An atomic absorption spectroscopy method for the measurment of magnesium in seaw

Question

An atomic absorption spectroscopy method for the measurment of magnesium in seawater specifies a standard addictions approach. A stock solution of magnesium was made by adding 6.3810 g of dry MgCl2 into 500.00 mL of ultrapure water. Standard addition solutions were made by adding 0.00, 1.00, 2.00, 3.00, 4.00, 5.00 mL of stock solution to 5.00 mL of seawater and diluting the mixture to a total of 1000.00 mL using a volumetric flask. The concentration of added Mg2+ and measured absorbance of each sample are shown above.

a.) report the concentration of Mg2+ in the seawater sample in units of ppm Mg.

b.) seven successive blank corrected measurements of the absorbance of a very dilute Mg2+ standard had an average value of 0.0023 absorbance units with a standard deviation of 0.0041. What is the quantification limit of Mg2+ in solution using this instrument? Is this value greateer or less than the lowest amount of added internal standard? What does that result mean for our measurment?

measured vol. of vol. of Mg2+ stock seawater absorbance so n (mL) (mL) by flame AAS 0.085 none 0.131 0.162 0.206 0.258 0.289 Note: total volume 1000 mL

Explanation / Answer

a) ppm of Mg2+ in stock solution = 6.3810 x 1000/0.5 L = 12762 ppm

concentration of stock solution (Mg2+) in diluted samples,

when 1 ml of 12672 ppm stock standard added,

Mg2+ = 12762 x 1/1000 = 12.762 ppm

when 2 ml of 12672 ppm stock standard added,

Mg2+ = 12762 x 2/1000 = 25.524 ppm

when 3 ml of 12672 ppm stock standard added,

Mg2+ = 12762 x 3/1000 = 38.286 ppm

when 4 ml of 12672 ppm stock standard added,

Mg2+ = 12762 x 4/1000 = 51.048 ppm

when 5 ml of 12672 ppm stock standard added,

Mg2+ = 12762 x 5/1000 = 63.810 ppm

let x be the concentration of Mg2+ in solution,

then,

x/(x + 63.810) = 0.085/0.289

0.289x = 0.085x + 5.42385

x = 5.42385/0.204 = 26.5875 ppm is concentration of Mg2+ in sea water (5 ml)

b) average of blank = 0.0023

standard deviation = 0.0041

Limit of quantification = 0.0023 + 2 x 1.645(0.0041)

                                   = 0.015789

We can accurately measure the concentration of Mg2+ at this level for any quantification of the given experiment.

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