MCQ
If $4.0 \,gm $ $NaOH$ is present in $1$ litre solution, then its $pH$ will be
- A$6$
- ✓$13$
- C$18$
- D$24$
$[O{H^ - }] = {10^{ - 1}}M$, $[{H^ + }] = {10^{ - 13}}M$, $pH = 13$
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$xMnO_4^ - \, + \,y{C_2}O_4^{2 - }\, + \,z{H^ + }\, \rightarrow $$\,xM{n^{2 + }} + 2yC{O_2} + \frac{z}{2}{H_2}O$
The value's of $x, y$ and $z$ in the reaction are, respectively :
$\left( C _6 H _5\right)_3 C - Cl \frac{ OH ^{-}}{\text {Pyridine }}\left( C _6 H _5\right)_3 C - OH$