Calculate the pH and percent ionization of a 0.25 M solution of hydrazine, N2H4 , Kb...

Calculate the pH and percent ionization of a 0.25 M solution of hydrazine, N2H4 , Kb = 1.3x10-6 .

Solutions

Expert Solution

1)
N2H4 dissociates as:

N2H4 +H2O     ----->     N2H5+   +   OH-
0.25                   0         0
0.25-x                 x         x

Kb = [N2H5+][OH-]/[N2H4]
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 ((1.3*10^-6)*0.25) = 5.701*10^-4

since c is much greater than x, our assumption is correct
so, x = 5.701*10^-4 M

so.[OH-] = x = 5.701*10^-4 M

use:
pOH = -log [OH-]
= -log (5.701*10^-4)
= 3.2

use:
PH = 14 - pOH
= 14 - 3.2
= 10.8

Answer: 10.8

2)

% dissociation = (x*100)/c
= 5.701*10^-4*100/0.25
= 0.23 %

Answer: 0.23 %

queen_honey_blossom answered 1 year ago

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