Analytical Science: Standard Additions Calibration

Video explaining standard additions calibration, including how it is performed and how to calculate the concentration of an unknown solution. Presented by Dr Daniel Belton, University Teaching Fellow, University of Huddersfield. Made with support from the Analytical Chemistry Trust Fund.

[@youdontwannaknow3397]

You’re a life saver. Thank you. This was so helpful.

[@iremcakar6545]

best and cleariest and absolute explanation ever, thank you so much! :D

[@hilarylee20]

Very helpful, needed this refresher. Much Thanks! :)

[@Whiteblack171]

Thank you , this is very helpful !

[@the-state-of-the-art4967]

This makes sense! good explanation.

[@randeepkaur9881]

Helpful video.
U tell me meaning of spike....
thank u very much

[@brianapomales3775]

Thank youuu!!

[@NGUYENQUANGKHAIINVEST]

thank you!

[@mijanrahaman2808]

Thank you very much. It is very helpful

[@lyn4713]

Why is it multiplied by 10?

[@MAngelier]

Because each time he added 10mL of the sample in a total volume of 100mL, so it gets 10 times diluted. The measured value is therefore actually 10x to low, to compensate for this, multiply by 10 in the end.

[@drdeepaksoni6639]

i found more area of analyte than standard in egg matrix ( LC-UV analysis)
sample 5 g
reconstitute volume 2ml
spiked conc 1 ppm
how do i calculate conc of analyte to find out recovery?
area of sample 57289
area of standard 42380

[@igsr1869]

Thank you. It was very useful. How can we find the percentage recovery?

[@ChemEngTutor]

To do this, you need to know the concentration of analyte you were expecting to get. You can then work out the experimentally determined concentration as a percentage of the expected concentration. I hope this helps.

[@mimosaya4462]

Hi, just wanna ask. Why do we need to multiply 10 to the calculated concentration? (As per the example calculation)

[@ChemEngTutor]

We need to multiply by 10 because the initial specimen solution was diluted by a factor of ten when the standards were made.

[@LOLA-yj9rr]

Why do we extrapolate it to when the instrument response is zero? Doesn't a zero response mean the absence of analyte?

[@ChemEngTutor]

We are trying to get the difference between when there is zero analyte (y = 0) and when we just at have the instrument response for the specimen (x = 0). I hope this helps.

[@omerakrawi8729]

from where 0.506 ???

[@funnysundayfunnys8537]

Use y=mx+c formula, 0.506 in this case is the gradient/slope of the sample.

[@matthulvey8615]

All values of the calibration plot are from least squares regression which finds the algebraic formula that best fits the raw data. The constants are merely the slope and intercept with x and y as your variables. Typically you would use software to calculate, but you can do so manually.

[@ericdominic8382]

Wondering too