A drop of solution volume 0-05 mL contains 3- 10 -6 mole H - ions- If the rate of disappearance of the H - ions is 1- 10 7 mol litre -1 sec -1- how long would it take for H - ions in the drop to disappear-

A drop of solution (volume = 0.050 mL) contains $6 \times 10^{- 7} m o l o f H_{3} O^{+}$ ions. If $H_{3} O^{+}$ ions disappear at a constant rate of $6.0 \times 10^{5}$ mol $L^{- 1} S^{- 1}$ , how long will take for the $H_{3} O^{+}$ ions in the drop to disappear ?

Q. A drop of a solution (volume

= 0.05 m l

) contains

6 \times 10^{- 7} m o l e

H^{+}

If the rate of disappearance of

H^{+}

6.0 \times 10^{5} m o l L^{- 1} s^{- 1}

How long will it take for the

H^{+}

in the drop to disappear ?

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A drop of solution (volume 0.05 mL) contains 3×10−6 mole H+ ions. If the rate of disappearance of the H+ ions is 1×107 mol litre−1 sec−1, how long would it take for H+ ions in the drop to disappear?

Concentration of drop =molevolumeinmL×1000 =3×10−60.05×1000=0.06 mol litre−1Rate of disappearance =conc.changetime1×107=0.06timeTime =6×10−9sec

A drop of solution (volume $0.05 m L$ ) contains $3 \times 10^{- 6} m o l e$ $H^{+}$ ions. If the rate of disappearance of the $H^{+}$ ions is $1 \times 10^{7}$ $m o l$ ${l i t r e}^{- 1}$ ${s e c}^{- 1}$ , how long would it take for $H^{+}$ ions in the drop to disappear?

Concentration of drop $= \frac{m o l e}{v o l u m e i n m L} \times 1000$
$= \frac{3 \times 10^{- 6}}{0.05} \times 1000 = 0.06$ $m o l$ ${l i t r e}^{- 1}$
Rate of disappearance $= \frac{c o n c . c h a n g e}{t i m e}$
$1 \times 10^{7} = \frac{0.06}{t i m e}$
Time $= 6 \times 10^{- 9} s e c$