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$2 \times 10^{-9}$
- ✓$2 \times 10^{9}$
✓
Answer
Correct option: D.
$2 \times 10^{9}$
d
$\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{k}_{\mathrm{w}}}{\mathrm{k}_{\mathrm{b}}}=\frac{10^{-14}}{2 \times 10^{-5}}=\frac{1}{2} \times 10^{-9}$
$\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{k}_{\mathrm{w}}}{\mathrm{k}_{\mathrm{b}}}=\frac{10^{-14}}{2 \times 10^{-5}}=\frac{1}{2} \times 10^{-9}$
$\therefore \mathrm{K}_{\mathrm{N}}=\frac{1}{\mathrm{K}_{\mathrm{h}}}=\frac{1}{\left(1 / 2 \times 10^{-9}\right)}=2 \times 10^{9}$
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