MCQ
$0.01\,\,M\,HA\,(aq.)$ is $2\%$ ionized $; [OH^-]$ of solution is
- A$2 \times 10^{-4}$
- B$ 10^{-8}$
- ✓$5 \times 10^{-11}$
- D$5 \times 10^{-12}$
$\Rightarrow \frac{\left[\mathrm{H}^{+}\right]}{[\mathrm{HA}]} \times 100=2$
$\Rightarrow \frac{\left[\mathrm{H}^{+}\right]}{0.01} \times 100=2$
$\Rightarrow\left[\mathrm{H}^{+}\right]=\frac{0.02}{100}=2 \times 10^{-4} \mathrm{M}$
$\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{2 \times 10^{-4}}=0.5 \times 10^{-10} \mathrm{m}$
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$X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q,$
respectively are in the ratio of $1 : 4.$ If the degree of dissociation of $X$ is $2$ times that of $Z,$ then the ratio of total pressure $(P_1 : P_2)$ at these equilibria is : (Assume degree of dissociation for both reactions are very very small)