To perform the above analysis, you need to prepare a standardization curve, with

ID: 931443 • Letter: T

Question

To perform the above analysis, you need to prepare a standardization curve, with a set of known Na standards. The following standards were prepared in 100 mL Volumetric Flasks: 0.0 ppm, 0.50 ppm, 1.00 ppm, 1.50 ppm, 2.00 ppm. The instrument was set at 589.6 nm, and these solutions were analyzed with the following %T values:

0.00 ppm 100.0%T

0.50 ppm 82.8%T

1.00 ppm 68.5%T

1.50 ppm 57.1%T

2.00 ppm 47.6%T

Assume that the pathlength was 5.0 cm.

Graph the data, giving the slope, the intercept, and the correlation coefficient, r.

What is the epsilon value for this absorption?

You are to now finally anaylze an unknown urine sample. You take 250 uL of the urine unknown, dilute it in a 100.0 mL volumetric flask and then analyze it. It produces a %T = 84.1%T.

How many ppm of Na are in the original sample of urine?

Explanation / Answer

Conc, ppm

0.000

0.5

0.828

0.082

0.685

0.164

1.5

0.571

0.243

0.476

0.322

slope

SLOPE

0.161236

intercept

INTERCEPT

0.001171

CORREL

0.9999

A = 0.163 C + 0.001

C = (A-0.001)/0.163

For the unknown, A= log(1/T) = log(1/0.841) = 0.075

C = (0.075+0.001)/0.163 = 0.45 ppm

Correct for dilution

0.45 ppm x 1000mL/0.250mL = 1800 g/mL = 1.8 mg Na/mL urine