Substrate solution (PNPP solution) In the measurements, we will use para-nitroph

Question

Substrate solution (PNPP solution)

In the measurements, we will use para-nitrophenyl phosphate (PNPP) as substrate.

The molecular weight is 371.14 as the di-sodium salt with 6 crystal water molecules, and this is the version of PNPP that we will use.

We will dissolve a tablet containing 5 mg PNPP in 260 µl of water (final volume will be 270 l).

When the phosphate group is cleaved off, the remaining compound, para-nitrophenol (PNP), becomes yellow in an alkaline solution. It is this yellow color development we can measure at 405 nm by using a spectrophotometer.

Find the concentration of PNPP (in molarity) when 5 mg PNPP is dissolved in a final volume of 270 µl?

Explanation / Answer

Molar mass of PNPP = 371.14 g/mol.

We took 5 mg PNPP.

Moles of PNPP taken = (5 mg)*(1 g/1000 mg)*(1 mole/371.14 g) = 1.347*10-5 mole.

Volume of the solution = 270 µL = (270 µL)*(1 L/106 µL) = 2.7*10-4 L.

Concentration of PNPP in the solution = (moles of PNPP)/(volume of solution in L) = (1.347*10-5 mole)/(2.7*10-4 L) = 0.0499 mol/L 0.050 M = (0.050 M)*(1000 mM/1 M) = 50 mM (ans).

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