(a) [10 pts] Using an electrochemical method, you measured the ascorbic acid of a 30.0 mL watermelon juice sample. A detector signal of 2.12 uA was observed. A standard addition of 1.00 mL of 24.9 mM ascorbic acid standard increased the current to 3.09 uA. Find the concentration of vitamin C in the watermelon juice.

Explanation:

We assume that the concentration of unknown sample is x M.

As the volume is given as 30.0 ml or 0.030 L (as 1 L = 1000 ml) therefore, moles of ascorbic acid will be calculated as follows.

                    =

                    = 0.030x mole

Therefore, standard moles of ascorbic acid added are as follows.

          = mole

As the detector signal is additive in nature so, signal due to mole of ascorbic acid alone will be calculated as follows.

             = 0.97

This shows that for mol, the signa;l observed is 0.97 . Therefore, 1 will be equal to as follows.

       1 =

and, moles that will observe 2.12 signal will be calculated as follows.

           =

Therefore, 0.030 x =

                          x = 1.814 mM

Thus, we can conclude that the concentration of vitamin C in the watermelon juice is 1.814 mM.