Question

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

• A. $6\times {10}^{-8}sec$
• B. $6\times {10}^{-9}sec$
• C. $6\times {10}^{-10}sec$
• D. $6\times {10}^{-12}sec$

Answer 1

The correct option is B $6\times {10}^{-9}sec$

$k={10}^7{molL}^{-1}{sec}^{-1}\left[H^{+}\right]=\frac{3\times {10}^{-6}}{0.05\times {10}^{-3}}{molL}^{-1}=\frac{3}{5}\times {10}^{-1}mol{L}^{-1}\therefore \ln \left(\frac{A_o}{A}\right)=kt{t}_{99\%}=\frac{\ln \left(\frac{A_o}{A}\right)}{k}=\frac{\ln \left(\frac{3/5\times {10}^{-1}}{3/5\times {10}^{-3}}\right)}{{10}^7}=9.212\times {10}^{-3}secSo,v=k\left[H^{+}\right]=0.6\times {10}^{-1}\times {10}^7=0.6\times {10}^6=6\times {10}^5{mol}^2{L}^{-1}{s}^{-1}\therefore t=6\times {10}^{-9}sec$ for $\frac{3}{5}\times {10}^{-1}mol{L}^{-1}$