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Question

Calculate [OH] of 0.01 M ammonium hydroxide solution. The ionization constant for NH4OH is 1.8×105 M.

A
7.84×104M
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B
0.88×105M
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C
1.56×105M
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D
4.24×104M
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Solution

The correct option is D 4.24×104M
NH4OH(aq.)NH+4(aq.)+OH(aq.)Initial: C 0 0Equilibrium:CCα Cα CαKb=[NH+4][OH][NH4OH]=[Cα][Cα]C(1α)Kb=Cα21α1.8×105=0.01×α21α

Since NH4OH is a weak base α<<11α1

0.01×α2=1.8×105
α=1.8×103=4.24×102
[OH]=Cα=0.01×4.24×102[OH]=4.24×104M

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