The $\mathrm{pH}$ of a $0.02 \mathrm{M} \mathrm{NH}$ [given $\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=10^{-5}$ and $\log 2=0.301$ ]
$4.65$
$5.35$
$4.35$
$2.65$
Correct Option: , 2
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