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In the Fluorometric analysis of Quinine in Tonic Water Lab Prepare four standard

ID: 928388 • Letter: I

Question

In the Fluorometric analysis of Quinine in Tonic Water Lab

Prepare four standards of 0.15, 0.30, 0.45, and 0.60 g/mL by dilution of the stock standard solution. Start with the lowest concentration standard. Use the appropriate micropipette to introduce 150 L 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 L, 450 L and 600 L of stock standard.

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

Q: Calculate the actual concentrations of these standards.

Explanation / Answer

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