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Question

Consider a solution of ammonium hydroxide NH4OH (a weak base), having a concentration of 0.05 M. Then the pH value of this solution is :
The ionization constant (Kb) for NH4OH is 1.8×105 M.

A
12.66
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B
9.05
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C
12.01
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D
10.98
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Solution

The correct option is D 10.98
Given, [NH4OH]=0.005M,Kb=1.8×105
The dissociation is given by :

NH4OH(aq.) NH+4(aq)+OH(aq)Initial: C 0 0Equilibrium: C(1α) Cα Cα

At equilibrium [OH]=Cα

Kb=[NH+4][OH]NH4OHKb=(Cα)2C(1α)
Since, NH4OH is a weak base.
α<<1 1α1

Kb×C=(Cα)2Kb×C=[OH]2[OH]=Kb×C=1.8×105×0.05[OH]=9×107=9.5×104pOH=log[OH]pOH=log(9.5×104)=(4log(9.5))=(40.98)pOH=3.02pH+pOH=14pH=143.02pH=10.98




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