The number of grams/weight of NH4Cl required to be added to 3litres of 0.01MNH3 to prepare the buffer of pH=9.45 at temperature 298K (Kb for NH3 is 1.85×10−5).
A
0.354gm
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B
4.55gm
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C
0.455gm
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D
3.55gm
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Solution
The correct option is A0.354gm ∵pOH=pKb+log[NH4Cl][NH3] pOH=−logKb+log[NH4Cl][NH3] (∵pKb=−logKb) Also, pOH=14−pH=14−9.45=4.55 and log[NH4Cl][NH3].Kb=log3≈0.470 Thus, log1014−(log109+log3)≈log[NH4Cl][NH3].Kb ⇒log1014109×3=log[NH4Cl][NH3].Kb ⇒1014109×3≈[NH4Cl][NH3].Kb or 1014×[NH3].Kb109×3 =[NH4Cl]≈1014109×3×0.01×1.85×10−5 =[NH4Cl]≈0.354gm.