Calculation of p-nitrophenol - (Oct/01/2006 )

Hi there

I'm unsure how to work out this question

in my test tube I have 1.9mL of glycine buffer and 0.1mL of p-nitrophenol and then I am suppose to calculate how much p-nitrophenol (in nmol) is present!!! can someone please help

Info.
0.05m glycine buffer, pH 9.5
0.25mM p-nitrophenol, in 0.05M glycine buffer, pH 9.5
0.2M NaOH

Thanks
biology_06r

-biology_06er-

QUOTE (biology_06er @ Oct 1 2006, 01:11 AM)

You start with 0.1 ml of 0.25mM (250 uM, or 250,000 nM) p-NP, final volume is 2 ml, so you diluted it 20-fold. The final concentration will be 250,000 nm/20=12,500 nM.

-genehunter-1-

Is the 0.25 mM the final concetration or the concentration before dilution in glycine?

If it's the final concentration then the amount in nanomoles (nmol) is:

cV/1000 = 0.00025 * 2 / 1000 = 0.5 µmol

If it's the original concentration then it's:

0.00025 * 0.1 / 1000 = 25 nmol

-Doc_Martin-