Calculation of p-nitrophenol

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    • #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.

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