Question

When a groundwater sample was analyzed for the pe:rchlorate ion it gave a signal of 16.8 mV. When 2.00 ml of
0.0500 M pe:rchlorate standard (standard addition) adde,d to 150.0 ml of the groundwater sample, the: signal increased
to 2761 mV. Find the: concentration of perchlorate ion in the original groundwater sample.

Answer 1

Ans. Given,

            Signal of original groundwater sample = 16.8 mV

            Signal of spiked groundwater sample = 2761 mV

Now,

            Increase in signal due to spiking = 2761 mV – 16.8 mV

                                                            = 2744.2 mV

# Given, 2.00 mL of 0.0500 M perchlorate is added to 150 mL ground water sample.

Final volume of spiked solution = 2.0 mL (perchlorate) + 150.0 mL (groundwater sample)

                                                            = 152.0 mL

Using the following equation-                    C1V1 = C2V2            - equation 1

            Where, C1 = concentration; V1 = Volume           - Std. perchlorate solution

                        C2 = Concertation;   V2 = Volume           - spiked sample

            C1 = 0.0500 M          ; V1 = 2.00 mL

            C2 = ?                         ; V2 = 152 mL

Putting the values in equation 1-

            C2 = (0.0500 M x 2.00 mL) / 152.00 mL = 6.579 x 10^-4 M

So, addition of 2.0 mL of 0.050M perchlorate to 150 mL groundwater sample increase [perchlorate] by 6.579 x 10^-4 M.

# So far, we have-

            Spiking increases signal by 2744.2 mV.

            Spiking increase [perchlorate] by 6.579 x 10^-4 M.

Thus, a signal of 2744.2 mV is equivalent to a [perchlorate] of 6.579 x 10^-4 M.

Now,

[Perchlorate] in original water sample =

(6.579 x 10^-4 M / 2744.2 mV) x signal of pure sample

                                                            = (6.579 x 10^-4 M / 2744.2 mV) x 16.8 mV

                                                            = 4.027 x 10^-6 M

Therefore, [perchlorate] in original water sample = 4.027 x 10^-6 M