Question

What is the ratio of [a–]/[ha] at ph 2.75? the pka of formic acid (methanoic acid, h–cooh) is 3.75?

Answer 1

The dissociation of formic acid is:

`HCOOH \rightleftharpoons HCOO^(-) + H^(+)`

The acid dissociation constant of formic acid,
`k_a` is:

`k_a = ([HCOO^(-)]  [H^(+)])/(HCOOH)`

Rearranging the equation:

`([HCOO^(-)])/([HCOOH]) = (k_a)/([H_+])`

pH = 2.75

`pH = -log[H^(+)]`

`[H^(+)]= 10^(-2.75) = 1.78 * 10^(-3)`

`pk_a = 3.75`

`k_a = 10^(-3.75) = 1.78* 10^(-4)`

Substituting the values in the equation:

`([HCOO^(-)])/([HCOOH]) = (k_a)/([H_+])`

`([HCOO^(-)])/([HCOOH]) = (1.78* 10^(-4))/(1.78* 10^(-3))`

Hence, the ratio is
`(1)/(10)`.