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

User CBBSpike
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1 Answer

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Answer:

1.8 x 10⁻⁵

Step-by-step 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

User Sorianiv
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