Question

The amount of sodium can be determined in an apple using atomic absorption spectroscopy. 0.4326 g of apple was dried, heated at 700oC and chemically treated with a concentrated HCl solution and diluted to 250 mL. 10 mL aliquots were transferred to 50 mL flasks. Each was treated with a standard solution of Na+ at 25.0 ppm and analyzed using flame AAS. Construct a standard addition graph and calculate the weight percentage of Na in the apple.

Na+ standard (mL)	Absorbance
0.00	0.2633
1.00	0.3313
5.00	0.4553
10.00	0.5613
15.00	0.7373
20.00	0.8513
25.00	1.1053
30.00	1.4073

Answer 1

Ans. Part A: Determining [Na⁺] in 10.0 mL aliquot using standard addition method.

Prepare a standard addition graph of final [Na⁺] contributed by addition of standard Na⁺ solution vs their respective absorbance as shown in picture.   

The trendline equation for the Abs vs concertation graph is “y = 0.0699x + 0.2508” in the form of y = m x + c. where, y-axis = Absorbance, and x-axis = [Na⁺] in ppm; Y-Intercept = c ; Slope = m

Now,

[Na⁺] in un-spiked aliquot = (y-Intercept / Slope) concertation units

= (0.2508 / 0.0699) ppm

Hence, [Na⁺] in 50.0 mL un-spiked aliquot (aliquot 1) = 3.5880 ppm

# The un-spiked aliquot (aliquot) is prepared by taking 10.0 mL of original digested and diluted (to 250.0 mL) sample and making the final volume to 50.0 mL.

Now, using    C1V1 (digested sample) = C2V2 (un-spiked aliquot)

                        Or, C1 x 10.0 mL = 3.5880 ppm x 50.0 mL

                        Or, C1 = (3.5880 ppm x 50.0 mL) / 10.0 mL = 17.9400 ppm

Therefore, [Na⁺] in 10.0 mL of of original digested and diluted sample = 17.9400 ppm

Part B: Calculate total Na+ content in original 250.0 mL digest.

Since 10.0 mL aliquot is taken from original digested and diluted (to 250.0 mL) sample, both the solutions have same concentrations.

So,

            [Na⁺] in original 250.mL digest = 17.9400 ppm = 17.9400 mg/ L solution

Note: 1 ppm = 1 mg of solute per liter of solution

Now,

            Na⁺ content of original 250 mL digest = [Na⁺] x volume of digest in liter

                                                                        = (17.9400 mg/ L) x 0.250 L

                                                                        = 4.4850 mg

                                                                        = 0.004485 g

Part C: Calculate mass % of Na⁺ in apple sample

Since 250 mL original digest is prepared from 0.4326 g apple sample, both the samples must have equal amount of total Na⁺ in them.

So,

            Amount of Na⁺ in 0.4326 g apple = 0.004485 g

Now,

            % Na⁺ (w/w) = (Mass of Na⁺ / Mass of sample) x 100

                                    = (0.004485 g / 0.4326 g) x 100

                                    = 1.0367 %

Hence, % Na⁺ (w/w) in apple sample = 1.0367 %