Question

In aqueous solution, the azide ion, N3−, is a weak base that accepts a proton from water to form the hydroxide ion, OH −, and hydrazoic acid, HN3, according to the following equation.
N3−(aq) + H2O(l) equilibrium reaction arrow OH −(aq) + HN3(aq)
The base-dissociation constant (Kb) for this base is 4.04 ✕ 10−10. If a 0.082 M solution of azide ions is prepared, what is the final pH of the solution? (Assume that the temperature is 25°C.)

Answer 1

N3- dissociates as:

N3- +H2O     ----->     HN3   +   OH-
8.2*10^-2                   0         0
8.2*10^-2-x                 x         x


Kb = [HN3][OH-]/[N3-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((4.04*10^-10)*8.2*10^-2) = 5.756*10^-6

since c is much greater than x, our assumption is correct
so, x = 5.756*10^-6 M



So, [OH-] = x = 5.756*10^-6 M


use:
pOH = -log [OH-]
= -log (5.756*10^-6)
= 5.2399


use:
PH = 14 - pOH
= 14 - 5.2399
= 8.7601
Answer: 8.76