The PH solution of NH3 of 0.950molar solution is 11.612. find the Kb

1.8 x 10⁻⁵

Explanation:

 NH3(aq)  +  H2O(l)  ⇄ NH4⁺(aq)  +   OH⁻(aq)

I  0.95                              0                    0

C -x                                 +x                  +x

E 0.95-x                           x                    x

Kb= [NH₄⁺] [OH⁻] / (  NH₃) = x²/ (0.95-x )

P(OH) = 14-PH = 14-11.612 = 2.388

(OH)⁻¹ = 10⁻²°³⁸⁸ = 4.09 x 10⁻³ = x

Kb = (4.09 x 10⁻³)²/ (0.95-4.09 x 10⁻³)

= 1.8 x 10⁻⁵

false

because it does not have the potential to be entirely used up