a) Phosphate can be determined spectrophotometrically by using the molybdenum bl

Question

a) Phosphate can be determined spectrophotometrically by using the molybdenum blue method. A 2.00 mL urine sample is treated with molybdenum blue reagents and diluted to 100. mL. An aliquot is removed and placed in a 1.0 cm cell and the absorbance at 820 nm is recorded as 0.428.  To a second 2.00 mL urine sample, 1.00 mL of a 5.0 X10-4 M phosphate standard is added and then diluted to 100. mL.  Again an aliquot is removed and the absorbance at 820 nm is recorded as 0.517.  Calculate the concentration of phosphate in the urine sample in M

Explanation / Answer

Ans. #Step 1: Let the phosphate concertation in original urine sample = X

# 2.00 mL original urine sample is diluted to 100.0 mL. Let’s label is as solution 1 - that give absorbance of 0.428.

Now, using    C1V1 (original urine sample) = C2V2 (diluted solution)

            Or, X x 2.0 mL = C2 x 100.0 mL

            Or, C2 = (X x 2.0 mL) / 100.0 mL = 0.02X

So, an phosphate concertation of 0.02X gives an abs of 0.428.

# Step 2: 1.00 mL standard phosphate solution is added to 2.00 mL urine sample and finally diluted to 100.0 mL.

# Increase in phosphate concertation due to spiking, C2 =

(5.0 x 10-4 M x 1.0 mL) / 100 mL

= 5.0 x 10-6 M

# Step 3: Increase in absorbance = Abs of spiked aliquot – Abs of un-spiked aliquot

                                                = 0.517 – 0.428

                                                = 0.089

# Step 4: Since Abs is proportional to concertation, an increase in phosphate concertation of 5.0 x 10-6 M must be equivalent to an abs of 0.089.

Concentration of un-spiked aliquot =

Spiking (Increase in conc. / Increase in Abs) x Abs of un-spiked aliquot

                                                = (5.0 x 10-6 M / 0.089) x 0.428

                                                = 2.4045 x 10-5 M

# Step 5: As in # step 1, the concertation of phosphate in u-spiked aliquot is equal to 0.02X.

So,

            0.02X = 2.4045 x 10-5 M

            Or, X = 2.4045 x 10-5 M / 0.02 = 0.00120 M

Therefore, [Phosphate] in original urine sample = 0.00120 M = 1.20 x 10-3 M

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