Question

# A drop of solution (volume 0.05 mL) contains $3×{10}^{-6}$ mole of ${\mathrm{H}}^{+}$. If the rate constant of the disappearance of ${\mathrm{H}}^{+}$ is ${10}^{7}\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{sec}}^{-1}$, how long would it take for ${\mathrm{H}}^{+}$ in the drop to disappear?

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Solution

## Step 1: Given dataVolume of solution= $0.05\mathrm{mL}$Mole of ${\mathrm{H}}^{+}$= $3×{10}^{-6}$ moleThe rate constant of the disappearance of ${\mathrm{H}}^{+}$= ${10}^{7}\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{sec}}^{-1}$Step 2: Calculating the concentration of dropWe know that the concentration of the dop is $\mathrm{Concentration}\mathrm{of}\mathrm{drop}=\frac{\mathrm{Moles}\mathrm{of}\mathrm{the}\mathrm{solution}}{\mathrm{Volume}\left(\mathrm{in}\mathrm{mL}\right)}×1000\phantom{\rule{0ex}{0ex}}⇒\mathrm{Concentration}\mathrm{of}\mathrm{drop}=\frac{3×{10}^{-6}}{0.05}×1000\phantom{\rule{0ex}{0ex}}⇒\mathrm{Concentration}\mathrm{of}\mathrm{drop}=0.06\mathrm{mol}{\mathrm{L}}^{-1}$Step 3: Calculating the timeLet us assume the time taken for the${\mathrm{H}}^{+}$ in the drop to disappear be TNow we know that the rate of disappearance is $\mathrm{Rate}\mathrm{of}\mathrm{disappearance}=\frac{\mathrm{Change}\mathrm{in}\mathrm{concentration}}{\mathrm{Time}\left(\mathrm{T}\right)}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Time}\left(\mathrm{T}\right)=\frac{\mathrm{Change}\mathrm{in}\mathrm{concentration}}{\mathrm{Rate}\mathrm{of}\mathrm{disappearance}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Time}\left(\mathrm{T}\right)=\frac{0.06}{{10}^{7}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Time}\left(\mathrm{T}\right)=6×{10}^{-9}\mathrm{sec}$Therefore, the time taken for the ${\mathrm{H}}^{+}$ in the drop to disappear is $6×{10}^{-9}\mathrm{sec}$.

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