Hey there!:
ka = kw/ kb
Ka = 1.0*10⁻¹⁴ / 1.8* 10⁻⁵
Ka = 5.56*10⁻¹⁰
Now Cl⁻ is spectator ion
NH⁴⁺ ---> NH3 + H⁺
Kb = [NH⁴⁺ / [NH3] [H⁺]
at equilibrium the concentration of products is taken as x each and that of ammonium ion will be ( 0.10 -x )
5.56*10⁻¹⁰ = x² / [0.10 -x]
x = 7.54* 10⁻⁶ M
pH = - log [ H⁺ ]
pH = - log [ 7.54*10⁻⁶ ]
pH = 5.13
Answer C
Hope that helps!
