MCQ
$pH$ of $ 0.005\, M\,\,{H_2}S{O_4}$ solution will be
- A$0.005$
- ✓$2$
- C$1$
- D$0.01$
$H _2 SO _{4( aq )} \longrightarrow 2 H ^{+}+ SO _4^{2-}$
Since the equation tells us that each molecule of acid will produce $2$ hydrogen ions, the concentration of the $H ^{+}$ion must be $2 \times 0.0050 \,M$ or $0.010 \,M$
Using the definition of $pH =-\log H ^{+}=-\log (0.010\, M )=2$
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$K + {H_2}O + $Water $ \to KOH(aq) + \frac{1}{2}{H_2};\,\Delta H = - 48\,kcal$
$KOH + $Water $ \to KOH(aq);\,\Delta H = - 14\,kcal$
The heat of formation of $KOH$ is (in kcal)
$(I)$ $C{H_2} = CH\mathop C\limits^ + HC{H_3}$
$\begin{array}{*{20}{c}} {{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\ {(II)\,\,\,\,\,\,\,\,\,\,C{H_2} = C - \mathop {{\text{ }}C}\limits^ + {H_2}} \end{array}$
$(III)$ $C{H_3}CH = CH\mathop C\limits^ + {H_2}$