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Question

When 0.1 mole solid NaOH is added in 1 lt of 0.1 M NH3(aq) then which statement is wrong? (Kb=2×105,log2=0.3)

A
Degree of dissociation of NH3 approaches to zero.
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B
Change in pH by adding NaOH would be 1.85
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C
In solution, [Na+]=0.1M,[NH3]=0.1M,[OH]=0.2M.
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D
On addition of OHKb of NH3 does not changes.
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Solution

The correct option is C In solution, [Na+]=0.1M,[NH3]=0.1M,[OH]=0.2M.
Ionisation of NH3 follow as
NH3H2ONH4OHNH+4+OH
Initially 1
After dissociation 1Cα Cα Cα
Kb=C2α2CCα
{NH3 weak base, hence dissociation very less. So, we can write CCαC}
α=KbC=2×1050.1=2×102
(1) α is very less (2×102) for NH3. Hence, degree of dissociation for NH3 is nearly zero
(2) Initial pH of weak base NH3
{pKb=logKb=log(2×105)=4.7}pOH=12[PKblogC]pOH=12[4.7log(0.1)]pOH=2.85pH=142.85=11.15
If we add NaOH in NH3 solution, NaOH is a strong base so dissociation 100 and if we compare to NaOH with NH3 the dissociation of NH3 is very less. So, we leave it.
pH of 0.1 M NaOH solution
[OH]=101MpOH=log[OH]=1pH=141=13
Change in pH=1311.15=1.85
(3) In solution
NaOHNa++OH0.1M0.1M0.1MNH4OHNH+4+OH 1- Cα Cα Cα =0.1×2×102
[OH]=Ca=2×103
Hence, [Na+]=0.1M,[NH3]=0.12×1030.1M[OH]=0.1+2×1030.1M
Hence, concentration of [OH]=0.2M not possible
(4) Addition of NaOH Kb of NH3 doesn't change.

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