A 0.110 M solution of NH4OH is 13% ionised at 25°C. Calculate the
dissociation constant of NH4OH.
Answers
We have to find dissociation constant of NH₄OH if concentration of solution of NH₄OH is 0.11 M and ionised 13% at 25°C.
solution : dissociation reaction is..
NH₄OH ⇔ NH₄⁺ + OH¯
at t = 0 0.11 0 0
at t = t₀. 0.11(1 - 0.13) 0.13 0.13
0.11 × 0.87 0.13 0.13
as degree of dissociation is 13 % or 0.13
so dissociation constant = [NH₄⁺][OH¯]/[NH₄OH]
= (0.13)²/(0.11 × 0.87)
= 0.17659 ≈ 0.18
Therefore the dissociation constant of reaction is 0.18
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