Answer:
- Methanoic acid, Ka = 1.8 × 10⁻⁴
Explanation:
At buffering capacity, the pH = pKa. This means that we can take the -log of each substance's Ka value to get the pKa and whichever of these values is closest to 4.15, will be the most appropriate acid.
Solving:
[tex]\section*{pKa:}\[\text{pKa} = -\log(K_a)\]\subsection*{Hydrofluoric Acid}\[K_a = 6.6 \times 10^{-4} \implies \text{pKa} = -\log(6.6 \times 10^{-4}) \approx \boxed{3.18}\]\subsection*{Methanoic Acid}\[K_a = 1.8 \times 10^{-4} \implies \text{pKa} = -\log(1.8 \times 10^{-4}) \approx \boxed{3.74}\]\subsection*{Ascorbic Acid}\[K_a = 4.3 \times 10^{-7} \implies \text{pKa} = -\log(4.3 \times 10^{-7}) \approx \boxed{6.37}\][/tex]
Since Methanol acid has a pKa value of 3.74, that means the pH in a buffer solution would also be 3.74 which is the closest to 4.15.