Question

the pH of 0.04M hydrazine solution is 9.7.Calculate ionization constant kb and PKb.

Answer 1

 

NH2NH2 + H2O -----> NH2NH3+ + OH-

From   the  given pH   the Hion concentration can be measured.

So, we have
[H+] = antilog (–pH) = antilog (–9.7) = 1.67 × 10^-10
Now, [OH-]  = Kw / [H+] = 1 × 10^-14/ 1.67 × 10^-10                         = 5.98 × 10^-5//

The concentration of the corresponding hydrazine ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004 M

Thus, Kb= [NH2NH3+][OH-] / [NH2NH2]= (5.98 × 10^-5)^2/ 0.004 = 8.96 × 10^-7//

So, pKb = –logKb = –log(8.96 × 10–7) = 6.04//

 

Answer 2

Answer:

NH2 NH2 + H2O ====> NH2NH3+ + OH-

From   the  given pH   the Hion concentration can be measured.

So, we have

[H+] = antilog (–pH) = antilog (–9.7) = 1.67 × 10^-10

Now, [OH-]  = Kw / [H+] = 1 × 10^-14/ 1.67 × 10^-10                         = 5.98 × 10^-5//

The concentration of the corresponding hydrazine ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004 M

Thus, Kb= [NH2NH3+][OH-] / [NH2NH2]= (5.98 × 10^-5)^2/ 0.004 = 8.96 × 10^-7//

So, pKb = –logKb = –log(8.96 × 10–7) = 6.04

Explanation: