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natta225 [31]
3 years ago
6

What is the ratio of [a–]/[ha] at ph 3.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Chemistry
2 answers:
amid [387]3 years ago
6 0

Answer:

\frac{[A^-]}{[HA]}=1

Explanation:

Hello,

In this case, one state the following relationship among the pH, pKa and the [a–]/[ha] ratio for the formic acid:

\frac{[A^-]}{[HA]}=\frac{Ka}{[H^+]}

In such a way, we compute both the concentration of hydrogen ions and the acid's dissociation constant as:

[H^+]=10^{-pH}=10^{-3.75}=1.78x10^{-4}M

Ka=10^{-Ka}=10^{-3.75}=1.78x10^{-4}

Thus, the [a–]/[ha] ratio becomes:

\frac{[A^-]}{[HA]}=\frac{1.78x10^{-4}}{1.78x10^{-4}}\\\frac{[A^-]}{[HA]}=1

Best regards.

Vikentia [17]3 years ago
3 0

The formula for pH given the pKa and the concentrations are:

pH = pKa + log [a–]/[ha]

<span>
Therefore calculating:</span>

3.75 = 3.75 + log [a–]/[ha]

log [a–]/[ha] = 0

[a–]/[ha] = 10^0

<span>[a–]/[ha] = 1</span>

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