Question

Calculate the [H^+] and pH of a 0.000429 M hydrazoic acid solution. Keep in mind that the K_a of hydrazoic acid is 2.20 times 10^-5 [H^+] = pH =

Answer 1

Let the acid be HA for simplicity.

HA dissociates as:

HA -----> H+ + A-

0.000429 0 0

0.000429-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

since ka is small, x will be small and it can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.2*10^-5)*0.000429) = 9.715*10^-5

So,

[H+] = 9.715*10^-5 M

use:

pH = -log [H+]

= -log (9.715*10^-5)

= 4.013