A drop (0.05 ml) of solution contains 3 × 10^-7 moles of H^+. If the rate constant of disappearance of H^+ is 1.0 × 10^-7 mol l^-1 s^-1, how long (in min) would it take for H^+ in drop to disappear?

Question

Prashant Kotian · Accepted Answer

Correct answer: 1000. Initial concentration [H^+]0 = (3 × 10^-7 mol) / (0.05 × 10^-3 L) = 6 × 10^-3 M. For a zero-order reaction (indicated by rate constant units), complete disappearance time t = [A0] / k = (6 × 10^-3) / (1.0 × 10^-7) = 60000 seconds. In minutes, 60000 / 60 = 1000 min. Therefore, correct answer is 1000.

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