MCQ
Assuming complete ionisation, the $pH$ of $0.1\,M\,HCl,$ is $1$. The molarity of ${H_2}S{O_4}$ with the same $pH$ is
- ✓$0.05$
- B$0.2$
- C$0.1$
- D$2.0$
$\therefore\left[\mathrm{H}^{+}\right]=0.1 \mathrm{M}$
The acid dissociation reaction of $\mathrm{H}_2 \mathrm{SO}_4$ is :
$\mathrm{H}_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{H}^{+}+\mathrm{HSO}_4^{-} $
$\mathrm{HSO}_4^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{SO}_4^{2-}$
$\therefore 1 \mathrm{H}_2 \mathrm{SO}_4$ ionizes to give $2 \mathrm{H}^{+}$ions.
$2 \mathrm{MH}^{+} \text {ions }=1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4 $
$1 \mathrm{MH}^{+} \text {ions }=\frac{1}{2} \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4 $
$0.1 \mathrm{MH}^{+} \text {ions }=\frac{0.1}{2} \mathrm{M} \mathrm{H}_2 \mathrm{SO}$$_4=0.05 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
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$N{H_3}(g)\, \rightleftharpoons \,\frac{1}{2}{N_2}(g)\, + \,\frac{3}{2}{H_2}(g)$
What is the $K_P$ for following reaction ?
$\,\,{N_2}(g)\, + \,3{H_2}(g) \rightleftharpoons 2N{H_3}(g)$