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When 0.1 mole of solid NaOH is added in 1 L of 0.1 M NH3(aq) then which statement is going to be wrong ?

Kb = 2 × 10-5, log 2 = 0.3

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degree of dissociation of NH3 approaches to zero.
change in pH would be 1.85
concentration of Na+ = 0.1 M, NH3 = 0.1 M,  OH- = 0.2 M
on addition of OH-,  Kb  of  NH3 does not changes

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detailed solution

Correct option is C

Before adding 0.1 M NaOH pH  of weak base 0.1M NH3 ispH=12pkb-log10C=124.7+1=2.85 When 0.1 M NaOH is added due to common ion effect dissociation of NH3 decreases.NH3+H2O⇌NH4++OH- Kb=NH4+OH-NH3=0.1OH-0.1+x=2×10-5 x=OH- combining with NH4+ to form ammonia solution in backward direction OH-=2×10-5  since 0.1+x≅0.1 pOH=4.7 Change in pOH=4.7-2.85=1.85 For a given basic solution change in pOH is also equal to change in pH

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