Question

Gas chromatography and the quantitative method of standard addition to constant volume were used to investigate the concentration of 2,4‑dichlorophenoxyacetic acid (a chlorinated phenoxy herbicide) in a water sample from Lake Eildon.

Liquid-liquid extraction using dichloromethane was used to extract 1.00 L of water sample and the combined extracts were made up to 25 mL in a volumetric flask. A 5.00 mL aliquot of the combined extract was added to each of two 10 mL volumetric flasks. To one flask was added 50.00 μL of a 1.150 x 10^-4 M standard solution of 2,4‑dichlorophenoxyacetic acid. Both flasks were then made up to the final volume. The GC analysis of 10.00 μL of injected sample from each flask resulted in peak areas of 245 mV.s and 290 mV.s for the signal in the chromatograms due to 2,4‑dichlorophenoxyacetic acid.

(Relative molar mass of 2,4-dichlorophenoxyacetic acid = 221.04 g.mol^-1)

(i). Calculate the concentration (mol/L) of 2,4-dichlorophenoxyacetic acid in the working standard solution used for quantitative analysis of the analyte

(ii). Write the standard addition equation and identify the terms to be applied

(iii). Apply the standard addition equation and determine the concentration in mg/L of 2,4‑dichlorophenoxyacetic acid in the original water sample from Lake Eildon

(iv). Identify the likely sources of experimental error in the laboratory method and calculate the percentage error in the value for the concentration of 2,4‑dichlorophenoxyacetic acid determined in (iii) (2 + 4 = 6 marks). Briefly, discuss how you would confirm the validity of the value determined in (iii)

(v). Discuss the likely health risks associated with a concentration of 30 μg/L of 2,4‑dichlorophenoxyacetic acid as compared to typical values found in Australian drinking waters

(vi) Discuss the effectiveness of dissolved oxygen to other common water treatment processes to effectively remove 2,4‑dichlorophenoxyacetic acid from Australian drinking waters

Answer 1

A long question. Let me go step by step:

i) The concentration of 2,4-dichlorophenoxyacetic acid of added standard is calculated like this:

Cf = Co x (Vo/Vf) where Cf: Final Analyte concentration Co = innitial analyte concentration; Vo = innitial volume Vf = final volume

Cf = 1.15x10^-4 x (50x10^-6 / 10 x 10^-3) = 5.75 x 10^-7 M

ii) The standard addition equation is:

Xi / Cf + Xf = Sx / Sx+c Where:

Xi: innitial concentration of the analyte in the original sample.

Cf: Final Analyte concentration

Xf: final concentration of the analyte in the original sample.

Sx = disolution innitial signal (without the added pattern)

Sx+c = disolution final signal. (With the added pattern)

iii) If we apply the previous equation we have:

Xi / (5.75x10^-7) + (5/10)Xi = 245/290 IF you solve for Xi:

0.8448 (5.75x10^-7 + 0.5Xi) = Xi

4.86x10^-7 + 0.4224Xi = Xi

Xi - 0.4224 = 4.86x10^-7

Xi = 8.4141x10^-7 mol/L

Xi (mg/L) = 8.4141x10^-7 x 221.04 x 10³ = 0.1859 mg/L

iv) Let me tell you first the method of standard addition definition:

"The method of standard addition is a type of quantitative analysis approach whereby the standard is added directly to the aliquots of analyzed sample. This method is used in situations where sample matrix also contributes to the analytical signal, a situation known as the matrix effect, thus making it impossible to compare the analytical signal between sample and standard using the traditional calibration curve approach"

According to this, the major experimental error in this method is the fact that it didn't use more added patrons. With more patterns, we should be able to graph a calibration curve, and with that, the concentration of the acid, would be more exact, and accurate. That's the major experimental error in this.

I don't quite understand well the question of percentage error, because I need more data to use the next equation:

% = [E(Xi - Xs)²/ n-1)^1/2 / Xs x 100 where Xs is the medium of the concentration.

Now if you use the only two data you have: Xf = 8.4141x10^-7 x (5/10) = 4.2074x10^-7

Xs = 4.2074x10^-7 + 8.4141x10^-7/ 2 = 6.3108x10^-7

% = [(8.4141x10^-7 - 6.3108x10^-7)² + (4.2074x10^-7 - 6.3108x10^-7)²]^1/2 / 6.3108x10^-7 x 100

% = 47.13 %