Help with a pre lab question! Prepare two weighing bottles (clean, rinse with DI

ID: 1051728 • Letter: H

Question

Help with a pre lab question!

Prepare two weighing bottles (clean, rinse with DI water, dry in the oven)

2. Mark one weighing bottle – “HPLC unknown sample”, your name, obtain the unknown sample from the Instructor.

3. Using syringe filter (45 mm) filter the unknown sample to the second clean, dry weighing bottle.

5. Preparation of sample for injection: 50 mL of the filtered unknown sample + 100 mL of tyrosine standards + 850 mL of DI water into 1 mL HPLC vial.

6. From Mix of standards (1mg/mL) using micropipettes and HPLC vials prepare following solutions for calibration curve: 0.010 mg/mL, 0.020 mg/mL, 0.030 mg/mL, 0.040 mg/mL and 0.050 mg/mL, each solution containing 0.1 mg/mL of tyrosine (internal standard). WRITE A PROCEDURE (IN A TABLE) HOW TO PREPARE SOLUTIONS FOR CALIBRATION CURVE AND HAVE IT APPROVED BY INSTRUCTOR.

I need help with making the table for part six. Thank you!

Explanation / Answer

The standard mix of solutions is 1 mg/mL while the standard solutions used have concentration of the absorbing species as low as 0.010 mg/mL.

The dilution factor is (1 mg/mL)/(0.010 mg/mL) = 100.

This would mean that if you are using a 1 mL micropipette, you will need to pipette out 0.010 mL of 1 mg/mL standard mix into a 1 mL vial and make up the volume. This is clearly cumbersome, so we will do a two step dilution.

1) We will prepare a 100 mL 0.10 mg/mL stock solution first.

2) We shall dilute the 0.10 mg/mL stock solution as required.

To prepare the 0.10 mg/mL stock solution from 1 mg/mL stock solution: The volume is 100 mL.

Use the dilution equation. Let V mL of 1 mg/mL standard mix be required.

(V mL)*(1 mg/mL) = (100 mL)*(0.10 mg/mL)

=====> V = 100*0.10/1 = 10

Take 10 mL of 1 mg/mL standard mix and dilute with tyrosine standards and D.I. water in a 100 mL volumetric flask till the 100 mL mark.

To prepare the required standard solutions for Beer’s law:

Let us see the first example. We want to prepare 1 mL of 0.01 mg/mL solution.

Let V’ mL of 0.10 mg/mL stock solution from the last step be required.

Therefore, use the dilution law to obtain

(V’ mL)*(0.10 mg/mL) = (1 mL)*(0.010 mg/mL)

====> V’ = 0.1

Therefore the volume required = 0.1 mL = 100 µL.

This volume is easily measurable. A micropipette can easily measure volumes from 100-1000 µL and we shall employ a micropipette here.

So, to prepare 0.010 mg/mL standard solution, we deliver 100 µL stock solution (0.10 mg/mL) into a 1 mL flask and make up the volume to the mark with tyrosine and D.I. water.

The rest of the table can be filled in as below:

Trial

Concentration of standard for Beer’s law determination

Volume of 0.10 mg/mL stock solution required (mL)

Volume of 0.10 mg/mL stock solution required (µL)

0.010

0.1

100

0.020

0.2

200

0.030

0.3

300

0.040

0.4

400

0.050

0.5

500