Answer :
[tex]Our \ goal \ here \ is \ to \ find \ an \acid-base \conjugate \pairs \ that \ has \ a \\\$pK_a-p H$. This is because, we assume that the acid-base conjugate pair have equal concentration to be considered an effective buffer.$p H-p K_a+\log \frac{\left[A^{-}\right]}{[H A]}$[/tex]
[tex]$p H-p K_a+\log \frac{\left[A^{-}\right]}{[H A]}$[/tex]
[tex]Where $\left[A^{-} \mid-[H A]\right.$, so $\log \frac{\left[A^{-}\right]}{|H A|}-\log 1-0$ so...$[/tex]
[tex]pH-p K_a[/tex]
[tex]Solve for $K_a[/tex]
[tex]$\begin{aligned}p H-p K_a &--\log K_a \\K_a &-10^{-p H} \\&-10^{-7.0} \\K_a &-1 \times 10^{-7}\end{aligned}$[/tex]
[tex]The \ nearest \ pairs \ are \ $\mathrm{H}_3 \mathrm{AsO}_4 / \mathrm{H}_2 \mathrm{AsO}_4^{-}$with a $\mathrm{K}_a-1.1 \times 10^{-7}$, and $\mathrm{H}_2 \mathrm{PO}_3^{-} / \mathrm{HPO}_3^{2-}$ with a $\mathrm{K}_a-1.7 \times$ $10^{-7}$. We can also check for the $K_b$. First solve for $p K_b$ then the $K_b$.[/tex]
[tex]$\begin{aligned}p K_w &-p K_b+p K_a \\p K_b &-p K_w-p K_a \\&-14-7 \\p K_b &-7 \\p K_b &--\log K_b \\K_b &-10^{-p K_b} \\&-10^{-7} \\K_b &-1 \times 10^{-7}\end{aligned}$[/tex]
[tex]There \ are \ no\ pairs \ near \ this \ $K_b$ value.[/tex]
What is a buffer?
Buffer is a solution which resists change of pH upon addition of a small amount of acid or base. Many chemical reactions are carried out at a constant pH. Numerous pH regulating systems in nature employ buffering. For instance, the pH of blood is controlled by the bicarbonate buffering system, and bicarbonate also serves as a buffer in the ocean. There are two types of buffer: acidic and basic buffer.
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