Question

A solution of HCOOH has 0.22M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?

Answer 1

Answer : The pH of the solution is, 2.21

Solution : Given,

Concentration (c) = 0.22 M

Acid dissociation constant =
`k_a=1.8* 10^(-4)`

The equilibrium reaction for dissociation of
`HCOOH` (weak acid) is,

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

initially conc. c 0 0

At eqm.
`c(1-\alpha)`
`c\alpha`
`c\alpha`

First we have to calculate the concentration of value of dissociation constant
`(\alpha)`.

Formula used :

`k_a=((c\alpha)(c\alpha))/(c(1-\alpha))`

Now put all the given values in this formula ,we get the value of dissociation constant
`(\alpha)`.

`1.8* 10^(-4)=((0.22\alpha)(0.22\alpha))/(0.22(1-\alpha))`

By solving the terms, we get

`\alpha=0.0282`

Now we have to calculate the concentration of hydrogen ion.

`[H^+]=c\alpha=0.22* 0.0282=6.204* 10^(-3)M`

Now we have to calculate the pH.

`pH=-\log [H^+]`

`pH=-\log (6.204* 10^(-3))`

`pH=2.21`

Therefore, the pH of the solution is, 2.21