Question

A drop of a solution (volume = 0.05 mL) contains $6\times {10}^{-7}$ moles of $H^{+}$. If the rate of disappearance of $H^{+}$ is $6.0\times {10}^5{\text{mol L}}^{-1}{\text{s}}^{-1}$, how long will it take for the $H^{+}$ in the drop to disappear?

• A. $8.0\times {10}^{-8}\text{s}$
• B. $2.0\times {10}^{-8}\text{s}$
• C. $6.0\times {10}^{-6}\text{s}$
• D. $2.0\times {10}^{-2}\text{s}$

Answer 1

The correct option is B $2.0\times {10}^{-8}\text{s}$

$\left[H^{+}\right]=\frac{{\text{moles of H}}^{+}}{\text{Volume}}=\frac{6\times {10}^{-7}\text{mol}}{0.05\times {10}^{-3}\text{L}}$
$=1.2\times {10}^{-2}\text{M}$
Now, $r=\frac{-d\left[H^{+}\right]}{dt}=6.0\times {10}^5{\text{mol L}}^{-}{\text{s}}^{-1}$
$=\frac{1.2\times {10}^{-2}\text{M}}{dt}=6.0\times {10}^5{\text{mol L}}^{-1}{\text{s}}^{-1}$
$\therefore dt=\frac{1.2\times {10}^{-2}\text{M}}{6.0\times {10}^5{\text{Ms}}^{-1}}=2\times {10}^{-8}\text{s}$