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When 0.1 mole solid NaOH is added to 1 L...

When 0.1 mole solid NaOH is added to 1 L of 0.01 M `NH_3(aq)` , then which statement is wrong ? (`K_b=2xx10^(-5)`, log 2 =0.3)

A

Degree of dissociation of `NH_3` approaches to zero

B

Change in pH by adding NaOH would be 1.85

C

In solution , `[Na^+]` =0.1M , `[NH_3]`=0.1 M , `[OH^-]` =0.2 M

D

on adding of `OH^-, K_b` of `NH_3` does not change .

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The correct Answer is:
c
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