 \mathrm{COO}^{-} / \mathrm{OOCCH}_2 \mathrm{CH}_2 \mathrm{COO}^{2-}$ with a $\mathrm{K}_a-2.3 \times 10^{-6}$. We can also check for the $\mathrm{K}_b$. First solve for $p K_b$ then the $K_b$ -$[/tex]
[tex]$\begin{gathered}p K_w-p K_b+p K_a \\p K_b-p K_w-p K_a \\-14-5.5 \\p K_b-8.5\end{gathered}$$[/tex]
[tex]$\begin{gathered}p K_b=-\log K_b \\K_b-10^{-p K_b} \\\quad-10^{-8.5} \\K_b-3.2 \times 10^{-9}\end{gathered}$The nearest pair is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}^{+} / \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}$ with a $\mathrm{K}_6-1.7 \times 10^{-9}$. .[/tex]
What is a conjugate acid and base pair?
A pair of molecules that differ by one proton is referred to as a conjugate acid-base pair. Conjugate acid-base pair refers to a group of chemicals that can accept and give hydrogen ions to one another. A proton is added to create the conjugate base of the chemical, while a proton is taken out to create the conjugate acid. A proton given to a base creates a conjugate acid, while a proton taken away from an acid creates a conjugate base.
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