Question

What is the pH of a 0.231 M solution of hydrazine, N2H4? The pKb of hydrazine is 6.02.

Answer 1

Given

Hydrazine concentration = [N₂H₄] = 0.231 M

pKb = 6.02

We know that

pKb = -log(Kb)

hence

Kb = 10^-pKb = 10^-6.02 = 9.55 $\times$ 10^-7

Now lets see the dissocition nreaction of hydrazine in water

N₂H₄ + H₂O $\rightleftharpoons$ N₂H₅⁺ + OH^-  

lets see the ICE table for above reaction

......................N₂H_4........+....H₂O...... $\rightleftharpoons$.....N₂H₅⁺ ....+.....OH^-  

Initial.............0.231 M.................................0.....................0.

Change.............-x...................................... +x..................+x

Equilibrium....0.231-x ....................................x...................x

the equilibrium constant for above reaction is

Kb = [N₂H₅⁺][OH^-] / [N₂H₄]

9.55 $\times$ 10^-7 = (x)(x) /(0.231-x) = x² / (0.231-x)

As x is very small than 0.231 , 0.231-x = 0.231

then

9.55 $\times$ 10^-7 = x² / 0.231

x² = 0.231 $\times$ 9.55 $\times$ 10^-7 = 2.20 $\times$ 10^-7

by taking the square roots on both sides we get

x = 0.000469 M

hence

[OH^-] = 0.000469 M

Now we know that

pOH = -log[OH^-]

= -log(0.000469)

= 3.33

Now

pH + pOH = 14

the

pH = 14 - pOH = 14 - 3.33 = 10.67

so our answer is

pH = 10.67