Question

A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution

Answer 1

Answer: 2.75%

Step-by-step explanation:

`pH=-log [H+]`

`3.26 = -log [H+]`

`[H+] = 5.495* 10^(-4) M`

`HA\rightleftharpoons H^++A^-`

initial 0.020 0 0

eqm 0.020 -x x x

`K_a=([H+][A-])/([HA])`

`K_a=([x][x])/([0.020-x])`

`x=5.495* 10^(-4)`

`K_a=([5.495* 10^(-4)]^2)/([0.020-5.495* 10^(-4)])`

`K_a =1.553* 10^(-5)`

percent dissociation =
`([H^+_eqm])/([Acid_(initial)])* 100`

percent dissociation=
`(5.495* 10^(-4))/(0.020)* 100`

Thus percent dissociation= 2.75 %