A drop of solution (Volume \(0.05 mL\)) contains \(3.0 × 10^{-6}\) mole of \(H^+\) ions. If the rate constant of disappearance of \(H^+\) is \(1.0 × 10^{-7}\, \ mol L^{-1} s^{-1}\), how long will take \(H^+\) ions to disappear?

\(6 × 10^{-9} s\)

The correct answer is option 2. \(6 × 10^{-9} s\).

Given:

Rate constant (\(k\)) = \(1.0 \times 10^{-7} \, \text{mol L}^{-1} \text{ s}^{-1}\)

Number of moles of \(H^+\) ions = \(3.0 \times 10^{-6}\) mol

Volume of the solution = \(0.05 \, \text{mL}\) = \(0.05 \times 10^{-3} \, \text{L}\)

Thus,

Concentration of drop \(= \frac{\text{mole}}{\text{Volume in mL}}× 1000\)

Concentration of drop\(= \frac{3.0 \times 10^{-6}}{0.05} = 0.06 \text{mol L}^{-1}\)

The rate of disappearance is \(1 × 10^7 \text{mol L}^{-1}\) per second

Therefore, The rate of disappearance of \(0.06 \text{mol L}^{-1}\) will be

\(\frac{0.06}{10^7} = 6 × 10^{-9} s\)