Biology Forum › Molecular Biology › Calculation of p-nitrophenol
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- October 1, 2006 at 4:41 am #5867
biology_06erParticipantHi 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 NaOHThanks
biology_06r - October 1, 2006 at 4:24 pm #55757canalonParticipant
Ci x Vi=Cf x Vf
What you know:
Ci: initial concentration (1M is 1mol/L)
Vi: initial volume
Vf: final volumeWhat you wantto know:
Cf: final concentrationSolve the equation for Cf (not too hard) and if you are careful with the units (mM vs M and mL and L) you have your answer with the volume and the concentration you have your number of moles (beware of the units, once again).
- October 2, 2006 at 8:20 am #55837sdekivitParticipantquote biology_06er: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 NaOHThanks
biology_06ryou added 0,1 mL 0,25 mM p-nitrophenol in 0.05 M glycine buffer, pH = 9.5 to 1,9 mL glycine buffer ?
if so, then the dilution factor is 20 and thus [PNP] = 0.0125 mM
–> in a volume of 2 mL, amount of mmol PNP = 0.0125 x 10^-3 * 2 = 2.5 x 10^-5 mmol
–> thus amount nmol PNP = 2.5 x 10^-5 * 10^6 = 25 nmol.
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