What is the volume of dichromate ion solution required in the following scenario:Ionic Equation : 3SO2+ Cr2O7 (2-) + 2H (+) ----> 3SO4 (2-) +...

The ionic equation given here is 3SO2 + Cr2O7 (2- ) + 2H
(+) --> 3SO4 (2- ) + 2Cr (3+) + H20.

The polluted
air contains an amount of SO2 that occupies 4.8% of 3.0 dm^3 of the air. For every three
molecules of SO2 we require 1 ion of Cr2O7 (2- ).

At STP,
using PV = nRT, the moles of SO2 in the air is given by:

n
= PV / RT

=> n = [4.8%*100*1000*(3/1000)]/
273*8.3145

=> n =
4.8*3/273*8.3145

=> n = 6.34 * 10^-3
mol.

So we need the dichromate ions for 6.34 * 10^-3 moles
of SO2.

We need 2.11* 10^-3 mol of dichromate
ion.

Now there is 1 mole of dichromate ions in 10 dm^3 of
solution or in 10/1000 m^3 of the solution.

2.11*10^-3
moles will be found in 2.11*10^-3*10^-2 m^3 of the
solution.

The minimum volume of dichromate
solution required is 2.11*10^-5 m^3.