Question

What is the approximate pH of a 0.06 M solution of CH3COOH (given that Ka = 1.78 x 10-5)? 1 3 6 7 9

Answer 1

The answer would come out as 3

Answer 2

Answer : The pH of the solution is, 3

Solution : Given,

Concentration (C) = 0.06 M

Acid dissociation constant =
`k_a=1.78* 10^(-5)`

The equilibrium reaction for dissociation of
`CH_3COOH` (weak acid) is,

`CH_3COOH+H_2O\rightleftharpoons CH_3COO^-+H_3O^+`

initially conc. c 0 0

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

First we have to calculate the concentration of
`[H^+]`

As, we know that

`\alpha=\sqrt{(k_a)/(c)}` (for weak electrolyte) ...........(1)

where,
`\alpha` is degree of dissociation

`[H^+]=c\alpha` ..................(2)

By equation both the equations (1) and (2), we get

`[H^+]=√(k_a* c)`

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

`[H^+]=\sqrt{1.78* 10^(-5)* 0.06}=1.033* 10^(-3)M`

Now we have to calculate the pH.

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

`pH=-\log (1.033* 10^(-3))`

`pH=2.8=3`

Therefore, the pH of the solution is, 3