Question

The herbicide 2,4-D is being analyzed in plant matter using triphenylphosphate (TPP) as an internal standard. A calibration sample is prepared by pipetting 10.0 µL of 119 µg/mL 2,4-D and 10.0 µL of 100 µg/mL TPP into 1.00 mL of acetonitrile (ACN). (Hint: What is the total volume?) Separation by HPLC gave the peaks shown in the table below. The sample was analyzed by homogenizing 5.196 g of plant material and extracting the 2,4-D into 2.4 mL of ACN. 10.0 µL of 100 µg/mL TPP was added to the ACN extract before HPLC gave the peak areas shown below.

Peak Areas    2,4-D TPP

Calibration Solution 28685    15865

Plant Extract    2217    5816

Calculate the concentration of 2,4-D in the plant material in units of ng/g (ppb).

ng/g

Answer 1

From the first data

10 uL = 0.01 ml

concentration of TPP in solution = 100 ug/ml x 0.01 ml/1.02 ml = 0.98 ug/ml

concentration of 2,4-D in solution = 119 ug/ml x 0.01 ml/1.02 ml = 1.16 ug/ml

Area for TPP = 15865

Area for 2,4-D = 28685

Response factor (F) = (Area/conc.)2,4-D/(Area/conc.)TPP

                                 = (28685/1.16)/(15865/0.98) = 1.52

Now plant material analysis

concentration of TPP in solution = 100 ug/ml x 0.01 ml/2.4 ml = 0.416 ug/ml

area for TPP = 5816

area for 2,4-D = 2217

Using response factor equation,

Response factor (F) = (Area/conc.)2,4-D/(Area/conc.)TPP

1.52 = (2217/conc.)2,4-D/(5816/0.416)

so,

concentration of 2,4-D in solution = 0.1043 ug

grams of plant extract = 5.196 g

1 ug = 1000 ng

So,

concentration of 2,4-D in the plant material = 0.1043 ug x 1000/5.196 g = 20.07 ng/g (ppb)