Question

1.(20%) The phosphorus content of soft drinks may be determined spectrophotometrically at 830 nm. The phosphoric acid in the sample is treated with ammonium molybdate followed by ascorbic acid in order to convert the phosphorus to a blue compound.

A series of standard solutions was prepared and their absorbances were measured at 830 nm:

Conc. Of P (ppm)	Absorbance
0.175	0.255
0.350	0.588
0.524	0.866
0.699	1.151
0.874	1.436

Three samples of the same soft drink were then prepared by allowing the sample to stand open to the atmosphere for 24 h, then taking 2.50-mL aliquots and diluting to 50.0-mL. After development of the blue color, the following results were obtained:

Sample Number	Absorbance
1	0.565
2	0.571
3	0.563

a. Determine the molar absorptivity of the blue substance.

b. Determine the average concentration of P in the soft drink. Include the standard deviation and the 95% confidence limits.

c. The manufacturer of the soft drink claims a P content of 7.87 ppm. Is the manufacturer being truthful? Explain your answer. No credit for a lucky guess.

Answer 1

Plot concentration of P vs absorption. Slope of the graph gives the value to molar absorptivity.

a. absorptivity = 0.5996

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A = *c*l

or, c = A/ * l

concentraion in sample 1 : 0.565/0.5996 = 0.9423

concentration in sample 2 :0.571/0.5996 = 0.9523

concentration in sample 3 : 0.563/0.5996 = 0.9389

average = 0.9445 ppm

Standard deviation = 0.006966

Standerd error = 0.006966/3 =0.004021

concentration of P = means + standerd error * t value

                            = 0.9445 + 0.004021 *2.353

                            =0.9445 + 0.0094

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in 50mL dilute solution P concentration is 0.9445 ppm. concentation of P in 2.5mL aliquote is

0.9445 * 50mL/2.5mL =18.89 ppm

So, the concentration of P is more than what is reported.