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Question

Calculate [OH] of a 0.01 M solution of ammonium hydroxide. The ionization constant for NH4OH (kb)=1.6×105 (Take log2=0.30)

A
4×104 M
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B
1.6×106 M
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C
1.6×104 M
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D
4×106 M
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Solution

The correct option is A 4×104 M
We know,
NH4OHNH+4+OH
At,t=0 C 0 0
At,t=teq CCα Cα Cα
Again, Kb=[NH+4][OH][NH4OH]
Kb=Cα21α
1.6×105=0.01×α2
α2=1.6×103
α=4×102
So,
[OH]=Cα=0.01×4×102=4×104 M

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