MCQ
The $pH$ of $0.01\; M \;NaOH (aq)$ solution will be
- A$7.01$
- B$2$
- ✓$12$
- D$9$
$\mathrm{So},\left[\mathrm{OH}^{-}\right]=\mathrm{N}=10^{-2} \mathrm{N}$
$\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log 10^{-2}=2$
$\mathrm{pH}=14-\mathrm{pOH}=14-2$
$\mathrm{pH}=12$
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$(I) 0.5\ mole$ of $O_3$ $(II) 0.5\ gm$ atom of oxygen $(III) 3.011 \times 10^{23}$ molecules of $O_2$ $(IV) 5.6\ litre$ of $CO_2$ at $STP$
Above reaction is an example of
(Atomic numbers : $Fe=26,Co=27,Mn=25,Cr=24$)