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nataly862011 [7]
3 years ago
14

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

Chemistry
1 answer:
Vesnalui [34]3 years ago
8 0

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

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

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And with solution...
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Answer:

The answer to your question is: V = 6.93 L

Explanation:

Data

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Reaction

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                                1 mol of N₂   ----------------  2 moles of NH₃

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