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

`([A^-])/([HA])=1`

Step-by-step explanation:

Hello,

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

`([A^-])/([HA])=(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:

`([A^-])/([HA])=(1.78x10^(-4))/(1.78x10^(-4))\\([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