Solute s has a partition coefficient of 6.2 between water (phase 1) and hexane (phase 2). 74.0 ml solution of s in water is extracted six times with 17.0 ml of hexane. calculate the fraction of s remaining in the aqueous phase.

Explanation:

The given data is as follows.

      Partition coefficient, K = 6.2

      Volume of phase 1, = 74.0 mL

      Volume of phase 2, = 17.0 mL

So, after one extraction fraction of solute remaining is given as follows.

              q =

After 3 times extraction, fraction of S remaining is as follows.

              q =

Putting the given values into the above formula as follows.

               q =

                  =

                  =

                  = 0.0699

Thus, we can conclude that the fraction of S remaining in the aqueous phase is 0.0699.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

it takes on an electronical energy