The PH solution of NH3 of 0.950molar solution is 11.612. find the Kb
1 answer:
Answer:
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
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