Answer
$CH_3COOH$ and $CH_3COO^-$ are capable of making a buffer with $8.3 \times 10^{-6}M$ of $H_3O^+$
This happens because the $K_a$ value for $CH_3COOH$ is close to this hydronium ion concentration value.
Work Step by Step
1. Analyze the $[H_3O^+] (8.3 \times 10^{-6}M)$, and find the conjugate acid-base pair with the closest $K_a$ value.
- Use Table $15-1$ on page $659$.
- $CH_3COOH$ and $CH_3COO^-$: $K_a$ = $1.8 \times 10^{-5}$
Therefore, this is the best pair that can be used to make a buffer with hydronium ion concentration equal to $8.3 \times 10^{-6}.$
