Spectrophotometric Determination of Nitrite

Dilution of test samples

Since the calibration standards prepared range between 1.00 and 4.00 mg L-1 it is necessary to dilute the sample. A suitable dilution factor is 1/5. Two diluted samples will be prepared in order to check reproducibility of results.

• Prepare two diluted test samples by pipetting an appropriate volume of the neat sample, 20mL into each of two clean 100.0 mL volumetric flasks.

• Dilute each flask to the mark with distilled water and mix well.

Visual observation

Visual interpretation of the result shows that the samples concentrations are values between (2) 2.00 mg L-1 and (4)4.00 mg L-1. The colour of sample 1 and sample 2 are very similar in appearance. So visually they can be arrange as 1-2-sample1-sample2 and 4.

Analytical data

Nitrite Solution (mg L-1) Absorbance @ 540 nm

0.00 0.0000

1.00 0.0872

2.00 0.1841

4.00 0.3671

S1 sample 2.42 0.2288

S2 sample 2.45 0.2229

Neat sample’s colour was intense which indicates that its concentration is higher than number 1 and 2 standard but much less than 4. That could be a value between 2 mg L-1 and 3 mg L-1.

Using the interpolation technique the set point equivalent to this in X for the nitrite in the unknown samples is:

• S1 2.42 mg L-1

• S2 2.45 mg L-1

The dilution process was a necessary process to measure samples on an appropriated scale.

Result

The nitrite level in the undiluted sample considering the dilution factor is:

Dilution factor = 1/5

2.45 mg L-1*5= 12.25 mg L-1

2.42 mg L-1*5=12.10 mg L-1

Average = 12.25 mg L-1 12.10 mg L-1 = 12.18 mg L-1 NO2 -N

2

So the concentration of the undiluted solution is 12.18 mg L-1 NO2 –N, according with the lab manual the test sample supplied has a nitrite level of approximately 12 mg L-1 a very close value to the one here.

Relative spread for the absorbance values of the two dilutes samples it can be calculate as follow:

.

Average= (0.2288+0.2229)/2 = 0.22585

RS = 0.2228- 0.2229 x100 = 2.61%...