# Calculation of p-nitrophenol

• Author
Posts
• #5867
biology_06er
Participant

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

• #55757
canalon
Participant

Ci x Vi=Cf x Vf

What you know:
Ci: initial concentration (1M is 1mol/L)
Vi: initial volume
Vf: final volume

What you wantto know:
Cf: final concentration

Solve 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).

• #55837
sdekivit
Participant
quote 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 NaOH

Thanks
biology_06r

you 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.