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 $$2$$ $$=$$ $$0.3$$

Question

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 $$2$$ $$=$$ $$0.3$$

Infinity Learn · Accepted Answer

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

When 0.1 mole of solid NaOH is added in 1 L of 0.1 M ${NH}_{3} (aq)$ then which statement is going to be wrong ?
$(K_{b} = 2 \times 10^{- 5}, \log 2 = 0.3)$

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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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When 0.1 mole of solid NaOH is added in 1 L of 0.1 M ${NH}_{3} (aq)$ then which statement is going to be wrong ?
$(K_{b} = 2 \times 10^{- 5}, \log 2 = 0.3)$