Question

In the Fluorometric analysis of Quinine in Tonic Water Lab

Prepare four standards of 0.15, 0.30, 0.45, and 0.60 μg/mL by dilution of the stock standard solution. Start with the lowest concentration standard. Use the appropriate micropipette to introduce 150 μL of the stock standard into the volumetric flask and dilute to 10.00 mL with 0.05 M H2SO4. Mix and transfer the solution to a vial. Repeat this process with 300 μL, 450 μL and 600 μL of stock standard.

Note:
- Acidic analysis solution: 0.05 M H2SO4 in distilled water.
- Stock quinine standard: 10 ppm (μg/ml) stock solution made in 0.05 M H2SO4.
- Quinine (C20H24N2O2, 324.43 g/mol)

Q: Calculate the actual concentrations of these standards.

Answer 1

Stock quinine standard = 10 ppm = 10 microgram/ml in 0.05 M H2SO4

To prepare the solutions with concentrations of,

0.15 microgram/ml (or ug/ml)

C1V1 = C2V2

with,

C1 = 10 ugram/ml

C2 = 0.15 ug/ml

V1 = ?

V2 = 10 ml

we get,

Volume of standard to be taken V1 = 0.15 x 10/10 = 0.15 ml

Transfer 0.15 ml by a pipetter and dilute to 10 ml by 0.05 M H2SO4

Final concentration of quinine in solution = 0.15 ug/ml

0.30 microgram/ml (or ug/ml)

C1V1 = C2V2

with,

C1 = 10 ugram/ml

C2 = 0.30 ug/ml

V1 = ?

V2 = 10 ml

we get,

Volume of standard to be taken V1 = 0.30 x 10/10 = 0.30 ml

Transfer 0.30 ml by a pipetter and dilute to 10 ml by 0.05 M H2SO4

Final concentration of quinine in solution = 0.30 ug/ml

0.45 microgram/ml (or ug/ml)

C1V1 = C2V2

with,

C1 = 10 ugram/ml

C2 = 0.45 ug/ml

V1 = ?

V2 = 10 ml

we get,

Volume of standard to be taken V1 = 0.45 x 10/10 = 0.45 ml

Transfer 0.45 ml by a pipetter and dilute to 10 ml by 0.05 M H2SO4

Final concentration of quinine in solution = 0.45 ug/ml

0.60 microgram/ml (or ug/ml)

C1V1 = C2V2

with,

C1 = 10 ugram/ml

C2 = 0.60 ug/ml

V1 = ?

V2 = 10 ml

we get,

Volume of standard to be taken V1 = 0.60 x 10/10 = 0.60 ml

Transfer 0.60 ml by a pipetter and dilute to 10 ml by 0.05 M H2SO4

Final concentration of quinine in solution = 0.60 ug/ml

Actual concentrations,

0.15 ug/ml

= 0.15 x 10^-3 mg x 324.43/0.010 = 4.9 x 10^-4 mg/L = 0.49 ppm

0.30 ug/ml

= 0.30 x 10^-3 mg x 324.43/0.010 = 0.97 x 10^-3 mg/L = 0.97 ppm

0.45 ug/ml

= 0.45 x 10^-3 mg x 324.43/0.010 = 1.47 x 10^-3 mg/L = 1.47 ppm

0.60 ug/ml

= 0.60 x 10^-3 mg x 324.43/0.010 = 1.96 x 10^-3 mg/L = 1.96 ppm