Autoionisation of liquid NH3 occurs as2NH3⇌NH4++NH2− with KNH3=NH4+NH2−=10−30 at −50∘CNumber of amide ions, NH2−, present per mm3 of pure liquid NH3 is

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602

300

200

100

answer is A.

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Detailed Solution

NH2−=NH4+=KNH3=10−15M=10−15molL−1 10−15molL−1=10−15mol1000cm3=10−15mol106mm3 =10−15×6.02×1023106NH2−ion /mm3=602

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