Question

The ph of 0.1n nh4oh is 11.27 calculate its dissociation constant

Answer 1

Kb=[NH+4][OH−][NH4OH]  

Given:

pH=11.27, [H+][OH−]=7.1×10−15

[H+]=5.37×10−12 M

⇒ [OH−]=1.322×10−3M

[NH+4]=[OH−]

∴[NH+4]=1.322×10−3M

[NH4OH]=0.1 M−1.322 mM

=98.678 mM

∴ Kb=[1.322×10−3][1.322×10−3][98.678×10−3]

=1.7477×10−398.678

Kb=17.72×10−6