Answer the following questions as TRUE/FALSE The limit of detection (LOD) is a s

Question

Answer the following questions as TRUE/FALSE The limit of detection (LOD) is a smaller value than the limit of quantitation (LOQ): _____ When using an internal standard, the standard added is the same as the analyte:_____ To make a spike sample, one needs to add a known amount of analyte to a sample:_____ When a groundwater sample was analyzed for the perchlorate ion it gave a signal of 12.5 mV. When 1.50 ml. of 0.0500 M perchlorate standard was added to 100.0 ml. of the groundwater sample, the signal increased to 25.1 mV. Find the concentration of petchlorate ion in the original groundwater sample. Show your work for full credit!

Explanation / Answer

Ans. 3A. True

            LOD = 3 x standard deviation

            LOQ = 10 x standard deviation

Thus, LOD is smaller than LOQ.

#3B. True

The standard being added is the same from the analyte (say, Zn2+ in river water) being analyzed. The standard is a solution consisting of pure molecule (Pure Zn2+ solution prepared separately) of a defined concentration.  

#3C. True

To make a spike sample, a known amount of standard analyte (solution) is added to the sample. Note: It’s assumed that the ‘analyte’ in the sentence means ‘standard analyte’.

Ans. 4. Preparation of spiked solution:

            Volume of standard perchlorate ion solution = 1.50 mL = 0.0015 L  

Molarity of solution = 0.500M                               ; [1 L = 1000 mL]

Moles of perchlorate ion = Molarity x Volume (L) of solution

                                                = 0.5 M x 0.0015 L = 0.00075 mol

Mass of perchlorate ion = moles x molar mass

                                                = 0.00075 mol x (99.45 g mol-1)

                                                = 0.0745875 g

                                                = 74.5875 mg

Final volume of spiked solution = 100.0 mL (sample) + 1.5 mL (standard)

                                                            = 101.5 mL

Now,

Concentration of standard perchlorate ion in spiked solution =

Mass of standard perchlorate ion / volume of spiked solution

= 74.5875 mg / 101.5 mL

= 0.73485 mg/mL

Note: this concentration of perchlorate ion 0.73485 mg/mL is solely due to addition of standard perchlorate ion.

Given,

            Signal of groundwater sample = 12.5 mV

Signal of spiked solution = 25.1 mV

Increase in signal due to spiking = (25.1 – 12.5) mV = 12.6 mV

The increase in signal of spied solution with respect to the sample is due to addition of spiked solution.

That is, 12.6 mV signal is equivalent to concentration of standard chlorate ion.

So,

            12.6 mV signal is equivalent to a concentration of 0.73485 mg/mL

            Or, 1 mV                     -           -           -           -           (0.73485 / 12.6) mg/mL

            12.5 mV                      -           -           -           -           (0.73485 / 12.6) x 12.5 mg/mL

                                                                                                = 0.7290 mg/ mL

Therefore, the 12.5 mV signal of the sample means 0.7290 mg/ mL perchlorate ion concentration in it.

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