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A drop of a solution (volume = 0.05 ml) contains ${6 5 10}^{-7} {mol of H}^{+}$ . If rate of disappearance of $H^{+}$ is how long will it take for the $H^{+}$ in the drop to disappear?

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$2.0 \times 10^{- 2} s$

$8.0 \times 10^{- 8} s$

$6.0 \times 10^{- 6} s$

$2.0 5 10^{- 8} s$

answer is B.

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Detailed Solution

$[H^{+}] = \frac{{6 ×10}^{-7}}{{0.05 × 10}^{-3}} {= 1.2 × 10}^{-2} M$

$Rate r = \frac{ΔX}{Δt}; Δt = \frac{ΔX}{r} = \frac{{1.2 × 10}^{-2} M}{{6 × 10}^{5} {Ms}^{-1}}$

$∴ {T = 2 × 10}^{-8} s$

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