Question

What is Kb for N2H4 if the pH of a 0.158M solution of N2H5Cl is 4.5?

Answer 1

Answer:

Kb$= 1.58 * 10^{-6}$

Explanation:

The correct question is

What is Kb for N2H4 if the pH of a 0.158M solution of N2H5Cl is 4.5?

Solution-

N2H4Cl hydrolyses on addition of water

The reaction equation is as follows -

N2H4+ + H2O ----> N2H4 + H3O+

$Ka = \frac{K_w}{K_b} = \frac{10^{-14}}{K_b}$

pH $= 4.5$

$[H+] = 10^{-pH} = 10^{-4.5}$

$K_b= \frac{10^{-14} }{K_b} = \frac{Y_2}{0.158} \\K_b= \frac{(10^{-4.5})^2}{0.158}\\=> Kb = 0.158 * 10-14 /(10-4.5)2$

 Kb$= 1.58 * 10^{-6}$