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Q. The pH of $ 0.1\,M $ aqueous solution of $ NH_{4}OH $ is $ (K_{b}=1.0\times 10^{-5}) $

Chhattisgarh PMTChhattisgarh PMT 2010

Solution:

$\left[O H^{-}\right]=\sqrt{K_{b} . C}$
$=\sqrt{1.0 \times 10^{-5} \times 0.1}=10^{-3}$
$p O H=-\log \left[O H^{-}\right]$
$=-\log 10^{-3}=3 p H$
$=14-p O H$
$=14-3=11$