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Question

What is [NH+4] in a solution that contain 0.02 M NH3 (Kb=1.8×105) and 0.01 M KOH?

A
9×106
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B
1.8×103
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C
3.6×105
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D
None of these
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Solution

The correct option is C 3.6×105
Given: Kb=1.8×105,NH3=0.02,KOH=0.01

Solution: KOHK++OH (KOH is a strong electrolyte)
At eq 0.01 0 0
(Ammonia is a weak electrolyte)

NH3+H2O(l)NH+4(aq)+OH(aq)
(0.02-x) x (0.01+ x) (at equillibrium)

Here the value of x is very less than 0.01. So we can neglect the value of x.

Then, (0.01-x) is changed to 0.01 and (0.02-x) is changed to 0.02

Kb=0.01×x0.02

x=1.8×105×2=3.6×105
Hence the correct option is C.

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