Question

What is the ratio [ A − ] / [ HA ] at pH 5.75 ? The p K a of formic acid (methanoic acid, H − COOH ) is 3.75.

Answer 1

Answer : The ratio of
`([A^-])/([HA])` at pH 5.75 is, 100

Explanation : Given,

`pK_a=3.75`

`pH=5.75`

The equilibrium reaction of methanoic acid is,

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

Using Henderson Hesselbach equation :

`pH=pK_a+\log ([Salt])/([Acid])`

`pH=pK_a+\log ([HCOO^-])/([HCOOH])`

or,

`pH=pK_a+\log ([A^-])/([HA])`

Now put all the given values in this expression, we get:

`5.75=3.75+\log ([A^-])/([HA])`

`([A^-])/([HA])=100`

Therefore, the ratio of
`([A^-])/([HA])` at pH 5.75 is, 100