Question

A drop of a solution (volume 0.05 mL)
contains 3x10-mole H+ ions. If the rate of
disappearance of the Hions is 1x10'mol litre
sec- how long would it take for Ht ions in
the drop to disappear? Time = 6 x 10-9 sec]​

Answer 1

Answer:

6×10^-9 sec

Explanation:

Concentration of drop =Mole ×1000  / Volume in ml

=3×10−6 ×1000/0.05 =0.06 mol L−1

Rate of disappearance =Concentration  change / Time

1×10^7=0.06 / Time

Time =6×10^-9 sec

Second method:

Units of k(mol L−1 s−1) suggest it is a zero order reaction.

∴ For zero order =t=x/k=Conc used/Rate constant

0.05mL has =3×10−6mol of H⊕

1000mL has =3×10−6×103 /0.05

t=0.6×10−1/1.0×10−7=6×10−9 sec