The correct option is C In solution, [Na+]=0.1M,[NH3]=0.1M,[OH−]=0.2M.
Ionisation of NH3 follow as
NH3H2O→NH4OH→NH+4+OH−
Initially 1
After dissociation 1−Cα Cα Cα
Kb=C2α2C−Cα
{NH3 weak base, hence dissociation very less. So, we can write C−Cα≡C}
∴α=√KbC=√2×10−50.1=√2×10−2
(1) ∵α is very less (√2×10−2) for NH3. Hence, degree of dissociation for NH3 is nearly zero
(2) Initial pH of weak base NH3
{pKb=−logKb=−log(2×10−5)=4.7}pOH=12[PKb−logC]⇒pOH=12[4.7−log(0.1)]⇒pOH=2.85⇒pH=14−2.85=11.15
If we add NaOH in NH3 solution, NaOH is a strong base so dissociation 100 and if we compare to NaOH with NH′3 the dissociation of NH3 is very less. So, we leave it.
pH of 0.1 M NaOH solution
[OH−]=10−1M⇒pOH=−log[OH−]=1⇒pH=14−1=13
Change in pH=13−11.15=1.85
(3) In solution
NaOH→Na++OH−0.1M0.1M0.1MNH4OH→NH+4+OH− 1- Cα Cα Cα =0.1×√2×10−2
⇒[OH−]=Ca=√2×10−3
Hence, ⇒[Na+]=0.1M,[NH3]=0.1−√2×10−3≡0.1M⇒[OH−]=0.1+√2×10−3≡0.1M
Hence, concentration of [OH−]=0.2M not possible
(4) Addition of NaOH Kb of NH3 doesn't change.