Question

Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:

Answer 1

The concentrations : 0.15 M

pH=11.21

Further explanation

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

$\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}$

$\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21$