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 129 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 4.874 g of plant material and extracting the 2,4-D into 2.9 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		27665		14827
Plant Extract		1313		6015

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 = 129 ug/ml x 0.01 ml/1.02 ml = 1.265 ug/ml

Area for TPP = 14827

Area for 2,4-D = 27665

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

                                 = (27665/1.265)/(14827/0.98) = 1.44

Now plant material analysis

concentration of TPP in solution = 100 ug/ml x 0.01 ml/2.91 ml = 0.344 ug/ml

area for TPP = 6015

area for 2,4-D = 1313

Using response factor equation,

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

1.44 = (1313/conc.)2,4-D/(6015/0.344)

so,

concentration of 2,4-D in solution = 0.05215 ug

grams of plant extract = 4.874 g

1 ug = 1000 ng

So,

concentration of 2,4-D in the plant material = 0.05215 ug x 1000/4.874 g = 10.70 ng/g (ppb)