Question

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

Answer 1

Step 1: Given data

Volume of solution= $0.05\mathrm{mL}$

Mole of ${\mathrm{H}}^{+}$= $3\times {10}^{-6}$ mole

The rate constant of the disappearance of ${\mathrm{H}}^{+}$= ${10}^7\mathrm{mol}{\mathrm{litre}}^{-1}{\sec }^{-1}$

Step 2: Calculating the concentration of drop

We know that the concentration of the dop is

$$\begin{gathered} \mathrm{Concentration}\mathrm{of}\mathrm{drop}=\frac{\mathrm{Moles}\mathrm{of}\mathrm{the}\mathrm{solution}}{\mathrm{Volume}(\mathrm{in}\mathrm{mL})}\times 1000 \\ \Rightarrow \mathrm{Concentration}\mathrm{of}\mathrm{drop}=\frac{3\times {10}^{-6}}{0.05}\times 1000 \\ \Rightarrow \mathrm{Concentration}\mathrm{of}\mathrm{drop}=0.06\mathrm{mol}{\mathrm{L}}^{-1} \end{gathered}$$

Step 3: Calculating the time

Let us assume the time taken for the${\mathrm{H}}^{+}$ in the drop to disappear be T

Now we know that the rate of disappearance is

$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{disappearance}=\frac{\mathrm{Change}\mathrm{in}\mathrm{concentration}}{\mathrm{Time}\left(\mathrm{T}\right)} \\ \Rightarrow \mathrm{Time}\left(\mathrm{T}\right)=\frac{\mathrm{Change}\mathrm{in}\mathrm{concentration}}{\mathrm{Rate}\mathrm{of}\mathrm{disappearance}} \\ \Rightarrow \mathrm{Time}\left(\mathrm{T}\right)=\frac{0.06}{{10}^7} \\ \Rightarrow \mathrm{Time}\left(\mathrm{T}\right)=6\times {10}^{-9}\sec \end{gathered}$$

Therefore, the time taken for the ${\mathrm{H}}^{+}$ in the drop to disappear is $6\times {10}^{-9}\sec$.