I have a question in the monograph of sodium chloride. In the limit test of bromides the limit is 100 ppm but the concentration of the standard is much lower than the limit. So I want to know why.


Maximum 100 ppm.

To 0.5 mL of solution S add 4.0 mL of water R, 2.0 mL of phenol red solution R2 and 1.0 mL of a 0.1 g/L solution of chloramine R and mix immediately. After exactly 2 min, add 0.15 mL of 0.1 M sodium thiosulfate, mix and dilute to 10.0 mL with water R. The absorbance (2.2.25) of the solution measured at 590 nm, using water R as the compensation liquid, is not greater than that of a standard prepared at the same time and in the same manner, using 5.0 mL of a 3.0 mg/L solution of potassium bromide R.


Without doing all the math I can hazard an educated guess.

First let's clear up the question.

For the sample its initial volume of 0.5 ml becomes 7.5 ml after all the reagents are added. So a 100 ppm max of $\ce{Br^-}$ becomes 6.67 ppm in the analyzed solution.

For the standard the solution is 67.1% $\ce{Br^-}$. I'm assuming it doesn't get diluted with the 4.5 ml of water, so the final volume would be 8 ml with the reagents. Thus the final solution would be 1.26 pmm.

So the difference is about a factor of 5.

Now typically one would create a calibration curve with multiple standards and calculate a best fit line. Around the best fit line the best precision will be at some point of line segment. Let's call that the "sweet spot" on the calibration curve.

Since absorbance is linear with concentration, the fitted line should go through the origin. Thus only one standard is really necessary. (Be careful here. Absorbance is linear with concentration, but the instrument is measuring transmission. So small differences in a lot of light are hard to measure, and small differences in very little light correspond to large differences in concentration.)

For all the error sources in the analysis the sweet spot must be around 1.26 ppm on the calibration curve.

| improve this answer | |
  • $\begingroup$ I thought about the answer and I found it. Thanks for you efforts. $\endgroup$ – abanoub Jul 28 '18 at 17:33
  • $\begingroup$ @abanoub - Was the answer something different? Bit of an explanation? Please add your own answer. $\endgroup$ – MaxW Jul 28 '18 at 17:44

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