MCQ
If $0.4\,gm\,\,NaOH$ is present in $ 1$ litre solution, then its $pH$ will be
- A$2$
- B$10$
- C$11$
- ✓$12$
$[{H^ + }] = {10^{ - 12}},\,\,pH = - \log [{H^ + }] = 12$
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|
$A$ $(mol/l)$ |
$B$ $(mol/l)$ |
Rate |
| $0.05$ | $0.05$ | $1.2\times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4\times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2\times 10^{-3}$ |