Calculation of p-nitrophenol

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biology_06er
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Calculation of p-nitrophenol

Post by biology_06er » Sun Oct 01, 2006 4:41 am

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

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canalon
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Post by canalon » Sun Oct 01, 2006 4:24 pm

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).
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

sdekivit
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Re: Calculation of p-nitrophenol

Post by sdekivit » Mon Oct 02, 2006 8:20 am

biology_06er wrote: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.

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