Question

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

Answer 1

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.

Answer 2

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

pH = pKa + log [a–]/[ha]

Therefore calculating:

3.75 = 3.75 + log [a–]/[ha]

log [a–]/[ha] = 0

[a–]/[ha] = 10^0

[a–]/[ha] = 1