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3 years ago

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]

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Therefore calculating:</span>

3.75 = 3.75 + log [a–]/[ha]

log [a–]/[ha] = 0

[a–]/[ha] = 10^0

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2 years ago

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Answer:

Explanation:

Just saw your request regarding answering this so here it is:

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Answer:

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azamat

Answer:

Explanation:

There are some mistakes in writing the equation. However, Please check the below one and contact me again if necessary.

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