Question

What is the kb expression for aniline (c6h5nh2)?

Answer 1

For C6H5NH2 + H2O >< C6H5NH3+ + OH-kb = 4.3×(10^-10) =[C6H5NH3+] [OH-] /[C6H5NH]

Answer 2

Answer: The $K_b$ expression for aniline is written as:

$K_b=\frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]}$

Explanation:

$C_6H_5NH_2+H_2O\rightleftharpoons C_6H_5NH_3^++OH^-$

The equilibrium constant of the reaction is written as:

$K=\frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2][H_2O]}$

$K_b=K\times [H_2O]$

The $K_b$ expression for aniline is written as:

$K_b=\frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]}$