MCQ
A certain weak base has a dissociation constant $2 \times 10^{-5}$. The equilibrium constant for its neutralisation reaction with strong acid is
- A$2 \times 10^{-5}$
- B$5 \times 10^{10}$
- C$2 \times 10^{-9}$
- ✓$2 \times 10^{9}$
$\therefore \mathrm{K}_{\mathrm{N}}=\frac{1}{\mathrm{K}_{\mathrm{h}}}=\frac{1}{\left(1 / 2 \times 10^{-9}\right)}=2 \times 10^{9}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $m-CH_3OPhCH_2^+$ $(B)$ $p-CH_3OPhCH_2^+$ $(C)$ $PhCH_2^+$ $(D)$ $p-NO_2PhCH_2^+$