Question

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

Answer 1

Kb
`= 1.58 * 10^(-6)`

Step-by-step 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 = (K_w)/(K_b) = (10^(-14))/(K_b)`

pH
`= 4.5`

`[H+] = 10^(-pH) = 10^(-4.5)`

`K_b= (10^(-14) )/(K_b) = (Y_2)/(0.158) \\K_b= ((10^(-4.5))^2)/(0.158)\\=> Kb = 0.158 * 10-14 /(10-4.5)2`

Kb
`= 1.58 * 10^(-6)`