Question

(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.

Answer 1

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.

                    = $x \times 0.030 L$

                    = 0.030x mole

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

            $24.9 \times 10^{-3} \times \frac{1}{1000} L$

          = $2.49 \times 10^{-5}$ mole

As the detector signal is additive in nature so, signal due to $2.49 \times 10^{-5}$ mole of ascorbic acid alone will be calculated as follows.

               $3.09 \muA - 2.12 \muA$        

             = 0.97 $\muA$

This shows that for $2.49 \times 10^{-5}$ mol, the signa;l observed is 0.97 $\muA$. Therefore, 1 $mol/\mu A$ will be equal to as follows.

       1 $mol/\mu A$ = $\frac{2.49 \times 10^{-5}}{0.97} \mu A$

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

             $\frac{2.49 \times 10^{-5}}{0.97} \times 2.12 mol$

           = $5.442 \times 10^{-5} mol$

Therefore, 0.030 x = $5.442 \times 10^{-5}$

                          x = 1.814 mM

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