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  4. Solution of 0.1NH4OH and 0.1N NH4Cl has pH 9.25. Then, pKb of NH4OH is

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Q. Solution of $ 0.1N{{H}_{4}}OH $ and $ 0.1N\,N{{H}_{4}}Cl $ has pH 9.25. Then, $ p{{K}_{b}} $ of $ N{{H}_{4}}OH $ is

MGIMS WardhaMGIMS Wardha 2009

Solution:

$ pOH=p{{K}_{b}}+\log \frac{[salt]}{[base]} $ $ \because $ $ [N{{H}_{4}}Cl]=[N{{H}_{4}}OH] $ $ \therefore $ $ pOH=p{{K}_{b}} $ $ p{{K}_{b}}=pOH=14-pH $ $ =14-9.25=4.75 $