MCQ
The $pH$ of a $0.001\,M\,NaOH$ will be
- A$3$
- B$2$
- ✓$11$
- D$12$
$ = {10^{ - 3}}\,M \Rightarrow pOH = 3$
$pH + pOH = 14 \Rightarrow pH = 14 - 3 = 11$
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| Exp | $[A]$ | $[B]$ | Rate |
| $1$ | $0.012$ | $0.035$ | $0.10$ |
| $2$ | $0.024$ | $0.070$ | $0.80$ |
| $3$ | $0.024$ | $0.035$ | $0.10$ |
| $4$ | $0.012$ | $0.070$ | $0.80$ |

$A$, (molecular formula $\left.\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\right)$ with a straight chain structure gives a $C_{4}$ carboxylic acid. $A$ is :
$A \frac{{Li} {A} {H} {H}_{4}}{{H}_{3} {O}^{+}} \longrightarrow B \stackrel{\text { Oxidation }}{\longrightarrow} {C}_{4}-$ carboxylic acid