Question

Question #68d45

Answer 1

`sf(p=0.19)`

Explanation:

In this case the octanol phase is the least dense so this forms the upper layer.

After shaking the propanoic acid distributes itself between the two layers. A dynamic equilibrium is established:

`sf(C_2H_5COOH_("oct")rightleftharpoonsC_2H_5COOH_(H_2O))`

So we can write:

`sf(([C_2H_5COOH]_(H_2O))/([C_2H_5COOH]_("oct"))=p)`

Where p is the partition coefficient.

The total no. of moles of propanoic acid ( which I will shorten to PA) is given by:

`sf(n"PA"=cxxv=1xx10/1000=0.01)`

During the titration the acid is neutralised:

`sf(C_2H_5COOH+NaOHrarrC_2H_5COONa+H_2O)`

We can see from this that 1 mole of PA reacts with 1 mole of `sf(NaOH)`

`:.` `sf(n_(OH^-)=cxxv=0.1xx3.2/1000=0.00032)`

`:.` `sf(n"PA"_(H_2O)=0.00032)` in 10 ml of the water phase

This means that the total number of moles of PA in 50 ml of the water phase = 0.00032 x 5 = 0.0016

`:.` `sf(n"PA"_("oct")=0.01-0.0016=0.0084)`

Since the volumes of both phases are the same we can say:

`sf(p=(n"PA"_(H_2O))/(n"PA"_("oct"))=0.0016/(0.0084)=0.19)`