Please complete this problem fully with details. Original undiluted standard gly

Question

Please complete this problem fully with details.

Original undiluted standard glycine = 7.5 mg mL-1

1mL of standard glycine is pipetted into 100mL volumetric flask and diluted. Six standards are produced.

Standard 1: 0 diluted std. glycine, 0.50 Distilled Water
Standard 2: 0.10 diluted std. glycine, 0.40 Distilled Water
Standard 3: 0.20 diluted std. glycine, 0.30 Distilled Water
Standard 4: 0.30 diluted std. glycine, 0.20 Distilled Water
Standard 5: 0.40 diluted std. glycine, 0.10 Distilled Water
Standard 6: 0.50 diluted std. glycine, 0 Distilled Water

2.0mL of ninhydrin reagent was added into each tube and put into water bath for 20min. When taken out and cooled, 8.0mL of 50%ethanol was added the absorbance was measured at a wavelength of 570nm.

Standard 1: 0
Standard 2: 0.0095
Standard 3: 0.0172
Standard 4: 0.0216
Standard 5: 0.0327
Standard 6: 0.042
Q: How do I find the amount of glycine in each standard?

Explanation / Answer

Use the dilution equation to arrive at the concentrations of glycin for the stock solution and the standards.

M1*V1 = M2*V2 where M1 = concentration of the concentrated solution; M2 = concentration of the dilute solution; V1 = volume of concentrated solution and V2 = volume of dilute solution.

Stock Solution:

(1 mL)*(7.5 mg mL-1) = Cstock*(100 mL)

===> Cstock = 0.075 mg mL-1

The stock solution is used to prepare the standard solutions.

Standard 1:

(0.0 mL)*(0.075 mg mL-1) = CS1*(0.50 mL)

===> CS1 = (0.0)*(0.075)/(0.50) mg mL-1 = 0.0 mg mL-1 (ans)

Standard 2:

(0.10 mL)*(0.075 mg mL-1) = CS2*(0.50 mL)

===> CS2 = (0.10)*(0.075)/(0.50) mg mL-1 = 0.015 mg mL-1 (ans)

Standard 3:

(0.20 mL)*(0.075 mg mL-1) = CS3*(0.50 mL)

===> CS3 = (0.20)*(0.075)/(0.50) mg mL-1 = 0.030 mg mL-1 (ans)

Standard 4:

(0.30 mL)*(0.075 mg mL-1) = CS4*(0.50 mL)

===> CS4 = (0.30)*(0.075)/(0.50) mg mL-1 = 0.045 mg mL-1 (ans)

Standard 5:

(0.40 mL)*(0.075 mg mL-1) = CS5*(0.50 mL)

===> CS5 = (0.40)*(0.075)/(0.50) mg mL-1 = 0.060 mg mL-1 (ans)

Standard 6:

(0.50 mL)*(0.075 mg mL-1) = CS6*(0.50 mL)

===> CS6 = (0.50)*(0.075)/(0.50) mg mL-1 = 0.075 mg mL-1 (ans)

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